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NCERT Solutions Class 12 Mathematics Chapter 4 – Determinants

NCERT Solutions Class 12 Mathematics Chapter 4 Determinants is available online for the students to refer to and study on Extramarks. The chapter introduces determinants, their properties, and their applications. In addition, the chapter also includes various problems and a step-by-step solution to each problem that aids students in understanding the concept better. Furthermore, the solutions are built on the CBSE syllabus of the latest academic year and with reference to NCERT.

A determinant is a number that can only be calculated with the help of square matrices. If the students want to score more in linear equations, then the determinant is essential to learn. It helps to check whether the system of equations has a unique solution or not. Besides, students will get a glimpse of the previous chapter, matrices. This chapter also carries essential elements and helps to improve their calculations.

In the NCERT Solutions Class 12 Mathematics Chapter 4, students can understand the complex theories and formulas in an easy to understand language. These sub-topics include basic principles of determinant, cofactors, adjoint, and the inverse of a matrix. Besides, this solution has proper explanations and clear concepts to help students understand better.

Students looking for step-by-step solutions to the problems mentioned in ch 4 mathematics class 12 may register in Extramarks. Extramarks not only offers other NCERT Solutions for Class 12 but also from Class 1 to class 11. In addition, students may log in to the website for updated study material, revision notes, CBSE sample papers, important questions, and the latest notifications and news regarding CBSE exams.

Key Topics Covered In NCERT Solutions Class 12 Mathematics Chapter 4

Extramarks NCERT Solutions for Class 12 Mathematics Chapter 4 covers all essential topics under determinants. Students will start by learning how to find the determinant of a matrix of order one and the properties of determinants. More precise calculations and solutions make them understand multiplications in positives instead of negatives. There are four exercises and three subtopics under NCERT Solutions for Class 12 Mathematics Chapter 4. In addition, students can refer to the study material offered by Extramarks as it offers better solutions.

Topics covered in Chapter 4 Determinant include

Exercise	Topics
4.1	Introduction
4.2	Determinant
4.3	Properties of Determinants
4.4	Area of Triangle
4.5	Minors and Cofactors
4.6	The Adjoint and Inverse of a Matrix
4.7	Applications of Determinants and Matrices

4.1 Introduction

NCERT Solutions Class 12 Mathematics Chapter 4 introduces the students to Determinants and their basic concepts and theorems, wherein they will be introduced to the properties of determinants such as minors, cofactors, and applications of determinants. Further, the students will get a hold of topics like the inconsistency of the system of linear equations and solutions of linear equations.

4.2 Determinant

This is one of the crucial sections. Students will understand the grassroots concepts; as learnt in the previous Chapter 3 Matrices, any matrix inside a mode is read as a determinant of a matrix. In addition, students can learn how to find the determinant of a matrix of order one, making it easy to comprehend. Students may refer to Extramarks NCERT Solutions Class 12 Mathematics Chapter 4 as it provides straightforward and easy-to-understand calculations.

4.3 Properties of Determinants

Under this topic, students will observe that prominent examples revolve around the properties of determinants. This section elaborates on how the statement properties are true for determinants of any order. The students will get to explore six different properties of determinants. Students may refer to the exercise problems to understand the properties better. Extramarks NCERT Solutions Class 12 Mathematics Chapter 4 offers step-by-step solutions for many exercises under this topic.

4.4 Area of Triangle

This section guides students on calculating the area of triangles and the application of vertices. It also elaborates on particular topics such as positive quantity and absolute values of determinants. It also helps the students understand the positive and negative values of the determinant.

4.5 Minors and Cofactors

In the NCERT Solutions Class 12 Mathematics Chapter 4, minors, cofactors and their main concepts are covered. In addition, this section will teach the students about five more expansion ways that help calculate the area of a triangle.

4.6 The Adjoint and Inverse of a Matrix

Further in Chapter 4 Class 12, students will learn about the inverse and existence. It contains detailed explanations in terms of finding the adjoint of the matrix. There is a total of five theorems that helps to verify the concepts.

4.7 Applications of Determinants and Matrices

NCERT Solutions Class 12 Mathematics Chapter 4 covers the essential concepts in applications of determinants and matrices. A student will learn how to check the consistency of the system of linear equations. Under this topic, students will learn and study the solution system of linear equations.

For detailed study material, students may refer to Extramarks, where step-by-step solutions on essential concepts, including consistent and inconsistent systems, are provided. With the help of Extramarks NCERT Solutions, the student can score more in determinants and matrices.

NCERT Solutions Class 12 Mathematics Chapter 4 Exercise & Answer Solutions

NCERT Solutions Class 12 Mathematics Chapter 4 Determinant is available for students to refer to on the Extramarks website. It contains step-by-step solutions with detailed explanations. The students can solve complex problems with the easiest and most time-saving methods. Referring to our study material, students can make their basics strong and easily solve advanced theorems.

Click on the below links to view exercise-specific questions and solutions for NCERT Solutions Class 12 Mathematics Chapter 4 Determinant.

Chapter 4: Exercise 4.1 Solutions – 8 Questions (3 Short Questions and 5 Long Questions)
Chapter 4: Exercise 4.2 Solutions – 16 Questions (4 Short Questions 10 Long Questions)
Chapter 4: Exercise 4.3 Solutions – 5 Questions (2 Short Questions and 3 Long Questions)
Chapter 4: Exercise 4.4 Solutions – 5 Questions (2 Short Questions and 3 Long Questions)
Chapter 4: Exercise 4.5 Solutions – 5 Questions (2 Short Questions and 3 Long Questions)
Chapter 4: Exercise 4.6 Solutions – 5 Questions (2 Short Questions and 3 Long Questions)

In addition to NCERT solutions class 12 mathematics chapter 4, students can also access other chapters of class 4. Furthermore, Extramarks offers NCERT solutions for other classes, which can be accessed by clicking on the respective link.

NCERT Solutions Class 1
NCERT Solutions Class 2
NCERT Solutions Class 3
NCERT Solutions Class 4
NCERT Solutions Class 5
NCERT Solutions Class 6
NCERT Solutions Class 7
NCERT Solutions Class 8
NCERT Solutions Class 9
NCERT Solutions Class 10
NCERT Solutions Class 11

NCERT Exemplar Class 12 Mathematics

The NCERT Exemplar Class 12 Mathematics covers the essential subject knowledge. Once the students clear all the necessary topics under determinants, they can start preparing with an exemplar. There is a lot of importance to the topic as it appears in the competitive exams. Students can start preparing for their first-term CBSE exam with the help of this exemplar. It provides a complete explanatory solution for students with good knowledge of the subject.

At Extramarks, our subject matter experts offer their best solution with a proper step-by-step explanation. The exemplar consists of all the six exercises, i.e., introduction, determinant, properties of determinants, area of a triangle, minors and cofactors. In addition, the NCERT Exemplar contains various questions, multiple-choice type questions, and long answer type questions.

Key Features of NCERT Solutions Class 12 Mathematics Chapter 4

Class 12 Mathematics Chapter 4 has some essential elements that are a part of competitive exams. The students will get a detailed explanation of the terms, definitions, rules, and properties of determinants. A basic understanding of these complex theories is essential for better understanding. Hence, NCERT Solutions Class 12 Mathematics Chapter 4 offers a precise solution for each example mentioned in the exercise.

Some of the key features of NCERT Solutions Class 12 Mathematics Chapter 4 include:

The solutions cover the entire syllabus of CBSE Class 12 Mathematics Chapter 4 and help to provide a detailed explanation of all topics.
The format of the NCERT Solutions is self-explanatory and clears all doubts a student may have while learning complex concepts.
To score better in the examination and competitive entrance exams, students must grasp all the topics. Hence NCERT Solutions Class 12 Mathematics Chapter 4 consists of all important aspects of Determinants needed for the competitive entrance exam.
The solution helps increase knowledge in main topics.

Q.1

$\begin{array}{l} Let A = [\begin{matrix} 1 & sinθ & 1 \\ - sinθ & 1 & sinθ \\ - 1 & - sinθ & 1 \end{matrix}], where 0 \leq θ \leq 2 π . Then \\ (A) Det (A) = 0 (B) Det (A) \in (2, \infty) \\ (C) Det (A) \in (2, 4) (D) Det (A) \in [2, 4] \end{array}$

Ans

$\begin{array}{l} Let A = [\begin{matrix} 1 & sinθ & 1 \\ - sinθ & 1 & sinθ \\ - 1 & - sinθ & 1 \end{matrix}], where 0 \leq θ \leq 2 π . Then \\ | A | = | \begin{matrix} 1 & sinθ & 1 \\ - sinθ & 1 & sinθ \\ - 1 & - sinθ & 1 \end{matrix} | \\ = 1 (1 + \sin^{2} θ) - sinθ (- sinθ + sinθ) + 1 (\sin^{2} θ + 1) \\ = 2 (1 + \sin^{2} θ) \\ Now, 0 \leq θ \leq 2 π \\ \Rightarrow 0 \leq sinθ \leq 1 \\ \Rightarrow 0 \leq \sin^{2} θ \leq 1 \\ \Rightarrow 1 \leq 1 + \sin^{2} θ \leq 2 \\ \Rightarrow 2 \leq 2 (1 + \sin^{2} θ) \leq 4 \\ ∴ Det (A) \in [2, 4] \\ Thus, correct option is D . \end{array}$

Q.2

$\begin{array}{l} If x, y, z are non zero real numbers, then the inverse of matrix \\ A = [\begin{matrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{matrix}] is \\ (A) [\begin{matrix} x^{- 1} & 0 & 0 \\ 0 & y^{- 1} & 0 \\ 0 & 0 & z^{- 1} \end{matrix}] \\ (B) xyz [\begin{matrix} x^{- 1} & 0 & 0 \\ 0 & y^{- 1} & 0 \\ 0 & 0 & z^{- 1} \end{matrix}] \\ (C) \frac{1}{xyz} [\begin{matrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{matrix}] \\ (D) \frac{1}{xyz} [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] \end{array}$

Ans

$\begin{array}{l} Given matrix is \\ A = [\begin{matrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{matrix}] \\ | A | = | \begin{matrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{matrix} | \\ = x (yz - 0) - 0 + 0 \\ = xyz \neq 0 \\ Thus, inverse of matrix A exists . \\ Now cofactors are \\ A_{11} = yz, A_{12} = 0, A_{13} = 0 \\ A_{21} = 0, A_{22} = xz, A_{23} = 0 \\ A_{31} = 0, A_{32} = 0, A_{33} = xy \\ Adj A = {[\begin{matrix} yz & 0 & 0 \\ 0 & xz & 0 \\ 0 & 0 & xy \end{matrix}]}^{‘} = [\begin{matrix} yz & 0 & 0 \\ 0 & xz & 0 \\ 0 & 0 & xy \end{matrix}] \\ ∴ A^{- 1} = \frac{1}{| A |} Adj A \\ = \frac{1}{xyz} [\begin{matrix} yz & 0 & 0 \\ 0 & xz & 0 \\ 0 & 0 & xy \end{matrix}] \\ = [\begin{matrix} \frac{yz}{xyz} & 0 & 0 \\ 0 & \frac{xz}{xyz} & 0 \\ 0 & 0 & \frac{xy}{xyz} \end{matrix}] \\ = [\begin{matrix} \frac{1}{x} & 0 & 0 \\ 0 & \frac{1}{y} & 0 \\ 0 & 0 & \frac{1}{z} \end{matrix}] = [\begin{matrix} x^{- 1} & 0 & 0 \\ 0 & y^{- 1} & 0 \\ 0 & 0 & z^{- 1} \end{matrix}] \\ Thus, the correct option is A . \end{array}$

Q.3

$\begin{array}{l} I f a, b, c a r e i n A . P ., t h e n t h e d e t e r m i n a n t \\ | \begin{matrix} x + 2 & x + 3 & x + 2 a \\ x + 3 & x + 4 & x + 2 b \\ x + 4 & x + 5 & x + 2 c \end{matrix} | i s \\ (A) 0 (B) 1 (C) x (D) 2 x \end{array}$

Ans

$\begin{array}{l} | \begin{matrix} x + 2 & x + 3 & x + 2 a \\ x + 3 & x + 4 & x + 2 b \\ x + 4 & x + 5 & x + 2 c \end{matrix} | \\ If a, b, c are in A . P ., so 2 b = a + c \\ Then, \\ | \begin{matrix} x + 2 & x + 3 & x + 2 a \\ x + 3 & x + 4 & x + 2 b \\ x + 4 & x + 5 & x + 2 c \end{matrix} | = | \begin{matrix} x + 2 & x + 3 & x + 2 a \\ x + 3 & x + 4 & x + (a + c) \\ x + 4 & x + 5 & x + 2 c \end{matrix} | \\ Applying R_{1} \to R_{1} - R_{2} and R_{3} \to R_{3} - R_{1}, we get \\ = | \begin{matrix} - 1 & - 1 & a - c \\ x + 3 & x + 4 & x + (a + c) \\ 1 & 1 & c - a \end{matrix} | \\ Applying R_{1} \to R_{1} + R_{3}, we get \\ = | \begin{matrix} 0 & 0 & 0 \\ x + 3 & x + 4 & x + (a + c) \\ 1 & 1 & c - a \end{matrix} | \\ = 0 [Since, all elements of R_{1} are zero .] \\ Hence, \\ | \begin{matrix} x + 2 & x + 3 & x + 2 a \\ x + 3 & x + 4 & x + 2 b \\ x + 4 & x + 5 & x + 2 c \end{matrix} | = 0 \\ Therefore, correct option is A . \end{array}$

Q.4

$\begin{array}{l} Solve the system of the following equations \\ \frac{2}{x} + \frac{3}{y} + \frac{10}{z} = 4 \\ \frac{4}{x} - \frac{6}{y} + \frac{5}{z} = 1 \\ \frac{6}{x} + \frac{9}{y} - \frac{20}{z} = 1 \end{array}$

Ans

$\begin{array}{l} \frac{2}{x} + \frac{3}{y} + \frac{10}{z} = 4 \\ \frac{4}{x} - \frac{6}{y} + \frac{5}{z} = 1 \\ \frac{6}{x} + \frac{9}{y} - \frac{20}{z} = 1 \\ Let p = \frac{1}{x} q = \frac{1}{y} and r = \frac{1}{z}, then \\ 2 p +3 q +10 r = 4 \\ 4 p - 6 q +5 r = 1 \\ 6 p +9 q - 20 r = 1 \\ The system can be written as AX = B, where \\ A = [\begin{matrix} 2 & 3 & 10 \\ 4 & - 6 & 5 \\ 6 & 9 & - 20 \end{matrix}], X = [\begin{array}{l} p \\ q \\ r \end{array}] and B = [\begin{array}{l} 4 \\ 1 \\ 1 \end{array}] \\ | A | = | \begin{matrix} 2 & 3 & 10 \\ 4 & - 6 & 5 \\ 6 & 9 & - 20 \end{matrix} | \\ = 2 (120 - 45) - 3 (- 80 - 30) + 10 (36 + 36) \\ = 150 + 330 + 720 \\ = 1200 \\ Thus, A is non - singular . Therefore, its inverse exists . \\ Now, \\ A_{11} = 75, A_{12} = 110, A_{13} = 72, \\ A_{21} = 150, A_{22} = - 100, A_{23} = 0, \\ A_{31} = 75, A_{32} = 30, A_{33} = - 24 \\ Adj A = {[\begin{matrix} 75 & 110 & 72 \\ 150 & - 100 & 0 \\ 75 & 30 & - 24 \end{matrix}]}^{‘} \\ = [\begin{matrix} 75 & 150 & 75 \\ 110 & - 100 & 30 \\ 72 & 0 & - 24 \end{matrix}] \\ A^{- 1} = \frac{1}{| A |} Adj A \\ = \frac{1}{1200} [\begin{matrix} 75 & 150 & 75 \\ 110 & - 100 & 30 \\ 72 & 0 & - 24 \end{matrix}] \\ Now, X = A^{- 1} B \\ \Rightarrow [\begin{array}{l} p \\ q \\ r \end{array}] = \frac{1}{1200} [\begin{matrix} 75 & 150 & 75 \\ 110 & - 100 & 30 \\ 72 & 0 & - 24 \end{matrix}] [\begin{array}{l} 4 \\ 1 \\ 2 \end{array}] \\ \Rightarrow [\begin{array}{l} p \\ q \\ r \end{array}] = \frac{1}{1200} [\begin{array}{l} 300 + 150 + 150 \\ 440 - 100 + 60 \\ 288 + 0 - 48 \end{array}] \\ \Rightarrow [\begin{array}{l} p \\ q \\ r \end{array}] = \frac{1}{1200} [\begin{array}{l} 600 \\ 400 \\ 240 \end{array}] \\ \Rightarrow [\begin{array}{l} p \\ q \\ r \end{array}] = [\begin{array}{l} \frac{1}{2} \\ \frac{1}{3} \\ \frac{1}{5} \end{array}] \\ \Rightarrow p = \frac{1}{2}, q = \frac{1}{3} and r = \frac{1}{5} \\ \Rightarrow x = \frac{1}{p} = 2, y = \frac{1}{q} = 3 and z = \frac{1}{r} = 5 \\ Hence, the required solution is \\ x = 2, y = 3 and z = 5 . \end{array}$

Q.5 Using properties of determinants prove that:

$|\begin{matrix} sinα & cosα & \cos (α + β) \\ sinβ & cosβ & \cos (β + δ) \\ sinγ & cosγ & \cos (γ + δ) \end{matrix}| = 0$

Ans

$\begin{array}{l} L . H . S . = | \begin{matrix} sinα & cosα & \cos (α + δ) \\ sinβ & cosβ & \cos (β + δ) \\ sinγ & cosγ & \cos (γ + δ) \end{matrix} | \\ Applying C_{1} \to {sinδC}_{1} and C_{2} \to {cosδC}_{2}, we get \\ = \frac{1}{sinδcosδ} | \begin{matrix} sinαsinδ & cosαcosδ & cosαcosδ - sinαsinδ \\ sinβsinδ & cosβcosδ & cosβcosδ - sinβsinδ \\ sinγsinδ & cosγcosδ & cosγcosδ - sinγsinδ \end{matrix} | \\ C_{1} \to C_{1} + C_{3}, we get \\ = \frac{1}{sinδcosδ} | \begin{matrix} cosαcosδ & cosαcosδ & cosαcosδ - sinαsinδ \\ cosβcosδ & cosβcosδ & cosβcosδ - sinβsinδ \\ cosγcosδ & cosγcosδ & cosγcosδ - sinγsinδ \end{matrix} | \\ = \frac{1}{sinδcosδ} \times 0 [\begin{array}{l} ∵ Δ = 0, if any two coloumns \\ are identical . \end{array}] \\ = 0 = R . H . S . \\ Thus, the given result is proved . \end{array}$

Q.6 Using properties of determinants prove that:

$|\begin{matrix} 1 & 1 + p & 1 + p + q \\ 2 & 3 + 2 p & 4 + 3 p + 2 q \\ 3 & 6 + 3 p & 10 + 6 p + 3 q \end{matrix}| = 1$

Ans

$\begin{array}{l} L . H . S . = | \begin{matrix} 1 & 1 + p & 1 + p + q \\ 2 & 3 + 2 p & 4 + 3 p + 2 q \\ 3 & 6 + 3 p & 10 + 6 p + 3 q \end{matrix} | \\ Applying R_{2} \to R_{2} - 2 R_{1} and R_{3} \to R_{3} - 3 R_{1}, we get \\ = | \begin{matrix} 1 & 1 + p & 1 + p + q \\ 0 & 1 & 2 + p \\ 0 & 3 & 7 + 3 p \end{matrix} | \\ Expanding along C_{1}, we get \\ = 1 | \begin{matrix} 1 & 2 + p \\ 3 & 7 + 3 p \end{matrix} | - 0 + 0 \\ = 7 + 3 p - 6 - 3 p \\ = 1 = R . H . S . \\ Thus, the given result is proved . \end{array}$

Q.7 Using properties of determinants prove that:

$|\begin{matrix} 3 a & - a + b & - a + c \\ - b + a & 3 b & - b + c \\ - c + a & - c + b & 3 c \end{matrix}| = 3 (a + b + c) (ab + bc + ca)$

Ans

$\begin{array}{l} L . H . S . = | \begin{matrix} 3 a & - a + b & - a + c \\ - b + a & 3 b & - b + c \\ - c + a & - c + b & 3 c \end{matrix} | \\ Applying C_{1} \to C_{1} + C_{2} + C_{3}, we get \\ = | \begin{matrix} a + b + c & - a + b & - a + c \\ a + b + c & 3 b & - b + c \\ a + b + c & - c + b & 3 c \end{matrix} | \\ = (a + b + c) | \begin{matrix} 1 & - a + b & - a + c \\ 1 & 3 b & - b + c \\ 1 & - c + b & 3 c \end{matrix} | \\ Apply R_{2} \to R_{2} - R_{1} and R_{3} \to R_{3} - R_{1}, we get \\ = (a + b + c) | \begin{matrix} 1 & - a + b & - a + c \\ 0 & a + 2 b & a - b \\ 0 & a - c & a + 2 c \end{matrix} | \\ Expanding along C_{1}, we get \\ = (a + b + c) {(a + 2 b) (a + 2 c) - (a - b) (a - c)} \\ = (a + b + c) {a^{2} + 2 ac + 2 ab + 4 bc - (a^{2} - ac - ba + bc)} \\ = (a + b + c) {a^{2} + 2 ac + 2 ab + 4 bc - a^{2} + ac + ba - bc} \\ = 3 (a + b + c) (ab + bc + ca) = R . H . S . \\ Thus, the given result is proved . \end{array}$

Q.8 Using properties of determinants prove that:

$|\begin{matrix} x & x^{2} & 1 + {px}^{3} \\ y & y^{2} & 1 + {py}^{3} \\ z & z^{2} & 1 + {pz}^{3} \end{matrix}| = (1 + pxyz) (x - y) (y - z) (z - x)$

Ans

$\begin{array}{l} L .H .S . = | \begin{matrix} x & x^{2} & 1 + p x^{3} \\ y & y^{2} & 1 + p y^{3} \\ z & z^{2} & 1 + p z^{3} \end{matrix} | \\ A p p l y i n g R_{3} \to R_{3} - R_{1} a n d R_{2} \to R_{2} - R_{1}, w e get \\ = | \begin{matrix} x & x^{2} & 1 + p x^{3} \\ y - x & y^{2} - x^{2} & (1 + p y^{3}) - (1 + p x^{3}) \\ z - x & z^{2} - x^{2} & (1 + p z^{3}) - (1 + p x^{3}) \end{matrix} | \\ = (y - x) (z - x) | \begin{matrix} x & x^{2} & 1 + p x^{3} \\ 1 & y + x & (p y^{3} - p x^{3}) \\ 1 & z + x & (p z^{3} - p x^{3}) \end{matrix} | \\ = (y - x) (z - x) | \begin{matrix} x & x^{2} & 1 + p x^{3} \\ 1 & y + x & p (y^{2} + y x + x^{2}) \\ 1 & z + x & p (z^{2} + z x + x^{2}) \end{matrix} | \\ A p p l y i n g R_{3} \to R_{3} - R_{2}, we get \\ = (y - x) (z - x) | \begin{matrix} x & x^{2} & 1 + p x^{3} \\ 1 & y + x & p (y^{2} + y x + x^{2}) \\ 0 & z - y & p (z^{2} - y^{2} + z x - y x) \end{matrix} | \\ = (y - x) (z - x) (z - y) | \begin{matrix} x & x^{2} & 1 + p x^{3} \\ 1 & y + x & p (y^{2} + y x + x^{2}) \\ 0 & 1 & p (z + y + x) \end{matrix} | \\ E x p a n d i n g {along R}_{3}, we get \\ = (y - x) (z - x) (z - y) {\begin{cases} 0 - 1 | \begin{matrix} x & 1 + p x^{3} \\ 1 & p (y^{2} + y x + x^{2}) \end{matrix} | \\ + p (z + y + x) | \begin{matrix} x & x^{2} \\ 1 & y + x \end{matrix} | \end{cases}} \\ = (y - x) (z - x) (z - y) [\begin{array}{l} - 1 {x p (y^{2} + y x + x^{2}) - (1 + p x^{3})} \\ + p (z + y + x) {x (y + x) - x^{2}} \end{array}] \\ = (y - x) (z - x) (z - y) {\begin{cases} - p x y^{2} - p x^{2} y - p x^{3} + 1 + p x^{3} \\ + p x^{2} y + p x y^{2} + p x y z \end{cases}} \\ = (y - x) (z - x) (z - y) (1 + p x y z) \\ = (1 + p x y z) (x - y) (y - z) (z - x) = R . H . S . \\ T h e r e f o r e, the given result is proved . \end{array}$

Q.9 Using properties of determinants prove that:

$|\begin{matrix} α & α^{2} & β + γ \\ β & β^{2} & γ + α \\ γ & γ^{2} & α + β \end{matrix}| = (β - γ) (γ - α) (α - β) (α + β + γ)$

Ans

$\begin{array}{l} L . H . S . = | \begin{matrix} α & α^{2} & β + γ \\ β & β^{2} & γ + α \\ γ & γ^{2} & α + β \end{matrix} | \\ Applying R_{2} \to R_{2} - R_{1} and R_{3} \to R_{3} - R_{1}, we get \\ = | \begin{matrix} α & α^{2} & β + γ \\ β - α & β^{2} - α^{2} & α - β \\ γ - α & γ^{2} - α^{2} & α - γ \end{matrix} | \\ = (β - α) (γ - α) | \begin{matrix} α & α^{2} & β + γ \\ 1 & β + α & - 1 \\ 1 & γ + α & - 1 \end{matrix} | \\ Applying R_{3} \to R_{3} - R_{2}, we get \\ = (β - α) (γ - α) | \begin{matrix} α & α^{2} & β + γ \\ 1 & β + α & - 1 \\ 0 & γ - β & 0 \end{matrix} | \\ Expanding along R_{3}, we get \\ = (β - α) (γ - α) {- (γ - β) (- α - β - γ)} \\ = (β - γ) (γ - α) (α - β) (α + β + γ) = R . H . S . \end{array}$

Q.10

$|Evaluate \begin{matrix} 1 & x & y \\ 1 & x + y & y \\ 1 & x & x + y \end{matrix}| .$

Ans

$\begin{array}{l} Δ = |\begin{matrix} 1 & x & y \\ 1 & x + y & y \\ 1 & x & x + y \end{matrix}| \\ Apply R_{2} \to R_{2} - R_{1} and R_{3} \to R_{3} - R_{1} \\ = |\begin{matrix} 1 & x & y \\ 0 & y & 0 \\ 0 & 0 & x \end{matrix}| \\ Expanding along C_{1}, we get \\ = 1 (xy - 0) - 0 + 0 \\ = xy \end{array}$

Q.11

$Evaluate | \begin{matrix} x & y & x + y \\ y & x + y & x \\ x + y & x & y \end{matrix} | .$

Ans

$\begin{array}{l} Δ = | \begin{matrix} x & y & x + y \\ y & x + y & x \\ x + y & x & y \end{matrix} | \\ Applying R_{1} \to R_{1} + R_{2} + R_{3}, we get \\ = | \begin{matrix} 2 x +2 y & 2 x +2 y & 2 x +2 y \\ y & x + y & x \\ x + y & x & y \end{matrix} | \\ = 2 (x + y) | \begin{matrix} 1 & 1 & 1 \\ y & x + y & x \\ x + y & x & y \end{matrix} | \\ Applying C_{2} \to C_{2} - C_{1} and C_{3} \to C_{3} - C_{1}, we get \\ = 2 (x + y) | \begin{matrix} 1 & 0 & 0 \\ y & x & x - y \\ x + y & - y & - x \end{matrix} | \\ Expanding along R_{1}, we get \\ = 2 (x + y) | \begin{matrix} x & x - y \\ - y & - x \end{matrix} | \\ = 2 (x + y) (- x^{2} + xy - y^{2}) \\ = - 2 (x + y) (x^{2} - xy + y^{2}) \\ = - 2 (x^{3} + y^{3}) \end{array}$

Q.12

$\begin{array}{l} Let A = [\begin{matrix} 1 & - 2 & 1 \\ - 2 & 3 & 1 \\ 1 & 1 & 5 \end{matrix}] . Verify that \\ (i) {[adj A]}^{- 1} = adj (A^{- 1}) (ii) {(A^{- 1})}^{- 1} = A \end{array}$

Ans

$\begin{array}{l} A = [\begin{matrix} 1 & - 2 & 1 \\ - 2 & 3 & 1 \\ 1 & 1 & 5 \end{matrix}] \\ | A | = | \begin{matrix} 1 & - 2 & 1 \\ - 2 & 3 & 1 \\ 1 & 1 & 5 \end{matrix} | \\ = 1 (15 - 1) + 2 (- 10 - 1) + 1 (- 2 - 3) \\ = 14 - 22 - 5 = - 13 \neq 0 \\ Now, cofactors are \\ A_{11} = 14, A_{12} = 11, A_{13} = - 5 \\ A_{21} = 11, A_{22} = 4, A_{23} = - 3 \\ A_{31} = - 5, A_{32} = - 3, A_{33} = - 1 \\ ∴ Adj A = {[\begin{matrix} 14 & 11 & - 5 \\ 11 & 4 & - 3 \\ - 5 & - 3 & - 1 \end{matrix}]}^{‘} = [\begin{matrix} 14 & 11 & - 5 \\ 11 & 4 & - 3 \\ - 5 & - 3 & - 1 \end{matrix}] \\ So, A^{- 1} = \frac{1}{| A |} Adj A \\ = \frac{1}{- 13} [\begin{matrix} 14 & 11 & - 5 \\ 11 & 4 & - 3 \\ - 5 & - 3 & - 1 \end{matrix}] \\ = \frac{1}{13} [\begin{matrix} - 14 & - 11 & 5 \\ - 11 & - 4 & 3 \\ 5 & 3 & 1 \end{matrix}] \\ (i) | Adj A | = | \begin{matrix} 14 & 11 & - 5 \\ 11 & 4 & - 3 \\ - 5 & - 3 & - 1 \end{matrix} | \\ = 14 (- 4 - 9) - 11 (- 11 - 15) - 5 (- 33 + 20) \\ = - 182 + 286 + 65 \\ = 169 \\ Cofactors of Adj A are \\ A_{11} = - 13, A_{12} = 26, A_{13} = - 13 \\ A_{21} = 26, A_{22} = - 39, A_{23} = - 13 \\ A_{31} = - 13, A_{32} = - 13, A_{33} = - 65 \\ Adj (Adj A) = {[\begin{matrix} - 13 & 26 & - 13 \\ 26 & - 39 & - 13 \\ - 13 & - 13 & - 65 \end{matrix}]}^{‘} \\ = [\begin{matrix} - 13 & 26 & - 13 \\ 26 & - 39 & - 13 \\ - 13 & - 13 & - 65 \end{matrix}] \\ ∴ {(Adj A)}^{- 1} = \frac{1}{| Adj A |} Adj (Adj A) \\ = \frac{1}{169} [\begin{matrix} - 13 & 26 & - 13 \\ 26 & - 39 & - 13 \\ - 13 & - 13 & - 65 \end{matrix}] \\ = \frac{1}{169} \times 13 [\begin{matrix} - 1 & 2 & - 1 \\ 2 & - 3 & - 1 \\ - 1 & - 1 & - 5 \end{matrix}] = \frac{1}{13} [\begin{matrix} - 1 & 2 & - 1 \\ 2 & - 3 & - 1 \\ - 1 & - 1 & - 5 \end{matrix}] \\ A^{- 1} = \frac{1}{13} [\begin{matrix} - 14 & - 11 & 5 \\ - 11 & - 4 & 3 \\ 5 & 3 & 1 \end{matrix}] = [\begin{matrix} - \frac{14}{13} & - \frac{11}{13} & \frac{5}{13} \\ - \frac{11}{13} & - \frac{4}{13} & \frac{3}{13} \\ \frac{5}{13} & \frac{3}{13} & \frac{1}{13} \end{matrix}] \end{array}$ $\begin{array}{l} A d j A^{- 1} = {[\begin{matrix} \frac{- 4}{169} - \frac{9}{169} & - (- \frac{11}{169} - \frac{15}{169}) & - \frac{33}{169} - \frac{20}{169} \\ - (\frac{- 11}{169} - \frac{15}{169}) & - \frac{14}{169} - \frac{25}{169} & - (- \frac{42}{169} + \frac{55}{169}) \\ - \frac{33}{169} + \frac{20}{169} & - (- \frac{42}{169} + \frac{55}{169}) & \frac{56}{169} - \frac{121}{169} \end{matrix}]}^{‘} \\ = \frac{1}{169} [\begin{matrix} - 13 & 26 & - 13 \\ 26 & - 39 & - 13 \\ - 13 & - 13 & - 65 \end{matrix}] \\ = \frac{13}{169} [\begin{matrix} - 1 & 2 & - 1 \\ 2 & - 3 & - 1 \\ - 1 & - 1 & - 5 \end{matrix}] \\ A d j A^{- 1} = {(A d j A)}^{- 1} \\ H e n c e, it is proved . \\ (i i) S i n c e, \\ A^{- 1} = \frac{1}{13} [\begin{matrix} - 14 & - 11 & 5 \\ - 11 & - 4 & 3 \\ 5 & 3 & 1 \end{matrix}] = [\begin{matrix} - \frac{14}{13} & - \frac{11}{13} & \frac{5}{13} \\ - \frac{11}{13} & - \frac{4}{13} & \frac{3}{13} \\ \frac{5}{13} & \frac{3}{13} & \frac{1}{13} \end{matrix}] \\ a n d A d j A = \frac{1}{13} [\begin{matrix} - 1 & 2 & - 1 \\ 2 & - 3 & - 1 \\ - 1 & - 1 & - 5 \end{matrix}] \\ | A^{- 1} | = | \begin{matrix} - \frac{14}{13} & - \frac{11}{13} & \frac{5}{13} \\ - \frac{11}{13} & - \frac{4}{13} & \frac{3}{13} \\ \frac{5}{13} & \frac{3}{13} & \frac{1}{13} \end{matrix} | \\ = {(\frac{1}{13})}^{3} | \begin{matrix} - 14 & - 11 & 5 \\ - 11 & - 4 & 3 \\ 5 & 3 & 1 \end{matrix} | \\ = {(\frac{1}{13})}^{3} {- 14 (- 4 - 9) + 11 (- 11 - 15) + 5 (- 33 + 20)} \\ = {(\frac{1}{13})}^{3} (182 - 286 - 65) \\ = {(\frac{1}{13})}^{3} (- 169) = - \frac{1}{13} \\ ∴ {(A^{- 1})}^{- 1} = \frac{A d j A^{- 1}}{| A^{- 1} |} \\ = \frac{1}{(- \frac{1}{13})} \frac{1}{13} [\begin{matrix} - 1 & 2 & - 1 \\ 2 & - 3 & - 1 \\ - 1 & - 1 & - 5 \end{matrix}] \\ = [\begin{matrix} 1 & - 2 & 1 \\ - 2 & 3 & 1 \\ 1 & 1 & 5 \end{matrix}] \Rightarrow {(A^{- 1})}^{- 1} = A \\ H e n c e, it is proved . \end{array}$

Q.13

${If A}^{- 1} = | \begin{matrix} 3 & - 1 & 1 \\ - 15 & 6 & - 5 \\ 5 & - 2 & 2 \end{matrix} | and B = [\begin{matrix} 1 & 2 & - 2 \\ - 1 & 3 & 0 \\ 0 & - 2 & 1 \end{matrix}], find {(AB)}^{- 1} .$

Ans

$\begin{array}{l} Since, {(AB)}^{- 1} = B^{- 1} A^{- 1} \\ B = [\begin{matrix} 1 & 2 & - 2 \\ - 1 & 3 & 0 \\ 0 & - 2 & 1 \end{matrix}] \\ ∴ | B | = | \begin{matrix} 1 & 2 & - 2 \\ - 1 & 3 & 0 \\ 0 & - 2 & 1 \end{matrix} | \\ = 1 (3 - 0) - 2 (- 1 - 0) - 2 (2 - 0) \\ = 3 + 2 - 4 = 1 \neq 0 \\ Thus, B^{- 1} exists . \\ Now, cofactors are \\ A_{11} = 3, A_{12} = 1, A_{13} = 2 \\ A_{21} = 2, A_{22} = 1, A_{23} = 2 \\ A_{31} = 6, A_{32} = 2, A_{33} = 5 \\ Adj B = {[\begin{matrix} 3 & 1 & 2 \\ 2 & 1 & 2 \\ 6 & 2 & 5 \end{matrix}]}^{‘} \\ = [\begin{matrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{matrix}] \\ B^{- 1} = \frac{1}{| B |} Adj B \\ = \frac{1}{1} [\begin{matrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{matrix}] = [\begin{matrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{matrix}] \\ Now, {(AB)}^{- 1} = B^{- 1} A^{- 1} \\ = [\begin{matrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{matrix}] [\begin{matrix} 3 & - 1 & 1 \\ - 15 & 6 & - 5 \\ 5 & - 2 & 2 \end{matrix}] \\ = [\begin{matrix} 9 - 30 + 30 & - 3 + 12 - 12 & 3 - 10 + 12 \\ 3 - 15 + 10 & - 1 + 6 - 4 & 1 - 5 + 4 \\ 6 - 30 + 25 & 2 - 12 + 10 & 2 - 10 + 10 \end{matrix}] \\ = [\begin{matrix} 9 & - 3 & 5 \\ - 2 & 1 & 0 \\ 1 & 0 & 2 \end{matrix}] \\ Thus, required solution is \\ {(AB)}^{- 1} = [\begin{matrix} 9 & - 3 & 5 \\ - 2 & 1 & 0 \\ 1 & 0 & 2 \end{matrix}] \end{array}$

Q.14

$\begin{array}{l} Prove that \\ | \begin{matrix} a^{2} & bc & ac + c^{2} \\ a^{2} + ab & b^{2} & ac \\ ab & b^{2} + bc & c^{2} \end{matrix} | = 4 a^{2} b^{2} c^{2} \end{array}$

Ans

$\begin{array}{l} L . H . S . = | \begin{matrix} a^{2} & bc & ac + c^{2} \\ a^{2} + ab & b^{2} & ac \\ ab & b^{2} + bc & c^{2} \end{matrix} | \\ Taking out common a, b and c from C_{1}, C_{2} and C_{3} respectively . \\ = abc | \begin{matrix} a & c & a + c \\ a + b & b & a \\ b & b + c & c \end{matrix} | \\ Applying R_{2} \to R_{2} - R_{1}, R_{3} \to R_{3} - R_{1} we get \\ = abc | \begin{matrix} a & c & a + c \\ b & b - c & - c \\ b - a & b & - a \end{matrix} | \\ Applying R_{2} \to R_{2} + R_{1}, we get \\ = abc | \begin{matrix} a & c & a + c \\ a + b & b & a \\ b - a & b & - a \end{matrix} | \\ Applying R_{3} \to R_{3} + R_{2}, we get \\ = abc | \begin{matrix} a & c & a + c \\ a + b & b & a \\ 2 b & 2 b & 0 \end{matrix} | \\ = 2 {ab}^{2} c | \begin{matrix} a & c & a + c \\ a + b & b & a \\ 1 & 1 & 0 \end{matrix} | \\ Applying C_{2} \to C_{2} - C_{1}, we get \\ = 2 {ab}^{2} c | \begin{matrix} a & c - a & a + c \\ a + b & - a & a \\ 1 & 0 & 0 \end{matrix} | \\ Expanding along R_{3}, we get \\ = 2 {ab}^{2} c {1 (ac - a^{2} + a^{2} + ac) - 0 + 0} \\ = 2 {ab}^{2} c (2 ac) \\ = 4 a^{2} b^{2} c^{2} = R . H . S . \end{array}$

Q.15

$\begin{array}{l} Solve thee quation \\ | \begin{matrix} x + a & x & x \\ x & x + a & x \\ x & x & x + a \end{matrix} | = 0, a \neq 0 . \end{array}$

Ans

$\begin{array}{l} Given : | \begin{matrix} x + a & x & x \\ x & x + a & x \\ x & x & x + a \end{matrix} | = 0, a \neq 0. \\ | \begin{matrix} x + a & x & x \\ x & x + a & x \\ x & x & x + a \end{matrix} | = 0 \\ Apply R_{1} \to R_{1} + R_{2} + R_{3}, we have \\ \Rightarrow | \begin{matrix} 3 x + a & 3 x + a & 3 x + a \\ x & x + a & x \\ x & x & x + a \end{matrix} | = 0 \\ \Rightarrow (3 x + a) | \begin{matrix} 1 & 1 & 1 \\ x & x + a & x \\ x & x & x + a \end{matrix} | = 0 \\ Apply C_{2} \to C_{2} - C_{1} and C_{3} \to C_{3} - C_{1}, we get \\ \Rightarrow (3 x + a) | \begin{matrix} 1 & 0 & 0 \\ x & a & 0 \\ x & 0 & a \end{matrix} | = 0 \\ Expanding along R_{1}, we get \\ \Rightarrow (3 x + a) {1 (a^{2} - 0) - 0 + 0} = 0 \\ \Rightarrow a^{2} (3 x + a) = 0 \\ \Rightarrow (3 x + a) = 0 \Rightarrow x = - \frac{a}{3} \\ Therefore, x = - \frac{a}{3} . \end{array}$

Q.16

$\begin{array}{l} If a, b and c are real numbers, and \\ Δ = | \begin{matrix} b + c & c + a & a + b \\ c + a & a + b & b + c \\ a + b & b + c & c + a \end{matrix} | = 0, \\ Show that either a + b + c = 0 or a = b = c . \end{array}$

Ans

$\begin{array}{l} Δ = | \begin{matrix} b + c & c + a & a + b \\ c + a & a + b & b + c \\ a + b & b + c & c + a \end{matrix} | \\ Applying R_{1} \to R_{1} + R_{2} + R_{3}, we get \\ = | \begin{matrix} 2 (a + b + c) & 2 (a + b + c) & 2 (a + b + c) \\ c + a & a + b & b + c \\ a + b & b + c & c + a \end{matrix} | \\ = 2 (a + b + c) | \begin{matrix} 1 & 1 & 1 \\ c + a & a + b & b + c \\ a + b & b + c & c + a \end{matrix} | \\ Apply C_{2} \to C_{2} - C_{1}, C_{3} \to C_{3} - C_{1} \\ = 2 (a + b + c) | \begin{matrix} 1 & 0 & 0 \\ c + a & b - c & b - a \\ a + b & c - a & c - b \end{matrix} | \\ Expanding along R_{1}, we get \\ = 2 (a + b + c) . [1 {(b - c) (c - b) - (b - a) (c - a)} - 0 + 0] \\ = 2 (a + b + c) (bc - b^{2} - c^{2} + cb - bc + ba + ac - a^{2}) \\ = 2 (a + b + c) (- a^{2} - b^{2} - c^{2} + cb + ba + ac) \\ But Δ = 0 \\ So, \\ 2 (a + b + c) (- a^{2} - b^{2} - c^{2} + cb + ba + ac) = 0 \\ Either a + b + c = 0 or - a^{2} - b^{2} - c^{2} + cb + ba + ac = 0 \\ If - a^{2} - b^{2} - c^{2} + cb + ba + ac = 0 \\ Then, \\ - 2 a^{2} - 2 b^{2} - 2 c^{2} + 2 cb + 2 ba + 2 ac = 0 \\ \Rightarrow 2 a^{2} + 2 b^{2} + 2 c^{2} - 2 cb - 2 ba - 2 ac = 0 \\ \Rightarrow {(a - b)}^{2} + {(b - c)}^{2} + {(c - a)}^{2} = 0 \\ \Rightarrow {(a - b)}^{2} = {(b - c)}^{2} = {(c - a)}^{2} = 0 [\begin{array}{l} {(a - b)}^{2}, {(b - c)}^{2}, {(c - a)}^{2} \\ are non - negative . \end{array}] \\ \Rightarrow a = b = c \\ Thus, either a + b + c = 0 or a = b = c . \end{array}$

Q.17

$\begin{array}{l} Evaluate : \\ |\begin{matrix} cosαcosβ & cosαsinβ & - sinα \\ - sinβ & cosβ & 0 \\ sinαcosβ & sinαsinβ & cosα \end{matrix}| . \end{array}$

Ans

$\begin{array}{l} Δ = | \begin{matrix} cosαcosβ & cosαsinβ & - sinα \\ - sinβ & cosβ & 0 \\ sinαcosβ & sinαsinβ & cosα \end{matrix} | \\ Expanding along C_{3}, we get \\ = - sinα | \begin{array}{l} - sinβ cosβ \\ sinαcosβ sinαsinβ \end{array} | - 0 | \begin{matrix} cosαcosβ & cosαsinβ \\ sinαcosβ & sinαsinβ \end{matrix} | \\ + cosα | \begin{matrix} cosαcosβ & cosαsinβ \\ - sinβ & cosβ \end{matrix} | \\ = - sinα (- {sinαsin}^{2} β - {sinαcos}^{2} β) - 0 + cosα ({cosαcos}^{2} β + {cosαsin}^{2} β) \\ = \sin^{2} {αsin}^{2} β + \sin^{2} {αcos}^{2} β + \cos^{2} {αcos}^{2} β + \cos^{2} {αsin}^{2} β \\ = \sin^{2} α (\sin^{2} β + \cos^{2} β) + \cos^{2} α (\cos^{2} β + \sin^{2} β) [∵ \sin^{2} θ + \cos^{2} θ = 1] \\ = \sin^{2} α (1) + \cos^{2} α (1) \\ = \sin^{2} α + \cos^{2} α \\ = 1 = R . H . S / \\ Hence, it is proved . \end{array}$

Q.18

$\begin{array}{l} Without expanding the determinant, prove that \\ | \begin{matrix} a & a^{2} & bc \\ b & b^{2} & ca \\ c & c^{2} & ab \end{matrix} | = | \begin{matrix} 1 & a^{2} & a^{3} \\ 1 & b^{2} & b^{3} \\ 1 & c^{2} & c^{3} \end{matrix} | . \end{array}$

Ans

$\begin{array}{l} L . H . S . = | \begin{matrix} a & a^{2} & bc \\ b & b^{2} & ca \\ c & c^{2} & ab \end{matrix} | \\ Apply R_{1} \to {aR}_{1}, R_{2} \to {bR}_{2} and R_{3} \to {cR}_{3}, we get \\ = \frac{1}{abc} | \begin{matrix} a^{2} & a^{3} & abc \\ b^{2} & b^{3} & abc \\ c^{2} & c^{3} & abc \end{matrix} | \\ = \frac{1}{abc} . abc | \begin{matrix} a^{2} & a^{3} & 1 \\ b^{2} & b^{3} & 1 \\ c^{2} & c^{3} & 1 \end{matrix} | [Taking out abc from C_{3} .] \\ = (- 1) | \begin{matrix} a^{2} & 1 & a^{3} \\ b^{2} & 1 & b^{3} \\ c^{2} & 1 & c^{3} \end{matrix} | [C_{2} \leftrightarrow C_{3}] \\ = {(- 1)}^{2} | \begin{matrix} 1 & a^{2} & a^{3} \\ 1 & b^{2} & b^{3} \\ 1 & c^{2} & c^{3} \end{matrix} | [C_{1} \leftrightarrow C_{2}] \\ = | \begin{matrix} 1 & a^{2} & a^{3} \\ 1 & b^{2} & b^{3} \\ 1 & c^{2} & c^{3} \end{matrix} | = R . H . S . \\ Hence, it is proved . \end{array}$

Q.19

$\begin{array}{l} Prove that the determinant | \begin{matrix} x & sinθ & cosθ \\ - sinθ & - x & 1 \\ cosθ & 1 & x \end{matrix} | is \\ independent of θ . \end{array}$

Ans

$\begin{array}{l} | \begin{matrix} x & sinθ & cosθ \\ - sinθ & - x & 1 \\ cosθ & 1 & x \end{matrix} | \\ Expanding along R_{1}, we get \\ = x (- x^{2} - 1) - sinθ (- xsinθ - cosθ) + cosθ (- sinθ + xcosθ) \\ = - x^{3} - x + {xsin}^{2} θ + sinθcosθ - cosθsinθ + {xcos}^{2} θ \\ = - x^{3} - x + x (\sin^{2} θ + \cos^{2} θ) \\ = - x^{3} - x + x (1) [\sin^{2} θ + \cos^{2} θ = 1] \\ = - x^{3}, which is independent of θ . \\ Therefore, determinant is independent of θ . \end{array}$

Q.20 The cost of 4 kg onion, 3 kg wheat and 2 kg rice is ₹60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is ₹90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is ₹70. Find cost of each item per kg by matrix method.

Ans

$\begin{array}{l} Let cost of 1 kg onions be ₹ x, 1 kg wheat be ₹ y and \\ 1 kg rice be ₹z . Then, the given situation can be represented \\ by a system of equations as: \\ 4x + 3y + 2z = 60 \\ 2x + 4y + 6z = 90 \\ 6x + 2y + 3z = 70 \\ This system of equations can be written in the form of \\ AX = B as given below: \\ A = [\begin{matrix} 4 & 3 & 2 \\ 2 & 4 & 6 \\ 6 & 2 & 3 \end{matrix}], X = [\begin{array}{l} x \\ y \\ z \end{array}] and B= [\begin{array}{l} 60 \\ 90 \\ 70 \end{array}] \\ |A| = |\begin{matrix} 4 & 3 & 2 \\ 2 & 4 & 6 \\ 6 & 2 & 3 \end{matrix}| \\ = 4 (12 - 12) - 3 (6 - 36) + 2 (4 - 24) \\ = 0 + 90 - 40 = 50 \neq 0 \\ N o w, \\ A_{11} = 0, A_{12} = 30, A_{13} = - 20 \\ A_{21} = - 5, A_{22} = 0, A_{23} = 10 \\ A_{31} = 10, A_{32} = - 20, A_{33} = 10 \\ ∴ A d j A = {[\begin{matrix} 0 & 30 & - 20 \\ - 5 & 0 & 10 \\ 10 & - 20 & 10 \end{matrix}]}^{‘} \\ = [\begin{matrix} 0 & - 5 & 10 \\ 30 & 0 & - 20 \\ - 20 & 10 & 10 \end{matrix}] \\ ∴ A^{- 1} = \frac{1}{|A|} A d j A \\ = \frac{1}{50} [\begin{matrix} 0 & - 5 & 10 \\ 30 & 0 & - 20 \\ - 20 & 10 & 10 \end{matrix}] \\ N o w, \\ X = A^{- 1} B \\ = \frac{1}{50} [\begin{matrix} 0 & - 5 & 10 \\ 30 & 0 & - 20 \\ - 20 & 10 & 10 \end{matrix}] [\begin{array}{l} 60 \\ 90 \\ 70 \end{array}] \\ = \frac{1}{50} [\begin{array}{l} 0 - 450 + 700 \\ 1800 + 0 - 1400 \\ - 1200 + 900 + 700 \end{array}] = \frac{1}{50} [\begin{array}{l} 250 \\ 400 \\ 400 \end{array}] \\ \Rightarrow [\begin{array}{l} x \\ y \\ z \end{array}] = [\begin{array}{l} 5 \\ 8 \\ 8 \end{array}] \\ \Rightarrow x = 5, y = 8 a n d z = 8. \\ Thus, the cost of 1 kg onions is ₹5, 1 kg wheat is ₹ 8 \\ and that of 1 kg rice is ₹8 . \end{array}$

Q.21

$\begin{array}{l} If A = |\begin{matrix} 2 & - 3 & 5 \\ 3 & 2 & - 4 \\ 1 & 1 & - 2 \end{matrix}|, find {A^{-}}^{1} . {UsingA}^{- 1} solve the system of \\ equations : \\ 2 x - 3 y + 5 z = 11 \\ 3 x + 2 y - 4 z = - 5 \\ x + y - 2 z = - 3 \end{array}$

Ans

$\begin{array}{l} A = [\begin{matrix} 2 & - 3 & 5 \\ 3 & 2 & - 4 \\ 1 & 1 & - 2 \end{matrix}] \\ | A | = | \begin{matrix} 2 & - 3 & 5 \\ 3 & 2 & - 4 \\ 1 & 1 & - 2 \end{matrix} | \\ = 2 (- 4 + 4) + 3 (- 6 + 4) + 5 (3 - 2) \\ = 0 - 6 + 5 \\ = - 1 \neq 0 \\ Now, \\ A_{11} = 0, A_{12} = 2, A_{13} = 1 \\ A_{21} = - 1, A_{22} = - 9, A_{23} = - 5 \\ A_{31} = 2, A_{32} = 23, A_{33} = 13 \\ Adj A = {[\begin{matrix} 0 & 2 & 1 \\ - 1 & - 9 & - 5 \\ 2 & 23 & 13 \end{matrix}]}^{‘} \\ = [\begin{matrix} 0 & - 1 & 2 \\ 2 & - 9 & 23 \\ 1 & - 5 & 13 \end{matrix}] \\ A^{- 1} = \frac{1}{| A |} Adj A \\ = \frac{1}{- 1} [\begin{matrix} 0 & - 1 & 2 \\ 2 & - 9 & 23 \\ 1 & - 5 & 13 \end{matrix}] \\ ∴ A^{- 1} = [\begin{matrix} 0 & 1 & - 2 \\ - 2 & 9 & - 23 \\ - 1 & 5 & - 13 \end{matrix}] . .. (i) \\ Now, the given system of equations can be written in the form \\ of AX = B, where \\ A = [\begin{matrix} 2 & - 3 & 5 \\ 3 & 2 & - 4 \\ 1 & 1 & - 2 \end{matrix}], X = [\begin{array}{l} x \\ y \\ z \end{array}] and B = [\begin{array}{l} 11 \\ - 5 \\ - 3 \end{array}] \\ The solution of the system of equations is given by \\ ∴ X = A^{- 1} B \\ = [\begin{matrix} 0 & 1 & - 2 \\ - 2 & 9 & - 23 \\ - 1 & 5 & - 13 \end{matrix}] [\begin{array}{l} 11 \\ - 5 \\ - 3 \end{array}] [Using equation (i)] \\ = [\begin{array}{l} 0 - 5 + 6 \\ - 22 - 45 + 69 \\ - 11 - 25 + 39 \end{array}] \\ [\begin{array}{l} x \\ y \\ z \end{array}] = [\begin{array}{l} 1 \\ 2 \\ 3 \end{array}] \\ Hence, x = 1, y = 2 and z = 3 . \end{array}$

Q.22 Solve system of linear equations, using matrix method,

x – y + 2z = 7
3x + 4y – 5z = – 5
2x – y + 3z = 12

Ans

$\begin{array}{l} The given system of equations is : \\ x - y + 2 z = 7 \\ 3 x + 4 y - 5 z = - 5 \\ 2 x - y + 3 z = 12 \\ The given system of equations can be written in the form of \\ AX = B, where \\ A = [\begin{matrix} 1 & - 1 & 2 \\ 3 & 4 & - 5 \\ 2 & - 1 & 3 \end{matrix}], X = [\begin{array}{l} x \\ y \\ z \end{array}] and B = [\begin{array}{l} 7 \\ - 5 \\ 12 \end{array}] \\ Now, \\ | A | = | \begin{matrix} 1 & - 1 & 2 \\ 3 & 4 & - 5 \\ 2 & - 1 & 3 \end{matrix} | \\ = 1 (12 - 5) + 1 (9 + 10) + 2 (- 3 - 8) \\ = 7 + 19 - 22 = 4 \neq 0 \\ ∴ A is a non - sigular matrix . \\ Therefore, A^{- 1} exists . \\ Now, \\ A_{11} = 7, A_{12} = - 19, A_{13} = - 11 \\ A_{21} = 1, A_{22} = - 1, A_{23} = - 1 \\ A_{31} = - 3, A_{32} = 11, A_{33} = 7 \\ Adj A = {[\begin{matrix} 7 & - 19 & - 11 \\ 1 & - 1 & - 1 \\ - 3 & 11 & 7 \end{matrix}]}^{‘} \\ = [\begin{matrix} 7 & 1 & - 3 \\ - 19 & - 1 & 11 \\ - 11 & - 1 & 7 \end{matrix}] \\ A^{- 1} = \frac{1}{| A |} Adj A \\ = \frac{1}{4} [\begin{matrix} 7 & 1 & - 3 \\ - 19 & - 1 & 11 \\ - 11 & - 1 & 7 \end{matrix}] \\ ∴ X = A^{- 1} B \\ = \frac{1}{4} [\begin{matrix} 7 & 1 & - 3 \\ - 19 & - 1 & 11 \\ - 11 & - 1 & 7 \end{matrix}] [\begin{array}{l} 7 \\ - 5 \\ 12 \end{array}] \\ = \frac{1}{4} [\begin{array}{l} 49 - 5 - 36 \\ - 133 + 5 + 132 \\ - 77 + 5 + 84 \end{array}] \\ [\begin{array}{l} x \\ y \\ z \end{array}] = \frac{1}{4} [\begin{array}{l} 8 \\ 4 \\ 12 \end{array}] = [\begin{array}{l} 2 \\ 1 \\ 3 \end{array}] \\ Hence, x = 2, y = 1 and z = 3 . \end{array}$

Q.23 Solve system of linear equations, using matrix method,

2x + 3y + 3z = 5
x – 2y + z = – 4
3x – y – 2z = 3

Ans

$\begin{array}{l} The given system of equations is : \\ 2 x + 3 y + 3 z = 5 \\ x - 2 y + z = - 4 \\ 3 x - y - 2 z = 3 \\ The given system of equations can be written in the form of \\ AX = B, where \\ A = [\begin{matrix} 2 & 3 & 3 \\ 1 & - 2 & 1 \\ 3 & - 1 & - 2 \end{matrix}], X = [\begin{array}{l} x \\ y \\ z \end{array}] and B = [\begin{array}{l} 5 \\ - 4 \\ 3 \end{array}] \\ Now, \\ | A | = | \begin{matrix} 2 & 3 & 3 \\ 1 & - 2 & 1 \\ 3 & - 1 & - 2 \end{matrix} | = 2 (4 + 1) - 3 (- 2 - 3) + 3 (- 1 + 6) \\ = 10 + 15 + 15 = 40 \neq 0 \\ ∴ A is a non - sigular matrix . \\ Therefore, A^{- 1} exists . \\ Now, \\ A_{11} = 5, A_{12} = 5, A_{13} = 5 \\ A_{21} = 3, A_{22} = - 13, A_{23} = 11 \\ A_{31} = 9, A_{32} = 1, A_{33} = - 7 \\ Adj A = {[\begin{matrix} 5 & 5 & 5 \\ 3 & - 13 & 11 \\ 9 & 1 & - 7 \end{matrix}]}^{‘} \\ = [\begin{matrix} 5 & 3 & 9 \\ 5 & - 13 & 1 \\ 5 & 11 & - 7 \end{matrix}] \\ A^{- 1} = \frac{1}{| A |} Adj A \\ = \frac{1}{40} [\begin{matrix} 5 & 3 & 9 \\ 5 & - 13 & 1 \\ 5 & 11 & - 7 \end{matrix}] \\ ∴ X = A^{- 1} B \\ = \frac{1}{40} [\begin{matrix} 5 & 3 & 9 \\ 5 & - 13 & 1 \\ 5 & 11 & - 7 \end{matrix}] [\begin{array}{l} 5 \\ - 4 \\ 3 \end{array}] \\ = \frac{1}{40} [\begin{array}{l} 25 - 12 + 27 \\ 25 + 52 + 3 \\ 25 - 44 - 21 \end{array}] \\ [\begin{array}{l} x \\ y \\ z \end{array}] = \frac{1}{40} [\begin{array}{l} 40 \\ 80 \\ - 40 \end{array}] \\ = [\begin{array}{l} 1 \\ 2 \\ - 1 \end{array}] \\ Hence, x = 1, y = 2 and z = - 1 . \end{array}$

Q.24 Solve system of linear equations, using matrix method,

x – y + z = 4
2x + y – 3z = 0
x + y + z = 2

Ans

$\begin{array}{l} The given system of equations is : \\ x - y + z = 4 \\ 2 x + y - 3 z = 0 \\ x + y + z = 2 \\ The given system of equations can be written in the form of \\ AX = B, where \\ A = [\begin{matrix} 1 & - 1 & 1 \\ 2 & 1 & - 3 \\ 1 & 1 & 1 \end{matrix}], X = [\begin{array}{l} x \\ y \\ z \end{array}] and B = [\begin{array}{l} 4 \\ 0 \\ 2 \end{array}] \\ Now, \\ | A | = | \begin{matrix} 1 & - 1 & 1 \\ 2 & 1 & - 3 \\ 1 & 1 & 1 \end{matrix} | \\ = 1 (1 + 3) + 1 (2 + 3) + 1 (2 - 1) \\ = 4 + 5 + 1 = 10 \neq 0 \\ ∴ A is a non - sigular matrix . \\ Therefore, A^{- 1} exists . \\ Now, \\ A_{11} = 4, A_{12} = - 5, A_{13} = 1 \\ A_{21} = 2, A_{22} = 0, A_{23} = - 2 \\ A_{31} = 2, A_{32} = 5, A_{33} = 3 \\ Adj A = {[\begin{matrix} 4 & - 5 & 1 \\ 2 & 0 & - 2 \\ 2 & 5 & 3 \end{matrix}]}^{‘} \\ = [\begin{matrix} 4 & 2 & 2 \\ - 5 & 0 & 5 \\ 1 & - 2 & 3 \end{matrix}] \\ A^{- 1} = \frac{1}{| A |} [\begin{matrix} 4 & 2 & 2 \\ - 5 & 0 & 5 \\ 1 & - 2 & 3 \end{matrix}] \\ = \frac{1}{10} [\begin{matrix} 4 & 2 & 2 \\ - 5 & 0 & 5 \\ 1 & - 2 & 3 \end{matrix}] \\ ∴ X = A^{- 1} B \\ = \frac{1}{10} [\begin{matrix} 4 & 2 & 2 \\ - 5 & 0 & 5 \\ 1 & - 2 & 3 \end{matrix}] [\begin{array}{l} 4 \\ 0 \\ 2 \end{array}] \\ = \frac{1}{10} [\begin{array}{l} 16 + 0 + 4 \\ - 20 + 0 + 10 \\ 4 + 0 + 6 \end{array}] \\ [\begin{array}{l} x \\ y \\ z \end{array}] = \frac{1}{10} [\begin{array}{l} 20 \\ - 10 \\ 10 \end{array}] \\ = [\begin{array}{l} 2 \\ - 1 \\ 1 \end{array}] \\ Hence, x = 2, y = - 1 and z = 1 . \end{array}$

Q.25 Solve system of linear equations, using matrix method,

2x + y + z = 1
x – 2y – z = 3/2
3y – 5z = 9

Ans

$\begin{array}{l} T h e given system of equations is: \\ 2 x + y + z = 1 \\ x - 2 y - z = \frac{3}{2} \\ 3 y - 5 z = 9 \\ The given system of equations can be written in the form of \\ AX = B, where \\ A = [\begin{matrix} 2 & 1 & 1 \\ 1 & - 2 & - 1 \\ 0 & 3 & - 5 \end{matrix}], X = [\begin{array}{l} x \\ y \\ z \end{array}] and B= [\begin{array}{l} 1 \\ \frac{3}{2} \\ 9 \end{array}] \\ N o w, \\ | A | = | \begin{matrix} 2 & 1 & 1 \\ 1 & - 2 & - 1 \\ 0 & 3 & - 5 \end{matrix} | \\ = 2 (10 + 3) - 1 (- 5 - 0) + 1 (3 - 0) \\ = 26 + 5 + 3 = 34 \neq 0 \\ ∴ A is a non-sigular matrix . \\ {Therefore, A}^{-1} exists . \\ N o w, \\ A_{11} = 13, A_{12} = 5, A_{13} = 3 \\ A_{21} = 8, A_{22} = - 10, A_{23} = - 6 \\ A_{31} = 1, A_{32} = 3, A_{33} = - 5 \\ A d j A = {[\begin{matrix} 13 & 5 & 3 \\ 8 & - 10 & - 6 \\ 1 & 3 & - 5 \end{matrix}]}^{‘} \\ = [\begin{matrix} 13 & 8 & 1 \\ 5 & - 10 & 3 \\ 3 & - 6 & - 5 \end{matrix}] \\ A^{- 1} = \frac{1}{| A |} [\begin{matrix} 13 & 8 & 1 \\ 5 & - 10 & 3 \\ 3 & - 6 & - 5 \end{matrix}] \\ = \frac{1}{34} [\begin{matrix} 13 & 8 & 1 \\ 5 & - 10 & 3 \\ 3 & - 6 & - 5 \end{matrix}] \\ ∴ X = A^{- 1} B \\ = \frac{1}{34} [\begin{matrix} 13 & 8 & 1 \\ 5 & - 10 & 3 \\ 3 & - 6 & - 5 \end{matrix}] [\begin{array}{l} 1 \\ \frac{3}{\begin{array}{l} 2 \\ 9 \end{array}} \end{array}] \\ = \frac{1}{34} [\begin{array}{l} 13 + 12 + 9 \\ 5 - 15 + 27 \\ 3 - 9 - 45 \end{array}] \\ [\begin{array}{l} x \\ y \\ z \end{array}] = \frac{1}{34} [\begin{array}{l} 34 \\ 17 \\ - 51 \end{array}] = [\begin{array}{l} 1 \\ \frac{1}{2} \\ - \frac{3}{2} \end{array}] \\ H e n c e, x = 1, y = \frac{1}{2} and z = - \frac{3}{2} . \end{array}$

Q.26 Solve system of linear equations, using matrix method,

5x + 2y = 3
3x + 2y = 5

Ans

$\begin{array}{l} The given system of equations is : \\ 5 x + 2 y = 3 \\ 3 x + 2 y = 5 \\ The given system of equations can be written in the form of \\ AX = B, where \\ A = [\begin{matrix} 5 & 2 \\ 3 & 2 \end{matrix}], X = [\begin{array}{l} x \\ y \end{array}] and B = [\begin{array}{l} 3 \\ 5 \end{array}] \\ Now, \\ | A | = | \begin{matrix} 5 & 2 \\ 3 & 2 \end{matrix} | \\ = 10 - 6 \\ = 4 \neq 0 \\ ∴ A is a non - sigular matrix . \\ Therefore, A^{- 1} exists . \\ Now, \\ Adj A = {[\begin{matrix} 2 & - 3 \\ - 2 & 5 \end{matrix}]}^{‘} \\ = [\begin{matrix} 2 & - 2 \\ - 3 & 5 \end{matrix}] \\ A^{- 1} = \frac{1}{| A |} (Adj A) \\ = \frac{1}{4} [\begin{matrix} 2 & - 2 \\ - 3 & 5 \end{matrix}] \\ ∴ X = A^{- 1} B \\ = \frac{1}{4} [\begin{matrix} 2 & - 2 \\ - 3 & 5 \end{matrix}] [\begin{array}{l} 3 \\ 5 \end{array}] \\ = \frac{1}{4} [\begin{array}{l} 6 - 10 \\ - 9 + 25 \end{array}] \\ [\begin{array}{l} x \\ y \end{array}] = \frac{1}{4} [\begin{array}{l} - 4 \\ 16 \end{array}] \Rightarrow [\begin{array}{l} x \\ y \end{array}] = [\begin{array}{l} - 1 \\ 4 \end{array}] \\ ∴ x = - 1 and y = 4 . \end{array}$

Q.27 Solve system of linear equations, using matrix method,

4x – 3y = 3
3x – 5y = 7

Ans

$\begin{array}{l} T h e given system of equations is: \\ 4 x - 3 y = 3 \\ 3 x - 5 y = 7 \\ The given system of equations can be written in the form of \\ AX = B, where \\ A = [\begin{matrix} 4 & - 3 \\ 3 & - 5 \end{matrix}], X = [\begin{array}{l} x \\ y \end{array}] and B = [\begin{array}{l} 3 \\ 7 \end{array}] \\ N o w, \\ | A | = | \begin{matrix} 4 & - 3 \\ 3 & - 5 \end{matrix} | \\ = - 20 + 9 \\ = - 11 \neq 0 \\ ∴ A is a non-sigular matrix . \\ {Therefore, A}^{-1} exists . \\ Now, \\ Adj A = {[\begin{matrix} - 5 & - 3 \\ 3 & 4 \end{matrix}]}^{‘} \\ = [\begin{matrix} - 5 & 3 \\ - 3 & 4 \end{matrix}] \\ A^{-1} = \frac{1}{| A |} (Adj A) \\ = \frac{1}{- 11} [\begin{matrix} - 5 & 3 \\ - 3 & 4 \end{matrix}] \\ = \frac{1}{11} [\begin{matrix} 5 & - 3 \\ 3 & - 4 \end{matrix}] \\ ∴ X = A^{- 1} B \\ = \frac{1}{11} [\begin{matrix} 5 & - 3 \\ 3 & - 4 \end{matrix}] [\begin{array}{l} 3 \\ 7 \end{array}] \\ = \frac{1}{11} [\begin{array}{l} 15 - 21 \\ 9 - 28 \end{array}] \\ [\begin{array}{l} x \\ y \end{array}] = \frac{1}{11} [\begin{array}{l} - 6 \\ - 19 \end{array}] \\ [\begin{array}{l} x \\ y \end{array}] = [\begin{array}{l} \frac{- 6}{11} \\ \frac{- 19}{11} \end{array}] \\ ∴ x = \frac{- 6}{11} a n d y = \frac{- 19}{11} . \end{array}$

Q.29 Solve system of linear equations, using matrix method,

2x – y = –2
3x + 4y = 3

Ans

$\begin{array}{l} T h e given system of equations is: \\ 2 x - y = - 2 \\ 3 x + 4 y = 3 \\ The given system of equations can be written in the form of \\ AX = B, where \\ A = [\begin{matrix} 2 & - 1 \\ 3 & 4 \end{matrix}], X = [\begin{array}{l} x \\ y \end{array}] and B = [\begin{array}{l} - 2 \\ 3 \end{array}] \\ N o w, \\ | A | = | \begin{matrix} 2 & - 1 \\ 3 & 4 \end{matrix} | \\ = 8 + 3 \\ = 11 \neq 0 \\ ∴ A is a non-sigular matrix . \\ {Therefore, A}^{-1} exists . \\ Now, \\ Adj A = {[\begin{matrix} 4 & - 3 \\ 1 & 2 \end{matrix}]}^{‘} = [\begin{matrix} 4 & 1 \\ - 3 & 2 \end{matrix}] \\ A^{-1} = \frac{1}{| A |} (Adj A) \\ = \frac{1}{11} [\begin{matrix} 4 & 1 \\ - 3 & 2 \end{matrix}] \\ ∴ X = A^{- 1} B \\ = \frac{1}{11} [\begin{matrix} 4 & 1 \\ - 3 & 2 \end{matrix}] [\begin{array}{l} - 2 \\ 3 \end{array}] \\ = \frac{1}{11} [\begin{array}{l} - 8 + 3 \\ 6 + 6 \end{array}] \\ [\begin{array}{l} x \\ y \end{array}] = \frac{1}{11} [\begin{array}{l} - 5 \\ 12 \end{array}] \\ [\begin{array}{l} x \\ y \end{array}] = [\begin{array}{l} \frac{- 5}{11} \\ \frac{12}{11} \end{array}] \\ ∴ x = \frac{- 5}{11} a n d y = \frac{12}{11} . \end{array}$

Q.30 Solve system of linear equations, using matrix method,

5x + 2y = 4
7x + 3y = 5

Ans

$\begin{array}{l} The given system of equations is : \\ 5 x + 2 y = 4 \\ 7 x + 3 y = 5 \\ The given system of equations can be written in the form of \\ AX = B, where \\ A = [\begin{matrix} 5 & 2 \\ 7 & 3 \end{matrix}], X = [\begin{array}{l} x \\ y \end{array}] and B = [\begin{array}{l} 4 \\ 5 \end{array}] \\ Now, \\ | A | = | \begin{matrix} 5 & 2 \\ 7 & 3 \end{matrix} | \\ = 15 - 14 \\ = 1 \neq 0 \\ ∴ A is a non - sigular matrix . \\ Therefore, A^{- 1} exists . \\ Now, \\ Adj A = {[\begin{matrix} 3 & - 7 \\ - 2 & 5 \end{matrix}]}^{‘} = [\begin{matrix} 3 & - 2 \\ - 7 & 5 \end{matrix}] \\ A^{- 1} = \frac{1}{| A |} (Adj A) \\ = \frac{1}{1} [\begin{matrix} 3 & - 2 \\ - 7 & 5 \end{matrix}] \\ = [\begin{matrix} 3 & - 2 \\ - 7 & 5 \end{matrix}] \\ ∴ X = A^{- 1} B \\ = [\begin{matrix} 3 & - 2 \\ - 7 & 5 \end{matrix}] [\begin{array}{l} 4 \\ 5 \end{array}] \\ = [\begin{array}{l} 12 - 10 \\ - 28 + 25 \end{array}] \\ [\begin{array}{l} x \\ y \end{array}] = [\begin{array}{l} 2 \\ - 3 \end{array}] \\ ∴ x = 2 and y = - 3 . \end{array}$

Q.31 Examine the consistency of the system of equations.

5x – y + 4z = 5
2x + 3y + 5z = 2
5x –2y + 6z = –1

Ans

$\begin{array}{l} The given system of equations is : \\ 5 x - y + 4 z = 5 \\ 2 x + 3 y + 5 z = 2 \\ 5 x - 2 y + 6 z = - 1 \\ The given system of equations can be written in the form of \\ AX = B, where \\ A = [\begin{matrix} 5 & - 1 & 4 \\ 2 & 3 & 5 \\ 5 & - 2 & 6 \end{matrix}], X = [\begin{array}{l} x \\ y \\ z \end{array}] and B = [\begin{array}{l} 5 \\ 2 \\ - 1 \end{array}] \\ Now, \\ | A | = | \begin{matrix} 5 & - 1 & 4 \\ 2 & 3 & 5 \\ 5 & - 2 & 6 \end{matrix} | \\ = 5 (18 + 10) + 1 (12 - 25) + 4 (- 4 - 15) \\ = 140 - 13 - 76 = 51 \neq 0 \\ ∴ A is a non - sigular matrix . \\ Therefore, A^{- 1} exists . \\ Hence, the given system of equations is consistent . \end{array}$

Q.32 Examine the consistency of the system of equations.

3x – y – 2z = 2
2y – z = – 1
3x –5y = 3

Ans

$\begin{array}{l} The given system of equations is : \\ 3 x - y - 2 z = 2 \\ 2 y - z = - 1 \\ 3 x - 5 y = 3 \\ The given system of equations can be written in the form of \\ AX = B, where \\ A = [\begin{matrix} 3 & - 1 & - 2 \\ 0 & 2 & - 1 \\ 3 & - 5 & 0 \end{matrix}], X = [\begin{array}{l} x \\ y \\ z \end{array}] and B = [\begin{array}{l} 2 \\ - 1 \\ 3 \end{array}] \\ Now, \\ | A | = | \begin{matrix} 3 & - 1 & - 2 \\ 0 & 2 & - 1 \\ 3 & - 5 & 0 \end{matrix} | \\ = 3 (0 - 5) + 1 (0 + 3) - 2 (0 - 6) \\ = - 15 + 3 + 12 = 0 \\ ∴ A is a sigular matrix . \\ A_{11} = {(- 1)}^{1 + 1} | \begin{matrix} 2 & - 1 \\ - 5 & 0 \end{matrix} | = - 5 \\ A_{12} = {(- 1)}^{1 + 2} | \begin{matrix} 0 & - 1 \\ 3 & 0 \end{matrix} | = - 3 \\ A_{13} = {(- 1)}^{1 + 3} | \begin{matrix} 0 & 2 \\ 3 & - 5 \end{matrix} | = - 6 \\ A_{21} = {(- 1)}^{2 + 1} | \begin{matrix} - 1 & - 2 \\ - 5 & 0 \end{matrix} | = 10 \\ A_{22} = {(- 1)}^{2 + 2} | \begin{matrix} 3 & - 2 \\ 3 & 0 \end{matrix} | = 6 \\ A_{23} = {(- 1)}^{2 + 3} | \begin{matrix} 3 & - 1 \\ 3 & - 5 \end{matrix} | = 12 \\ A_{31} = {(- 1)}^{3 + 1} | \begin{matrix} - 1 & - 2 \\ 2 & - 1 \end{matrix} | = 5 \\ A_{32} = {(- 1)}^{3 + 2} | \begin{matrix} 3 & - 2 \\ 0 & - 1 \end{matrix} | = 3 \\ A_{33} = {(- 1)}^{3 + 3} | \begin{matrix} 3 & - 1 \\ 0 & 2 \end{matrix} | = 6 \\ (Adj A) = {[\begin{matrix} - 5 & - 3 & - 6 \\ 10 & 6 & 12 \\ 5 & 3 & 6 \end{matrix}]}^{‘} = [\begin{matrix} - 5 & 10 & 5 \\ - 3 & 6 & 3 \\ - 6 & 12 & 6 \end{matrix}] \\ (Adj A) B = [\begin{matrix} - 5 & 10 & 5 \\ - 3 & 6 & 3 \\ - 6 & 12 & 6 \end{matrix}] [\begin{array}{l} 2 \\ - 1 \\ 3 \end{array}] \\ = [\begin{array}{l} - 10 - 10 + 15 \\ - 6 - 6 + 9 \\ - 12 - 12 + 18 \end{array}] \\ = [\begin{array}{l} - 5 \\ - 3 \\ - 6 \end{array}] \neq O \\ Thus, the solution of the given system of equations does \\ not exist . Hence, the system of equations is inconsistent . \end{array}$

Q.33 Examine the consistency of the system of equations.

x + y + z = 1
2x + 3y + 2z = 2
ax + ay + 2az = 4

Ans

$\begin{array}{l} The given system of equations is : \\ x + y + z = 1 \\ 2 x + 3 y + 2 z = 2 \\ ax + ay + 2 az = 4 \\ The given system of equations can be written in the form of \\ AX = B, where \\ A = [\begin{matrix} 1 & 1 & 1 \\ 2 & 3 & 2 \\ a & a & 2 a \end{matrix}], X = [\begin{array}{l} x \\ y \\ z \end{array}] and B = [\begin{array}{l} 1 \\ 2 \\ 4 \end{array}] \\ Now, \\ | A | = | \begin{matrix} 1 & 1 & 1 \\ 2 & 3 & 2 \\ a & a & 2 a \end{matrix} | \\ = 1 (6 a - 2 a) - 1 (4 a - 2 a) + 1 (2 a - 3 a) \\ = 4 a - 2 a - a = a \neq 0 \\ ∴ A is a non - sigular matrix . \\ Therefore, A^{- 1} exists . \\ Hence, the given system of equations is consistent . \end{array}$

Q.34 Examine the consistency of the system of equations.

x + 3y = 5
2x + 6y = 8

Ans

$\begin{array}{l} The given system of equations is : \\ x + 3 y = 5 \\ 2 x + 6 y = 8 \\ The given system of equations can be written in the form of \\ AX = B, where \\ A = [\begin{matrix} 1 & 3 \\ 2 & 6 \end{matrix}], X = [\begin{array}{l} x \\ y \end{array}] and B = [\begin{array}{l} 5 \\ 8 \end{array}] \\ Now, \\ | A | = | \begin{matrix} 1 & 3 \\ 2 & 6 \end{matrix} | = 6 - 6 \\ = 0 \\ ∴ A is a sigular matrix . \\ (Adj A) = {[\begin{matrix} 6 & - 2 \\ - 3 & 1 \end{matrix}]}^{‘} = [\begin{matrix} 6 & - 3 \\ - 2 & 1 \end{matrix}] \\ (Adj A) B = [\begin{matrix} 6 & - 3 \\ - 2 & 1 \end{matrix}] [\begin{array}{l} 5 \\ 8 \end{array}] \\ = [\begin{array}{l} 30 - 24 \\ - 10 + 8 \end{array}] \\ = [\begin{array}{l} 6 \\ - 2 \end{array}] \neq O \\ Thus, the solution of the given system of equations does \\ not exist . Hence, the system of equations is inconsistent . \end{array}$

Q.35 Examine the consistency of the system of equations

2x – y = 5
x + y = 4

Ans

$\begin{array}{l} The given system of equations is : \\ 2 x - y = 5 \\ x + y = 4 \\ The given system of equations can be written in the form of \\ AX = B, where \\ A = [\begin{matrix} 2 & - 1 \\ 1 & 1 \end{matrix}], X = [\begin{array}{l} x \\ y \end{array}] and B = [\begin{array}{l} 5 \\ 4 \end{array}] \\ Now, \\ | A | = | \begin{matrix} 2 & - 1 \\ 1 & 1 \end{matrix} | \\ = 2 + 1 \\ = 3 \neq 0 \\ ∴ A is a non - sigular matrix . \\ Therefore, A^{- 1} exists . \\ Hence, the given system of equations is consistent . \end{array}$

Q.36 Examine the consistency of the system of equations.
x + 2y = 2
2x + 3y = 3

Ans

$\begin{array}{l} The given system of equations is : \\ x + 2 y = 2 \\ 2 x + 3 y = 3 \\ The given system of equations can be written in the form of \\ AX = B, where \\ A = [\begin{matrix} 1 & 2 \\ 2 & 3 \end{matrix}], X = [\begin{array}{l} x \\ y \end{array}] and B = [\begin{array}{l} 2 \\ 3 \end{array}] \\ Now, \\ | A | = | \begin{matrix} 1 & 2 \\ 2 & 3 \end{matrix} | \\ = 3 - 4 \\ = - 1 \neq 0 \\ ∴ A is a non - sigular matrix . \\ Therefore, A^{- 1} exists . \end{array}$

Hence, the given system of equations is consistent.

Q.37

$\begin{array}{l} If A is aninvertible matrix of order 2, then \det (A^{- 1}) is \\ equalto (A) Det (A) (B) \frac{1}{Det (A)} (C) 1 (D) 0 \end{array}$

Ans

$\begin{array}{l} Since, A is an invertible matrix, then A^{- 1} exists and A^{- 1} = \frac{1}{| A |} adjA . \\ Let a matrix of order 2 be as A = [\begin{matrix} a & b \\ c & d \end{matrix}], \\ then | A | = ad - bc and Adj A = [\begin{matrix} d & - b \\ - c & a \end{matrix}] \\ Now, \\ A^{- 1} = \frac{1}{| A |} Adj A \\ = \frac{1}{| A |} [\begin{matrix} d & - b \\ - c & a \end{matrix}] \\ = [\begin{matrix} \frac{d}{| A |} & - \frac{b}{| A |} \\ - \frac{c}{| A |} & \frac{a}{| A |} \end{matrix}] \\ ∴ | A^{- 1} | = | \begin{matrix} \frac{d}{| A |} & - \frac{b}{| A |} \\ - \frac{c}{| A |} & \frac{a}{| A |} \end{matrix} | \\ = \frac{1}{{| A |}^{2}} | \begin{matrix} d & - b \\ - c & a \end{matrix} | \\ = \frac{1}{{| A |}^{2}} (ad - bc) \\ = \frac{1}{{| A |}^{2}} | A | \\ = \frac{1}{| A |} \\ ∴ \det (A^{- 1}) = \frac{1}{\det (A)} \\ Thus, the correct option is (B) . \end{array}$

Q.38 Let A be a nonsingular square matrix of order 3 x 3. Then |adj A| is equal to

(A) |A|
(B) |A|2
(C) |A|3
(D) 3|A|

Ans

$\begin{array}{l} Since, (Adj A) A = | A | I \\ = [\begin{matrix} | A | & 0 & 0 \\ 0 & | A | & 0 \\ 0 & 0 & | A | \end{matrix}] \\ \Rightarrow | (Adj A) A | = | \begin{matrix} | A | & 0 & 0 \\ 0 & | A | & 0 \\ 0 & 0 & | A | \end{matrix} | \\ \Rightarrow | (Adj A) | | A | = {| A |}^{3} | \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} | \\ \Rightarrow | (Adj A) | | A | = {| A |}^{3} (I) \\ ∴ | (Adj A) | = {| A |}^{2} \\ Hence, the correct option is (B) . \end{array}$

Q.39

$\begin{array}{l} If A = |\begin{matrix} 2 & - 1 & 1 \\ - 1 & 2 & - 1 \\ 1 & - 1 & 2 \end{matrix}|, verify that \\ A^{3} - 6 A^{2} + 9 A - 4 I = O and hence find A^{- 1} . \end{array}$

Ans

$\begin{array}{l} A = [\begin{matrix} 2 & - 1 & 1 \\ - 1 & 2 & - 1 \\ 1 & - 1 & 2 \end{matrix}] \\ A^{2} = [\begin{matrix} 2 & - 1 & 1 \\ - 1 & 2 & - 1 \\ 1 & - 1 & 2 \end{matrix}] [\begin{matrix} 2 & - 1 & 1 \\ - 1 & 2 & - 1 \\ 1 & - 1 & 2 \end{matrix}] \\ = [\begin{matrix} 4 + 1 + 1 & - 2 - 2 - 1 & 2 + 1 + 2 \\ - 2 - 2 - 1 & 1 + 4 + 1 & - 1 - 2 - 2 \\ 2 + 1 + 2 & - 1 - 2 - 2 & 1 + 1 + 4 \end{matrix}] \\ = [\begin{matrix} 6 & - 5 & 5 \\ - 5 & 6 & - 5 \\ 5 & - 5 & 6 \end{matrix}] \\ A^{3} = A^{2} A \\ = [\begin{matrix} 6 & - 5 & 5 \\ - 5 & 6 & - 5 \\ 5 & - 5 & 6 \end{matrix}] [\begin{matrix} 2 & - 1 & 1 \\ - 1 & 2 & - 1 \\ 1 & - 1 & 2 \end{matrix}] \\ = [\begin{matrix} 12 + 5 + 5 & - 6 - 10 - 5 & 6 + 5 + 10 \\ - 10 - 6 - 5 & 5 + 12 + 5 & - 5 - 6 - 10 \\ 10 + 5 + 6 & - 5 - 10 - 6 & 5 + 5 + 12 \end{matrix}] = [\begin{matrix} 22 & - 21 & 21 \\ - 21 & 22 & - 21 \\ 21 & - 21 & 22 \end{matrix}] \\ ∴ A^{3} - 6 A^{2} + 9 A - 4 I \\ = [\begin{matrix} 22 & - 21 & 21 \\ - 21 & 22 & - 21 \\ 21 & - 21 & 22 \end{matrix}] - 6 [\begin{matrix} 6 & - 5 & 5 \\ - 5 & 6 & - 5 \\ 5 & - 5 & 6 \end{matrix}] + 9 [\begin{matrix} 2 & - 1 & 1 \\ - 1 & 2 & - 1 \\ 1 & - 1 & 2 \end{matrix}] \\ - 4 [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] \\ = [\begin{matrix} 22 & - 21 & 21 \\ - 21 & 22 & - 21 \\ 21 & - 21 & 22 \end{matrix}] - [\begin{matrix} 36 & - 30 & 30 \\ - 30 & 36 & - 30 \\ 30 & - 30 & 36 \end{matrix}] + [\begin{matrix} 18 & - 9 & 9 \\ - 9 & 18 & - 9 \\ 9 & - 9 & 18 \end{matrix}] \\ - [\begin{matrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{matrix}] \\ = [\begin{matrix} 40 & - 30 & 30 \\ - 30 & 40 & - 30 \\ 30 & - 30 & 40 \end{matrix}] - [\begin{matrix} 40 & - 30 & 30 \\ - 30 & 40 & - 30 \\ 30 & - 30 & 40 \end{matrix}] \\ = [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}] = O \\ Thus, A^{3} - 6 A^{2} + 5 A + 11 I = 0 . \\ Hence, it is verified . \\ Now, \\ A^{3} - 6 A^{2} + 9 A - 4 I = O \\ \Rightarrow (AAA) A^{- 1} - 6 (AA) A^{- 1} + 9 {AA}^{- 1} - 4 {IA}^{- 1} = O \\ [Post - multiplying by A^{- 1} as | A | \neq 0] \\ \Rightarrow AA ({AA}^{- 1}) - 6 A ({AA}^{- 1}) + 9 I = 4 A^{- 1} \\ \Rightarrow A^{2} I - 6 AI + 9 I = 4 A^{- 1} \\ \Rightarrow \frac{1}{4} (A^{2} - 6 A + 9 I) = A^{- 1} \\ \Rightarrow A^{- 1} = \frac{1}{4} (A^{2} - 6 A + 9 I) . .. (i) \\ Now, \\ A^{2} - 6 A + 9 I = [\begin{matrix} 6 & - 5 & 5 \\ - 5 & 6 & - 5 \\ 5 & - 5 & 6 \end{matrix}] - 6 [\begin{matrix} 2 & - 1 & 1 \\ - 1 & 2 & - 1 \\ 1 & - 1 & 2 \end{matrix}] + 9 [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] \end{array}$ $\begin{array}{l} = [\begin{matrix} 6 & - 5 & 5 \\ - 5 & 6 & - 5 \\ 5 & - 5 & 6 \end{matrix}] - [\begin{matrix} 12 & - 6 & 6 \\ - 6 & 12 & - 6 \\ 6 & - 6 & 12 \end{matrix}] + [\begin{matrix} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{matrix}] \\ = [\begin{matrix} 15 & - 5 & 5 \\ - 5 & 15 & - 5 \\ 5 & - 5 & 15 \end{matrix}] - [\begin{matrix} 12 & - 6 & 6 \\ - 6 & 12 & - 6 \\ 6 & - 6 & 12 \end{matrix}] = [\begin{matrix} 3 & 1 & - 1 \\ 1 & 3 & 1 \\ - 1 & 1 & 3 \end{matrix}] \\ F r o m equation (i), we have \\ A^{-1} = [\begin{matrix} 3 & 1 & - 1 \\ 1 & 3 & 1 \\ - 1 & 1 & 3 \end{matrix}] \end{array}$

Q.40

$\begin{array}{l} For the matrix A = [\begin{matrix} 1 & 1 & 1 \\ 1 & 2 & - 3 \\ 2 & - 1 & 3 \end{matrix}], show that \\ A^{3} - 6 A^{2} + 5 A + 11 I = O . Hence find A^{- 1} . \end{array}$

Ans

$\begin{array}{l} A = [\begin{matrix} 1 & 1 & 1 \\ 1 & 2 & - 3 \\ 2 & - 1 & 3 \end{matrix}] \\ A^{2} = [\begin{matrix} 1 & 1 & 1 \\ 1 & 2 & - 3 \\ 2 & - 1 & 3 \end{matrix}] [\begin{matrix} 1 & 1 & 1 \\ 1 & 2 & - 3 \\ 2 & - 1 & 3 \end{matrix}] \\ = [\begin{matrix} 1 +1+2 & 1 +2 - 1 & 1 - 3 +3 \\ 1 +2 - 6 & 1 +4+3 & 1 - 6 - 9 \\ 2 - 1 +6 & 2 - 2 -3 & 2 + 3 +9 \end{matrix}] \\ = [\begin{matrix} 4 & 2 & 1 \\ - 3 & 8 & - 14 \\ 7 & - 3 & 14 \end{matrix}] \\ A^{3} = A^{2} A \\ = [\begin{matrix} 4 & 2 & 1 \\ - 3 & 8 & - 14 \\ 7 & - 3 & 14 \end{matrix}] [\begin{matrix} 1 & 1 & 1 \\ 1 & 2 & - 3 \\ 2 & - 1 & 3 \end{matrix}] \\ = [\begin{matrix} 4 + 2 + 2 & 4 + 4 - 1 & 4 - 6 + 3 \\ - 3 + 8 - 28 & - 3 + 16 + 14 & - 3 - 24 - 42 \\ 7 - 3 + 28 & 7 - 6 - 14 & 7 + 9 + 42 \end{matrix}] \\ = [\begin{matrix} 8 & 7 & 1 \\ - 23 & 27 & - 69 \\ 32 & - 13 & 58 \end{matrix}] \\ ∴ A^{3} - 6 A^{2} + 5 A + 11 I \\ = [\begin{matrix} 8 & 7 & 1 \\ - 23 & 27 & - 69 \\ 32 & - 13 & 58 \end{matrix}] - 6 [\begin{matrix} 4 & 2 & 1 \\ - 3 & 8 & - 14 \\ 7 & - 3 & 14 \end{matrix}] + 5 [\begin{matrix} 1 & 1 & 1 \\ 1 & 2 & - 3 \\ 2 & - 1 & 3 \end{matrix}] \\ + 11 [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] \\ = [\begin{matrix} 8 & 7 & 1 \\ - 23 & 27 & - 69 \\ 32 & - 13 & 58 \end{matrix}] - [\begin{matrix} 24 & 12 & 6 \\ - 18 & 48 & - 84 \\ 42 & - 18 & 84 \end{matrix}] + [\begin{matrix} 5 & 5 & 5 \\ 5 & 10 & - 15 \\ 10 & - 5 & 15 \end{matrix}] \\ + [\begin{matrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{matrix}] \\ = [\begin{matrix} 8 - 24 + 5 + 11 & 7 - 12 + 5 + 0 & 1 - 6 + 5 + 0 \\ - 23 + 18 + 5 + 0 & 27 - 48 + 10 + 11 & - 69 + 84 - 15 + 0 \\ 32 - 42 + 10 + 0 & - 13 + 18 - 5 + 0 & 58 - 84 + 15 + 11 \end{matrix}] \\ = [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}] = O \\ Thus, A^{3} - 6 A^{2} + 5 A + 11 I = 0 . \\ Now, \\ A^{3} - 6 A^{2} + 5 A + 11 I = O \\ \Rightarrow (AAA) A^{- 1} - 6 (AA) A^{- 1} + 5 {AA}^{- 1} + 11 {IA}^{- 1} = O \\ [Post - multiplying by A^{- 1} as | A | \neq 0] \\ \Rightarrow AA ({AA}^{- 1}) - 6 A ({AA}^{- 1}) + 5 I = - 11 A^{- 1} \\ \Rightarrow A^{2} I - 6 AI + 5 I = - 11 A^{- 1} \\ \Rightarrow \frac{1}{- 11} (A^{2} - 6 A + 5 I) = A^{- 1} \\ \Rightarrow A^{- 1} = \frac{1}{- 11} (A^{2} - 6 A + 5 I) . .. (i) \\ Now, \\ A^{2} - 6 A + 5 I = [\begin{matrix} 4 & 2 & 1 \\ - 3 & 8 & - 14 \\ 7 & - 3 & 14 \end{matrix}] - 6 [\begin{matrix} 1 & 1 & 1 \\ 1 & 2 & - 3 \\ 2 & - 1 & 3 \end{matrix}] + 5 [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] \\ = [\begin{matrix} 4 & 2 & 1 \\ - 3 & 8 & - 14 \\ 7 & - 3 & 14 \end{matrix}] - [\begin{matrix} 6 & 6 & 6 \\ 6 & 12 & - 18 \\ 12 & - 6 & 18 \end{matrix}] + [\begin{matrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{matrix}] \end{array}$ $\begin{array}{l} = [\begin{matrix} 9 & 2 & 1 \\ - 3 & 13 & - 14 \\ 7 & - 3 & 19 \end{matrix}] - [\begin{matrix} 6 & 6 & 6 \\ 6 & 12 & - 18 \\ 12 & - 6 & 18 \end{matrix}] \\ = [\begin{matrix} 3 & - 4 & - 5 \\ - 9 & 1 & 4 \\ - 5 & 3 & 1 \end{matrix}] \\ F r o m equation (i), we have: \\ A^{- 1} = \frac{1}{- 11} (A^{2} - 6 A + 5 I) \\ = \frac{1}{- 11} [\begin{matrix} 3 & - 4 & - 5 \\ - 9 & 1 & 4 \\ - 5 & 3 & 1 \end{matrix}] = \frac{1}{11} [\begin{matrix} - 3 & 4 & 5 \\ 9 & - 1 & - 4 \\ 5 & - 3 & - 1 \end{matrix}] \end{array}$

Q.41

$\begin{array}{l} For matrix A = [\begin{array}{l} 3 2 \\ 1 1 \end{array}], find the number a and b such that \\ A^{2} + aA + bI = O . \end{array}$

Ans

$\begin{array}{l} Given matrix A = [\begin{array}{l} 3 2 \\ 1 1 \end{array}], \\ A^{2} = [\begin{array}{l} 3 2 \\ 1 1 \end{array}] [\begin{array}{l} 3 2 \\ 1 1 \end{array}] \\ = [\begin{array}{l} 9 +2 6 +2 \\ 3 +1 2 +1 \end{array}] \\ = [\begin{matrix} 11 & 8 \\ 4 & 3 \end{matrix}] \\ Now, A^{2} + aA + bI = O \\ AA (A^{- 1}) + {aAA}^{- 1} = - {bIA}^{- 1} [Post - multiplying by A^{- 1} as | A | \neq 0] \\ \Rightarrow A ({AA}^{- 1}) + aI = - {bIA}^{- 1} \\ \Rightarrow A (I) + aI = - {bIA}^{- 1} \\ \Rightarrow A^{- 1} = \frac{1}{- b} (A + aI) \\ Now, \\ | A | = | \begin{matrix} 3 & 2 \\ 1 & 1 \end{matrix} | \\ = 3 - 2 = 1 \\ \Rightarrow A^{- 1} = \frac{1}{| A |} AdjA \\ = \frac{1}{1} [\begin{matrix} 1 & - 2 \\ - 1 & 3 \end{matrix}] \\ We have A^{- 1} = \frac{1}{- b} ([\begin{array}{l} 3 2 \\ 1 1 \end{array}] + a [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]) \\ \Rightarrow [\begin{matrix} 1 & - 2 \\ - 1 & 3 \end{matrix}] = \frac{1}{- b} [\begin{matrix} 3 + a & 2 \\ 1 & 1 + a \end{matrix}] \\ \Rightarrow [\begin{matrix} 1 & - 2 \\ - 1 & 3 \end{matrix}] = [\begin{matrix} \frac{3 + a}{- b} & \frac{2}{- b} \\ \frac{1}{- b} & \frac{1 + a}{- b} \end{matrix}] \\ Comparing the corresponding elements of two matrices, we have \\ - 1 = \frac{1}{- b} \Rightarrow b = 1 \\ and 1 = \frac{3 + a}{- b} \Rightarrow - b = 3 + a \\ \Rightarrow - 1 = 3 + a \\ \Rightarrow a = - 4 \\ Thus, the required value of a and b are - 4 and 1 . \end{array}$

Q.42

$If A = |\begin{array}{l} 3 1 \\ - 1 2 \end{array}|, show that A^{2} - 5 A + 7 I = O . Hence find A^{- 1} .$

Ans

$\begin{array}{l} Given, A = [\begin{array}{l} 3 1 \\ - 1 2 \end{array}], show that A^{2} - 5 A +7 I = O . Hence find A^{- 1} . \\ A^{2} = [\begin{array}{l} 3 1 \\ - 1 2 \end{array}] [\begin{array}{l} 3 1 \\ - 1 2 \end{array}] \\ = [\begin{array}{l} 9 - 1 3 + 2 \\ - 3 - 2 - 1 + 4 \end{array}] \\ = [\begin{array}{l} 8 5 \\ - 5 3 \end{array}] \\ ∴ A^{2} - 5 A +7 I = [\begin{array}{l} 8 5 \\ - 5 3 \end{array}] - 5 [\begin{array}{l} 3 1 \\ - 1 2 \end{array}] + 7 [\begin{array}{l} 1 0 \\ 0 1 \end{array}] \\ = [\begin{array}{l} 8 5 \\ - 5 3 \end{array}] - [\begin{array}{l} 15 5 \\ - 5 10 \end{array}] + [\begin{array}{l} 7 0 \\ 0 7 \end{array}] \\ = [\begin{array}{l} 8 - 15 + 7 5 - 5 + 0 \\ - 5 + 5 + 0 3 - 10 + 7 \end{array}] \\ = [\begin{array}{l} 0 0 \\ 0 0 \end{array}] \\ Hence, \\ A^{2} - 5 A +7 I = O \\ ∴ A . A - 5 A = - 7 I \\ \Rightarrow A . A (A^{- 1}) - 5 {AA}^{- 1} = - 7 I (A^{- 1}) \\ [Post - multiplying by A^{- 1} as | A | \neq 0] \\ \Rightarrow A ({AA}^{- 1}) - 5 I = - 7 A^{- 1} \\ \Rightarrow AI - 5 I = - 7 A^{- 1} \\ \Rightarrow A^{- 1} = \frac{1}{- 7} (A - 5 I) \\ \Rightarrow A^{- 1} = \frac{1}{7} (5 I - A) \\ \Rightarrow A^{- 1} = \frac{1}{7} (5 [\begin{array}{l} 1 0 \\ 0 1 \end{array}] - [\begin{array}{l} 3 1 \\ - 1 2 \end{array}]) \\ \Rightarrow A^{- 1} = \frac{1}{7} ([\begin{array}{l} 5 0 \\ 0 5 \end{array}] - [\begin{array}{l} 3 1 \\ - 1 2 \end{array}]) \\ \Rightarrow A^{- 1} = \frac{1}{7} [\begin{matrix} 5 - 3 & 0 - 1 \\ 0 + 1 & 5 - 2 \end{matrix}] \\ \Rightarrow A^{- 1} = \frac{1}{7} [\begin{matrix} 2 & - 1 \\ 1 & 3 \end{matrix}] \end{array}$

Q.43

$Let A= |\begin{array}{l} 3 1 \\ - 1 2 \end{array}| and B = |\begin{array}{l} 6 8 \\ 7 9 \end{array}| . Verify that {(AB)}^{- 1} = B^{- 1} A^{- 1} .$

Ans

$\begin{array}{l} L e t A = [\begin{matrix} 3 & 7 \\ 2 & 5 \end{matrix}] \\ | A | = | \begin{matrix} 3 & 7 \\ 2 & 5 \end{matrix} | \\ = 15 - 14 = 1 \\ N o w, \\ A_{11} = {(- 1)}^{1 + 1} 5 = 5 \\ A_{12} = {(- 1)}^{1 + 2} 2 = - 2 \\ A_{21} = {(- 1)}^{2 + 1} 7 = - 7 \\ A_{22} = {(- 1)}^{2 + 2} 3 = 3 \\ A d j A = {[\begin{matrix} 5 & - 2 \\ - 7 & 3 \end{matrix}]}^{‘} \\ = [\begin{matrix} 5 & - 7 \\ - 2 & 3 \end{matrix}] \\ A^{- 1} = \frac{1}{| A |} . A d j A \\ = 1 [\begin{matrix} 5 & - 7 \\ - 2 & 3 \end{matrix}] \\ = [\begin{matrix} 5 & - 7 \\ - 2 & 3 \end{matrix}] \\ L e t B = [\begin{matrix} 6 & 8 \\ 7 & 9 \end{matrix}] \\ | B | = | \begin{matrix} 6 & 8 \\ 7 & 9 \end{matrix} | \\ = 54 - 56 = - 2 \\ N o w, \\ B_{11} = {(- 1)}^{1 + 1} 9 = 9 \\ B_{12} = {(- 1)}^{1 + 2} 7 = - 7 \\ B_{21} = {(- 1)}^{2 + 1} 8 = - 8 \\ B_{22} = {(- 1)}^{2 + 2} 6 = 6 \\ A d j B = {[\begin{matrix} 9 & - 7 \\ - 8 & 6 \end{matrix}]}^{‘} \\ = [\begin{matrix} 9 & - 8 \\ - 7 & 6 \end{matrix}] \\ B^{- 1} = \frac{1}{| B |} . A d j B \\ = - \frac{1}{2} [\begin{matrix} 9 & - 8 \\ - 7 & 6 \end{matrix}] \\ = [\begin{matrix} - \frac{9}{2} & 4 \\ \frac{7}{2} & - 3 \end{matrix}] \\ B^{- 1} A^{- 1} = [\begin{matrix} - \frac{9}{2} & 4 \\ \frac{7}{2} & - 3 \end{matrix}] [\begin{array}{l} 5 - 7 \\ - 2 3 \end{array}] \\ = [\begin{array}{l} - \frac{45}{2} - 8 \frac{63}{2} + 12 \\ \frac{35}{2} + 6 - \frac{49}{2} - 9 \end{array}] \\ = [\begin{matrix} - \frac{61}{2} & \frac{87}{2} \\ \frac{47}{2} & - \frac{67}{2} \end{matrix}] … (1) \\ A B = [\begin{matrix} 3 & 7 \\ 2 & 5 \end{matrix}] [\begin{matrix} 6 & 8 \\ 7 & 9 \end{matrix}] \\ = [\begin{matrix} 18 + 49 & 24 + 63 \\ 12 + 35 & 16 + 45 \end{matrix}] \\ = [\begin{matrix} 67 & 87 \\ 47 & 61 \end{matrix}] \\ | A B | = | \begin{matrix} 67 & 87 \\ 47 & 61 \end{matrix} | \\ = 4087 - 4089 = - 2 \\ A d j (A B) = {[\begin{matrix} 61 & - 47 \\ - 87 & 67 \end{matrix}]}^{‘} \\ = [\begin{matrix} 61 & - 87 \\ - 47 & 67 \end{matrix}] \\ {(A B)}^{- 1} = - \frac{1}{2} [\begin{matrix} 61 & - 87 \\ - 47 & 67 \end{matrix}] \\ = [\begin{matrix} - \frac{61}{2} & \frac{87}{2} \\ \frac{47}{2} & - \frac{67}{2} \end{matrix}] … (2) \\ T h u s, from equation (1) and equation (2), we have \\ {(A B)}^{- 1} = B^{- 1} A^{- 1} \\ H e n c e proved . \end{array}$

Q.44 Find the inverse of each of the matrices (if it exists)

$|\begin{array}{l} 1 0 2 \\ 0 cosα s inα \\ 3 sinα - cosα \end{array}|$

Ans

$\begin{array}{l} L e t A = [\begin{matrix} 1 & 0 & 0 \\ 0 & \cos α & \sin α \\ 0 & \sin α & - \cos α \end{matrix}] \\ | A | = | \begin{matrix} 1 & 0 & 0 \\ 0 & \cos α & \sin α \\ 0 & \sin α & - \cos α \end{matrix} | \\ = 1 (- \cos^{2} α - \sin^{2} α) + 0 (0 - 0) + 0 (0 - 0) \\ = - 1 \neq 0 \\ S o, the inverse of A exists . \\ Now, \\ A_{11} = {(- 1)}^{1 + 1} | \begin{array}{l} \cos α \sin α \\ \sin α - \cos α \end{array} | \\ = - \cos^{2} α - \sin^{2} α \\ = - 1 \\ A_{12} = {(- 1)}^{1 + 2} | \begin{array}{l} 0 \sin α \\ 0 - \cos α \end{array} | \\ = 0 \\ A_{13} = {(- 1)}^{1 + 3} | \begin{array}{l} 0 \cos α \\ 0 s i n α \end{array} | = 0 \\ A_{21} = {(- 1)}^{2 + 1} | \begin{array}{l} 00 \\ s i n α - \cos α \end{array} | = 0 \\ A_{22} = {(- 1)}^{2 + 2} | \begin{array}{l} 1 0 \\ 0 - \cos α \end{array} | = - \cos α \\ A_{23} = {(- 1)}^{2 + 3} | \begin{array}{l} 10 \\ 0 \sin α \end{array} | \\ = - \sin α \\ A_{31} = {(- 1)}^{3 + 1} | \begin{array}{l} 0 0 \\ \cos α \sin α \end{array} | \\ = 0 \\ A_{32} = {(- 1)}^{3 + 2} | \begin{array}{l} 1 0 \\ 0 \sin α \end{array} | \\ = - s i n α \\ A_{33} = {(- 1)}^{3 + 3} | \begin{array}{l} 10 \\ 0 \cos α \end{array} | \\ = \cos α \\ A d j A = {[\begin{matrix} - 1 & 0 & 0 \\ 0 & - \cos α & - s i n α \\ 0 & - s i n α & \cos α \end{matrix}]}^{‘} \\ = [\begin{matrix} - 1 & 0 & 0 \\ 0 & - \cos α & - s i n α \\ 0 & - s i n α & \cos α \end{matrix}] \\ A^{- 1} = \frac{1}{| A |} A d j A \\ = - 1 [\begin{matrix} - 1 & 0 & 0 \\ 0 & - \cos α & - s i n α \\ 0 & - s i n α & \cos α \end{matrix}] \\ = [\begin{matrix} 1 & 0 & 0 \\ 0 & \cos α & s i n α \\ 0 & s i n α & - \cos α \end{matrix}] \end{array}$

Q.45 Find the inverse of each of the matrices (if it exists)

$|\begin{array}{l} 1 - 1 2 \\ 0 2 - 3 \\ 3 - 2 4 \end{array}|$

Ans

$\begin{array}{l} Let A = [\begin{matrix} 1 & - 1 & 2 \\ 0 & 2 & - 3 \\ 3 & - 2 & 4 \end{matrix}] \\ | A | = | \begin{matrix} 1 & - 1 & 2 \\ 0 & 2 & - 3 \\ 3 & - 2 & 4 \end{matrix} | \\ = 1 (8 - 6) + 1 (0 + 9) + 2 (0 - 6) \\ = 2 + 9 - 12 \\ = - 1 \neq 0 \\ So, the inverse of A exists . \\ Now, \\ A_{11} = {(- 1)}^{1 + 1} | \begin{array}{l} 2 - 3 \\ - 2 4 \end{array} | \\ = 2 \\ A_{12} = {(- 1)}^{1 + 2} | \begin{array}{l} 0 - 3 \\ 3 4 \end{array} | \\ = - 9 \\ A_{13} = {(- 1)}^{1 + 3} | \begin{array}{l} 0 2 \\ 3 - 2 \end{array} | \\ = 0 - 6 \\ = - 6 \\ A_{21} = {(- 1)}^{2 + 1} | \begin{array}{l} - 1 2 \\ - 2 4 \end{array} | \\ = - (- 4 + 4) = 0 \\ A_{22} = {(- 1)}^{2 + 2} | \begin{array}{l} 1 2 \\ 3 4 \end{array} | = - 2 \\ A_{23} = {(- 1)}^{2 + 3} | \begin{array}{l} 1 - 1 \\ 3 - 2 \end{array} | \\ = - (- 2 + 3) = - 1 \\ A_{31} = {(- 1)}^{3 + 1} | \begin{array}{l} - 1 2 \\ 2 - 3 \end{array} | = - 1 \\ A_{32} = {(- 1)}^{3 + 2} | \begin{array}{l} 1 2 \\ 0 - 3 \end{array} | \\ = 3 \\ A_{33} = {(- 1)}^{3 + 3} | \begin{array}{l} 1 - 1 \\ 0 2 \end{array} | \\ = 2 \\ Adj A = {[\begin{matrix} 2 & - 9 & - 6 \\ 0 & - 2 & - 1 \\ - 1 & 3 & 2 \end{matrix}]}^{‘} \\ = [\begin{matrix} 2 & 0 & - 1 \\ - 9 & - 2 & 3 \\ - 6 & - 1 & 2 \end{matrix}] \\ A^{- 1} = \frac{1}{| A |} Adj A \\ = - 1 [\begin{matrix} 2 & 0 & - 1 \\ - 9 & - 2 & 3 \\ - 6 & - 1 & 2 \end{matrix}] \\ = [\begin{matrix} - 2 & 0 & 1 \\ 9 & 2 & - 3 \\ 6 & 1 & - 2 \end{matrix}] \end{array}$

Q.46 Find the inverse of each of the matrices (if it exists)

$|\begin{array}{l} 2 1 3 \\ 4 - 1 0 \\ - 7 2 1 \end{array}|$

Ans

$\begin{array}{l} Let A = [\begin{array}{l} 2 1 3 \\ 4 - 1 0 \\ - 7 2 1 \end{array}] \\ | A | = | \begin{array}{l} 2 1 3 \\ 4 - 1 0 \\ - 7 2 1 \end{array} | \\ = 2 (- 1 - 0) - 1 (4 - 0) + 3 (8 - 7) \\ = - 2 - 4 + 3 \\ = - 3 \neq 0 \\ So, the inverse of A exists . \\ Now, \\ A_{11} = {(- 1)}^{1 + 1} | \begin{array}{l} - 1 0 \\ 2 1 \end{array} | \\ = - 1 \\ A_{12} = {(- 1)}^{1 + 2} | \begin{array}{l} 4 0 \\ - 7 1 \end{array} | \\ = - 4 \\ A_{13} = {(- 1)}^{1 + 3} | \begin{array}{l} 4 - 1 \\ - 7 2 \end{array} | \\ = 8 - 7 \\ = 1 \\ A_{21} = {(- 1)}^{2 + 1} | \begin{array}{l} 1 3 \\ 2 1 \end{array} | \\ = - (1 - 6) \\ = 5 \\ A_{22} = {(- 1)}^{2 + 2} | \begin{array}{l} 2 3 \\ - 7 1 \end{array} | \\ = 23 \\ A_{23} = {(- 1)}^{2 + 3} | \begin{array}{l} 2 - 1 \\ - 7 2 \end{array} | \\ = - (4 + 7) \\ = - 11 \\ A_{31} = {(- 1)}^{3 + 1} | \begin{array}{l} 1 3 \\ - 1 0 \end{array} | \\ = 3 \\ A_{32} = {(- 1)}^{3 + 2} | \begin{array}{l} 2 3 \\ 4 0 \end{array} | \\ = 12 \\ A_{33} = {(- 1)}^{3 + 3} | \begin{array}{l} 2 1 \\ 4 - 1 \end{array} | \\ = - 6 \\ Adj A = {[\begin{matrix} - 1 & - 4 & 1 \\ 5 & 23 & - 11 \\ 3 & 12 & - 6 \end{matrix}]}^{‘} \\ = [\begin{matrix} - 1 & 5 & 3 \\ - 4 & 23 & 12 \\ 1 & - 11 & - 6 \end{matrix}] \\ A^{- 1} = \frac{1}{| A |} Adj A \\ = - \frac{1}{3} [\begin{matrix} - 1 & 5 & 3 \\ - 4 & 23 & 12 \\ 1 & - 11 & - 6 \end{matrix}] \end{array}$

Q.47 Find the inverse of each of the matrices (if it exists)

$|\begin{array}{l} 1 0 0 \\ 3 3 0 \\ 5 2 - 1 \end{array}|$

Ans

$\begin{array}{l} L e t A = [\begin{array}{l} 1 0 0 \\ 33 0 \\ 52 - 1 \end{array}] \\ | A | = | \begin{array}{l} 1 0 0 \\ 33 0 \\ 5 2 - 1 \end{array} | \\ = 1 (- 3 - 0) - 0 (- 3 - 0) + 0 (6 - 15) \\ = - 3 \neq 0 \\ S o, the inverse of A exists . \\ Now, \\ A_{11} = {(- 1)}^{1 + 1} | \begin{array}{l} 3 0 \\ 2 - 1 \end{array} | = - 3 \\ A_{12} = {(- 1)}^{1 + 2} | \begin{array}{l} 3 0 \\ 5 - 1 \end{array} | = 3 \\ A_{13} = {(- 1)}^{1 + 3} | \begin{array}{l} 33 \\ 52 \end{array} | \\ = 6 - 15 = - 9 \\ A_{21} = {(- 1)}^{2 + 1} | \begin{array}{l} 0 0 \\ 2 - 1 \end{array} | = 0 \\ A_{22} = {(- 1)}^{2 + 2} | \begin{array}{l} 1 0 \\ 5 - 1 \end{array} | = - 1 \\ A_{23} = {(- 1)}^{2 + 3} | \begin{array}{l} 1 0 \\ 52 \end{array} | \\ = - (2 - 0) = - 2 \\ A_{31} = {(- 1)}^{3 + 1} | \begin{array}{l} 0 0 \\ 30 \end{array} | = 0 \\ A_{32} = {(- 1)}^{3 + 2} | \begin{array}{l} 10 \\ 30 \end{array} | = 0 \\ A_{33} = {(- 1)}^{3 + 3} | \begin{array}{l} 10 \\ 3 3 \end{array} | = 3 \\ A d j A = {[\begin{matrix} - 3 & 3 & - 9 \\ 0 & - 1 & - 2 \\ 0 & 0 & 3 \end{matrix}]}^{‘} \\ = [\begin{matrix} - 3 & 0 & 0 \\ 3 & - 1 & 0 \\ - 9 & - 2 & 3 \end{matrix}] \\ A^{- 1} = \frac{1}{| A |} A d j A \\ = - \frac{1}{3} [\begin{matrix} - 3 & 0 & 0 \\ 3 & - 1 & 0 \\ - 9 & - 2 & 3 \end{matrix}] \end{array}$

Q.48 Find the inverse of each of the matrices (if it exists)

$|\begin{array}{l} 1 2 3 \\ 0 2 4 \\ 0 0 5 \end{array}|$

Ans

$\begin{array}{l} Let A = [\begin{array}{l} 1 2 3 \\ 0 2 4 \\ 0 0 5 \end{array}] \\ | A | = | \begin{array}{l} 1 2 3 \\ 0 2 4 \\ 0 0 5 \end{array} | \\ = 1 (10 - 0) - 2 (0 - 0) + 3 (0 - 0) \\ = 10 \neq 0 \\ So, the inverse of A exists . \\ Now, \\ A_{11} = {(- 1)}^{1 + 1} | \begin{array}{l} 2 4 \\ 0 5 \end{array} | = 10 \\ A_{12} = {(- 1)}^{1 + 2} | \begin{array}{l} 0 4 \\ 0 5 \end{array} | = 0 \\ A_{13} = {(- 1)}^{1 + 3} | \begin{array}{l} 0 2 \\ 0 0 \end{array} | = 0 \\ A_{21} = {(- 1)}^{2 + 1} | \begin{array}{l} 2 3 \\ 0 5 \end{array} | = - 10 \\ A_{22} = {(- 1)}^{2 + 2} | \begin{array}{l} 1 3 \\ 0 5 \end{array} | = 5 \\ A_{23} = {(- 1)}^{2 + 3} | \begin{array}{l} 1 2 \\ 0 0 \end{array} | = 0 \\ A_{31} = {(- 1)}^{3 + 1} | \begin{array}{l} 1 2 \\ 0 2 \end{array} | = 2 \\ A_{32} = {(- 1)}^{3 + 2} | \begin{array}{l} 1 3 \\ 0 4 \end{array} | = - 4 \\ A_{33} = {(- 1)}^{3 + 3} | \begin{array}{l} 1 2 \\ 0 2 \end{array} | = 2 \end{array}$ $\begin{array}{l} A d j A = {[\begin{matrix} 10 & 0 & 0 \\ - 10 & 5 & 0 \\ 2 & - 4 & 2 \end{matrix}]}^{‘} \\ = [\begin{matrix} 10 & - 10 & 2 \\ 0 & 5 & - 4 \\ 0 & 0 & 2 \end{matrix}] \\ A^{- 1} = \frac{1}{|A|} A d j A \\ = \frac{1}{10} [\begin{matrix} 10 & - 10 & 2 \\ 0 & 5 & - 4 \\ 0 & 0 & 2 \end{matrix}] \end{array}$

Q.49 Find the inverse of each of the matrices (if it exists)

$|\begin{array}{l} - 1 5 \\ - 3 2 \end{array}|$

Ans

$\begin{array}{l} Let A = [\begin{array}{l} - 1 5 \\ - 3 2 \end{array}] \\ | A | = | \begin{array}{l} - 1 5 \\ - 3 2 \end{array} | \\ = - 2 + 15 \\ = 13 \neq 0 \\ So, the inverse of A exists . \\ Now, \\ A_{11} = {(- 1)}^{1 + 1} 2 = 2 \\ A_{12} = {(- 1)}^{1 + 2} (- 3) = 3 \\ A_{21} = {(- 1)}^{2 + 1} (5) = - 5 \\ A_{22} = {(- 1)}^{2 + 2} (- 1) = - 1 \\ Adj A = {[\begin{array}{l} 2 3 \\ - 5 - 1 \end{array}]}^{‘} \\ = [\begin{array}{l} 2 - 5 \\ 3 - 1 \end{array}] \\ A^{- 1} = \frac{1}{| A |} Adj A = \frac{1}{13} [\begin{array}{l} 2 - 5 \\ 3 - 1 \end{array}] \end{array}$

Q.50 Find the inverse of each of the matrices (if it exists)

$[\begin{array}{l} 2 - 2 \\ 4 3 \end{array}]$

Ans

$\begin{array}{l} Let A = [\begin{array}{l} 2 - 2 \\ 4 3 \end{array}] \\ | A | = | \begin{array}{l} 2 - 2 \\ 4 3 \end{array} | \\ = 6 + 8 \\ = 14 \neq 0 \\ So, the inverse of A exists . \\ Now, \\ A_{11} = {(- 1)}^{1 + 1} 3 = 3 \\ A_{12} = {(- 1)}^{1 + 2} 4 = - 4 \\ A_{21} = {(- 1)}^{2 + 1} (- 2) = 2 \\ A_{22} = {(- 1)}^{2 + 2} 2 = 2 \\ Adj A = {[\begin{array}{l} 3 - 4 \\ 2 2 \end{array}]}^{‘} \\ = [\begin{array}{l} 3 2 \\ - 4 2 \end{array}] \\ A^{- 1} = \frac{1}{| A |} Adj A = \frac{1}{14} [\begin{array}{l} 3 2 \\ - 4 2 \end{array}] \end{array}$

Q.51 Verify A(adj A)=(adj A)A = |A|I

$|\begin{array}{l} 1 - 1 2 \\ 3 0 - 2 \\ 1 0 3 \end{array}|$

Ans

$\begin{array}{l} L e t A = [\begin{array}{l} 1 - 1 2 \\ 3 0 - 2 \\ 1 0 3 \end{array}] \\ | A | = | \begin{array}{l} 1 - 1 2 \\ 3 0 - 2 \\ 1 0 3 \end{array} | \\ = 1 (0 - 0) + 1 (9 + 2) + 2 (0 - 0) \\ = 11 \\ A_{11} = {(- 1)}^{1 + 1} | \begin{array}{l} 0 - 2 \\ 0 3 \end{array} | = 0, \\ A_{12} = {(- 1)}^{1 + 2} | \begin{array}{l} 3 - 2 \\ 1 3 \end{array} | = - 11, \\ A_{13} = {(- 1)}^{1 + 3} | \begin{array}{l} 30 \\ 10 \end{array} | = 0, \\ A_{21} = {(- 1)}^{2 + 1} | \begin{array}{l} - 12 \\ 03 \end{array} | = 3 \\ A_{22} = {(- 1)}^{2 + 2} | \begin{array}{l} 12 \\ 13 \end{array} | = 1 \\ A_{23} = {(- 1)}^{2 + 3} | \begin{array}{l} 1 - 1 \\ 1 0 \end{array} | = - 1 \\ A_{31} = {(- 1)}^{3 + 1} | \begin{array}{l} - 1 2 \\ 0 - 2 \end{array} | = 2 \\ A_{32} = {(- 1)}^{3 + 2} | \begin{array}{l} 1 2 \\ 3 - 2 \end{array} | = 8 \\ A_{33} = {(- 1)}^{3 + 3} | \begin{array}{l} 1 - 1 \\ 3 0 \end{array} | = 3 \\ A d j A = {[\begin{array}{l} 0 - 11 0 \\ 3 1 - 1 \\ 2 8 3 \end{array}]}^{‘} \\ = [\begin{array}{l} 0 3 2 \\ - 11 1 8 \\ 0 - 1 3 \end{array}] \\ A (A d j A) = [\begin{array}{l} 0 3 2 \\ - 11 1 8 \\ 0 - 1 3 \end{array}] [\begin{array}{l} 1 - 1 2 \\ 3 0 - 2 \\ 1 0 3 \end{array}] \\ = [\begin{array}{l} 0 + 9 + 20 + 0 + 00 - 6 + 6 \\ - 11 + 3 + 8 11 + 0 + 0 - 2 2 - 2 + 24 \\ 0 - 3 + 30 + 0 + 00 + 2 + 9 \end{array}] \\ = [\begin{array}{l} 110 0 \\ 0 110 \\ 0011 \end{array}] \\ A l s o, \\ (A d j A) A = [\begin{array}{l} 1 - 1 2 \\ 3 0 - 2 \\ 1 0 3 \end{array}] [\begin{array}{l} 0 3 2 \\ - 11 1 8 \\ 0 - 1 3 \end{array}] \\ = [\begin{array}{l} 0 + 11 + 03 - 1 - 2 2 - 8 + 6 \\ 0 + 0 + 09 + 0 + 26 + 0 - 6 \\ 0 + 0 + 0 3 + 0 - 3 2 + 0 + 9 \end{array}] \\ = [\begin{array}{l} 1100 \\ 0 11 0 \\ 0011 \end{array}] \\ | A | I = 11 [\begin{array}{l} 1 0 0 \\ 0 1 0 \\ 001 \end{array}] \\ = [\begin{array}{l} 11 00 \\ 011 0 \\ 0011 \end{array}] \Rightarrow A (A d j A) = (A d j A) A = | A | I \end{array}$

Q.52 Verify A(adj A) = (adj A)A = |A|I

$|\begin{array}{l} 2 3 \\ - 4 - 6 \end{array}|$

Ans

$\begin{array}{l} Let A = [\begin{array}{l} 2 3 \\ - 4 - 6 \end{array}] \\ | A | = | \begin{array}{l} 2 3 \\ - 4 - 6 \end{array} | \\ = - 12 + 12 = 0 \\ A_{11} = {(- 1)}^{1 + 1} (- 6) = - 6, \\ A_{12} = {(- 1)}^{1 + 2} (- 4) = 4 \\ A_{21} = {(- 1)}^{2 + 1} 3 = - 3 \\ A_{22} = {(- 1)}^{2 + 2} (2) = 2 \\ Adj A = {[\begin{array}{l} - 6 4 \\ - 3 2 \end{array}]}^{‘} \\ = [\begin{array}{l} - 6 - 3 \\ 4 2 \end{array}] \\ A . (AdjA) = [\begin{array}{l} 2 3 \\ - 4 - 6 \end{array}] [\begin{array}{l} - 6 - 3 \\ 4 2 \end{array}] \\ = [\begin{array}{l} - 12 + 12 - 6 + 6 \\ 24 - 24 12 - 12 \end{array}] \\ = [\begin{array}{l} 0 0 \\ 0 0 \end{array}] = 0 \\ (AdjA) . A = [\begin{array}{l} - 6 - 3 \\ 4 2 \end{array}] [\begin{array}{l} 2 3 \\ - 4 - 6 \end{array}] \\ = [\begin{array}{l} - 12 + 12 - 18 + 18 \\ 12 - 12 12 - 12 \end{array}] \\ = [\begin{array}{l} 0 0 \\ 0 0 \end{array}] = 0 \\ | A | I = 0 [\begin{array}{l} 1 0 \\ 0 1 \end{array}] = 0 \\ Thus, \\ A . (AdjA) = (AdjA) . A = | A | I \end{array}$

Q.53 Find adjoint of each of the matrices

$|\begin{array}{l} 1 - 1 2 \\ 2 3 5 \\ - 2 0 1 \end{array}|$

Ans

$\begin{array}{l} Let A = [\begin{array}{l} 1 - 1 2 \\ 2 3 5 \\ - 2 0 1 \end{array}] \\ We have \\ A_{11} = {(- 1)}^{1 + 1} | \begin{array}{l} 3 5 \\ 0 1 \end{array} | \\ = 3 \\ A_{12} = {(- 1)}^{1 + 2} | \begin{array}{l} 2 5 \\ - 2 1 \end{array} | \\ = - (2 + 10) \\ = - 12 \\ A_{13} = {(- 1)}^{1 + 3} | \begin{array}{l} 2 3 \\ - 2 0 \end{array} | \\ = (0 + 6) \\ = 6 \\ A_{21} = {(- 1)}^{2 + 1} | \begin{array}{l} - 1 2 \\ 0 1 \end{array} | \\ = 1 \\ A_{22} = {(- 1)}^{2 + 2} | \begin{array}{l} 1 2 \\ - 2 1 \end{array} | \\ = 5 \\ A_{23} = {(- 1)}^{2 + 3} | \begin{array}{l} 1 - 1 \\ - 2 0 \end{array} | \\ = - 1 (0 - 2) \\ = 2 \\ A_{31} = {(- 1)}^{3 + 1} | \begin{array}{l} - 1 2 \\ 3 5 \end{array} | \\ = - 5 - 6 \\ = - 11 \\ A_{32} = {(- 1)}^{3 + 2} | \begin{array}{l} 1 2 \\ 2 5 \end{array} | \\ = - (5 - 4) \\ = - 1 \\ A_{33} = {(- 1)}^{3 + 3} | \begin{array}{l} 1 - 1 \\ 2 3 \end{array} | \\ = 3 + 2 \\ = 5 \\ Adj A = {[\begin{array}{l} 3 - 12 6 \\ 1 52 \\ - 11 - 1 5 \end{array}]}^{‘} = [\begin{array}{l} 3 1 - 11 \\ - 12 5 - 1 \\ 6 2 5 \end{array}] \end{array}$

Q.54 Find adjoint of each of the matrices

$|\begin{array}{l} 1 2 \\ 3 4 \end{array}|$

Ans

$\begin{array}{l} Let A = [\begin{array}{l} 1 2 \\ 3 4 \end{array}] \\ We have \\ A_{11} = {(- 1)}^{1 + 1} 4 \\ = 4 \\ A_{12} = {(- 1)}^{1 + 2} 3 \\ = - 3 \\ A_{21} = {(- 1)}^{2 + 1} 2 \\ = - 2 \\ A_{22} = {(- 1)}^{2 + 2} 1 \\ = 1 \\ Adj A = {[\begin{array}{l} 4 - 3 \\ - 2 1 \end{array}]}^{‘} \\ = [\begin{array}{l} 4 - 2 \\ - 3 1 \end{array}] \end{array}$

Q.55

$\begin{array}{l} If Δ = | \begin{array}{l} a_{11} a_{12} a_{13} \\ a_{21} a_{22} a_{23} \\ a_{31} a_{32} a_{33} \end{array} | and A_{ij} is Co factor so {fa}_{ij}, then value \\ of Δ is given by \\ (A) a_{11} A_{31} + a_{12} A_{32} + a_{13} A_{33} (B) a_{11} A_{11} + a_{12} A_{21} + a_{13} A_{31} \\ (C) a_{21} A_{11} + a_{22} A_{12} + a_{23} A_{13} (D) a_{11} A_{11} + a_{21} A_{21} + a_{31} A_{31} \end{array}$

Ans

$\begin{array}{l} Since, we know that \\ Δ = Sum of product of elements of C_{1} with their corresponding cofactors \\ = a_{11} A_{11} + a_{21} A_{21} + a_{31} A_{31} \\ Thus, the option D is correct . \end{array}$

Q.56

$\begin{array}{l} Using Co factors of element sof third column, evaluate \\ Δ = |\begin{array}{l} 1 x yz \\ 1 y zx \\ 1 z xy \end{array}| . \end{array}$

Ans

$\begin{array}{l} Given determinant is \\ Δ = | \begin{array}{l} 1 x yz \\ 1 y zx \\ 1 z xy \end{array} | \\ Minors of elements of third column are \\ M_{13} = | \begin{array}{l} 1 y \\ 1 z \end{array} | = z - y \\ M_{23} = | \begin{array}{l} 1 x \\ 1 z \end{array} | = z - x \\ M_{33} = | \begin{array}{l} 1 x \\ 1 y \end{array} | = y - x \\ Cofactors of elements of third column are \\ A_{13} = {(- 1)}^{1 + 3} M_{13} \\ = z - y \\ A_{23} = {(- 1)}^{2 + 3} M_{23} \\ = - (z - x) \\ = x - z \\ A_{33} = {(- 1)}^{3 + 3} M_{33} \\ = y - x \\ Expanding the determinant along C_{3}, we get \\ Δ = a_{13} A_{13} + a_{23} A_{23} + a_{33} A_{33} \\ = yz (z - y) + zx (x - z) + xy (y - x) \\ = {yz}^{2} - y^{2} z + {zx}^{2} - z^{2} x + {xy}^{2} - x^{2} y \\ = {zx}^{2} - y^{2} z - x^{2} y + {xy}^{2} + {yz}^{2} - z^{2} x \\ = z (x^{2} - y^{2}) + xy (y - x) + z^{2} (y - x) \\ = (x - y) {z (x + y) - xy - z^{2}} \\ = (x - y) (zx + zy - xy - z^{2}) \\ = (x - y) (zx - xy + zy - z^{2}) \\ = (x - y) {- x (y - z) + z (y - z)} \\ = (x - y) (y - z) (z - x) \end{array}$

Q.57

$\begin{array}{l} Using Co factors of element so f second row, evaluate \\ Δ = | \begin{array}{l} 5 3 8 \\ 2 0 1 \\ 1 2 3 \end{array} | \end{array}$

Ans

$\begin{array}{l} Δ = | \begin{array}{l} 5 3 8 \\ 2 0 1 \\ 1 2 3 \end{array} | \\ Minors corresponding to second row : \\ M_{21} = | \begin{array}{l} 3 8 \\ 2 3 \end{array} | \\ = 9 - 16 \\ = - 7 \\ ∵ Cofactor A_{ij} = {(- 1)}^{i + j} M_{ij} \\ ∴ A_{21} = {(- 1)}^{2 + 1} M_{21} \\ = (- 1) (- 7) \\ = 7 \\ M_{22} = | \begin{array}{l} 5 8 \\ 1 3 \end{array} | \\ = 15 - 8 \\ = 7 \\ A_{22} = {(- 1)}^{2 + 2} M_{22} \\ = 7 \\ M_{23} = | \begin{array}{l} 5 3 \\ 1 2 \end{array} | \\ = 10 - 3 \\ = 7 \\ A_{23} = {(- 1)}^{2 + 3} M_{23} \\ = (- 1) (7) \\ = - 7 \\ Expanding the determinant along R_{2}, we get \\ Δ = a_{21} A_{21} + a_{22} A_{22} + a_{23} A_{23} \\ = 2 \times 7 + 0 \times 7 + 1 \times - 7 \\ = 14 + 0 - 7 \\ = 7 \end{array}$

Q.58 Write Minors and Cofactors of the elements of following determinants:

$(i) |\begin{array}{l} 1 0 0 \\ 0 1 0 \\ 0 0 1 \end{array}|$ $(ii) |\begin{array}{l} 1 0 4 \\ 3 5 - 1 \\ 0 1 2 \end{array}|$

Ans

$\begin{array}{l} (i) | \begin{array}{l} 1 0 0 \\ 0 1 0 \\ 0 0 1 \end{array} | \\ Minor of the element a_{ij} is M_{ij} \\ ∴ Minor of element a_{11} = M_{11} \\ = | \begin{array}{l} 1 0 \\ 0 1 \end{array} | \\ = 1 \\ Minor of element a_{12} = M_{12} \\ = | \begin{array}{l} 0 0 \\ 0 1 \end{array} | \\ = 0 \\ Minor of element a_{13} = M_{13} \\ = | \begin{array}{l} 0 1 \\ 0 0 \end{array} | \\ = 0 \\ Minor of element a_{21} = M_{21} \\ = | \begin{array}{l} 0 0 \\ 0 1 \end{array} | \\ = 0 \\ Minor of element a_{22} = M_{22} \\ = | \begin{array}{l} 1 0 \\ 0 1 \end{array} | \\ = 1 \\ Minor of element a_{23} = M_{23} \\ = | \begin{array}{l} 1 0 \\ 0 0 \end{array} | \\ = 0 \\ Minor of element a_{31} = M_{31} \\ = | \begin{array}{l} 0 0 \\ 1 0 \end{array} | \\ = 0 \\ Minor of element a_{32} = M_{32} \\ = | \begin{array}{l} 1 0 \\ 0 0 \end{array} | \\ = 0 \\ Minor of element a_{33} = M_{33} \\ = | \begin{array}{l} 1 0 \\ 0 1 \end{array} | \\ = 1 \\ Now, cofactor of a_{ij} is A_{ij}, so \\ A_{ij} = {(- 1)}^{i + j} M_{ij} \\ A_{11} = {(- 1)}^{1 + 1} M_{11} \\ = 1 \\ A_{12} = {(- 1)}^{1 + 2} M_{12} \\ = 0 \\ A_{13} = {(- 1)}^{1 + 3} M_{13} \\ = 0 \\ A_{21} = {(- 1)}^{2 + 1} M_{21} \\ = 0 \end{array}$ $\begin{array}{l} A_{22} = {(- 1)}^{2 + 2} M_{22} \\ = 1 \\ A_{23} = {(- 1)}^{2 + 3} M_{23} \\ = 0 \\ A_{31} = {(- 1)}^{3 + 1} M_{31} \\ = 0 \\ A_{32} = {(- 1)}^{3 + 2} M_{32} \\ = 0 \\ A_{33} = {(- 1)}^{3 + 3} M_{33} \\ = 1 \\ (ii) | \begin{array}{l} 1 0 4 \\ 3 5 - 1 \\ 0 1 2 \end{array} | \\ M i n o r {of the element a}_{ij} {is M}_{ij} \\ ∴ M i n o r {of element a}_{11} = M_{11} \\ = | \begin{array}{l} 5 - 1 \\ 1 2 \end{array} | \\ = 10 + 1 \\ = 11 \\ M i n o r {of element a}_{12} = M_{12} \\ = | \begin{array}{l} 3 - 1 \\ 0 2 \end{array} | \\ = 6 \\ M i n o r {of element a}_{13} = M_{13} \\ = | \begin{array}{l} 3 5 \\ 0 1 \end{array} | \\ = 3 \\ M i n o r {of element a}_{21} = M_{21} \\ = | \begin{array}{l} 0 4 \\ 1 2 \end{array} | \\ = - 4 \\ M i n o r {of element a}_{22} = M_{22} \\ = | \begin{array}{l} 1 4 \\ 0 2 \end{array} | \\ = 2 \\ M i n o r {of element a}_{23} = M_{23} \\ = | \begin{array}{l} 1 0 \\ 0 1 \end{array} | \\ = 1 \\ M i n o r {of element a}_{31} = M_{31} \\ = | \begin{array}{l} 0 4 \\ 5 - 1 \end{array} | \\ = - 20 \\ M i n o r {of element a}_{32} = M_{32} \\ = | \begin{array}{l} 1 4 \\ 3 - 1 \end{array} | \\ = - 13 \\ M i n o r {of element a}_{33} = M_{33} \\ = | \begin{array}{l} 1 0 \\ 3 5 \end{array} | \\ = 5 \\ N o w, {cofactor of a}_{ij} {is A}_{ij}, so \\ A_{ij} = {(- 1)}^{i + j} M_{i j} \\ A_{11} = {(- 1)}^{1 + 1} M_{11} \\ = 11 \\ A_{12} = {(- 1)}^{1 + 2} M_{12} \\ = - 6 \\ A_{13} = {(- 1)}^{1 + 3} M_{13} \\ = 3 \\ A_{21} = {(- 1)}^{2 + 1} M_{21} \\ = (- 1) (- 4) \\ = 4 \\ A_{22} = {(- 1)}^{2 + 2} M_{22} \\ = 2 \end{array}$ $\begin{array}{l} A_{23} = {(- 1)}^{2 + 3} M_{23} \\ = - 1 \\ A_{31} = {(- 1)}^{3 + 1} M_{31} \\ = - 20 \\ A_{32} = {(- 1)}^{3 + 2} M_{32} \\ = (- 1) (- 1) \\ = 13 \\ A_{33} = {(- 1)}^{3 + 3} M_{33} \\ = 5 \end{array}$

Q.59 Write Minors and Cofactors of the elements of following determinants:

$(i) |\begin{array}{l} 2 - 4 \\ 0 3 \end{array}| (ii) |\begin{array}{l} a c \\ b d \end{array}|$

Ans

$\begin{array}{l} (i) | \begin{array}{l} 2 - 4 \\ 0 3 \end{array} | \\ Minor of the element a_{ij} is M_{ij} \\ ∴ Minor of element a_{11} = M_{11} = 3, \\ Minor of element a_{12} = M_{12} = 0, \\ Minor of element a_{21} = M_{21} = - 4, \\ Minor of element a_{22} = M_{22} = 2 . \\ Cofactor of the element a_{ij} is A_{ij} = {(- 1)}^{(i + j)} M_{ij} \\ ∴ A_{11} = {(- 1)}^{(1 + 1)} M_{11} = 3 \\ A_{12} = {(- 1)}^{(1 + 2)} M_{12} = 0 \\ A_{21} = {(- 1)}^{(2 + 1)} M_{21} = 4 \\ A_{22} = {(- 1)}^{(2 + 2)} M_{22} = 2 \\ (ii) | \begin{array}{l} a c \\ b d \end{array} | \\ Minor of the element a_{ij} is M_{ij} \\ ∴ Minor of element a_{11} = M_{11} = d, \\ Minor of element a_{12} = M_{12} = b, \\ Minor of element a_{21} = M_{21} = c, \\ Minor of element a_{22} = M_{22} = a . \\ Cofactor of the element a_{ij} is A_{ij} = {(- 1)}^{(i + j)} M_{ij} \\ ∴ A_{11} = {(- 1)}^{(1 + 1)} M_{11} = d \\ A_{12} = {(- 1)}^{(1 + 2)} M_{12} = - b \\ A_{21} = {(- 1)}^{(2 + 1)} M_{21} = - c \\ A_{22} = {(- 1)}^{(2 + 2)} M_{22} = a \end{array}$

Q.60 If area of triangle is 35 sq. units with vertices (2, -6), (5, 4) and (k, 4). Then k is
(A) 12 (B) –2 (C) –12, –2 (D) 12, –2

Ans

$\begin{array}{l} Given : area of triangle = \pm 35 (\begin{array}{l} ∵ Since area of triangle is absolute \\ value of Δ \end{array}) \\ Then the area of triangle formed by given vertices (2, - 6), \\ (5, 4), (k, 4) \\ Δ = \frac{1}{2} | \begin{array}{l} 2 - 6 1 \\ 5 4 1 \\ k 4 1 \end{array} | \\ \pm 35 = \frac{1}{2} {2 (4 - 4) - (- 6) (5 - k) + 1 (20 - 4 k)} \\ \pm 35 = \frac{1}{2} (0 + 30 - 6 k + 20 - 4 k) \\ \pm 70 = - 10 k + 50 \\ \Rightarrow - 10 k + 50 = 70 or - 10 k + 50 = - 70 \\ \Rightarrow k = \frac{20}{- 10} = - 2 or k = \frac{- 120}{- 10} = 12 \\ Thus, value of k is either 12 or - 2 . \\ Thus, option (D) is correct . \end{array}$

Q.61 (i) Find equation of line joining (1, 2) and (3, 6) using determinants.
(ii) Find equation of line joining (3, 1) and (9, 3) using determinants.

Ans

$\begin{array}{l} (i) Point A (1, 2) and B (3, 6) are joining each other and a point P (x, y) lies \\ on it, so A, P and B are collinear and area of triangle formed by these \\ points is zero . Then \\ A r e a o f triangle ABP = 0 \\ | \begin{array}{l} 121 \\ 36 1 \\ x y 1 \end{array} | = 0 \\ \Rightarrow \frac{1}{2} {1 (6 - y) - 2 (3 - x) + 1 (3 y - 6 x)} = 0 \\ \Rightarrow 6 - y - 6 + 2 x + 3 y - 6 x = 0 \\ \Rightarrow 2 y - 4 x = 0 \\ \Rightarrow y = 2 x \\ Thus, the equation of line joining two points (1, 2) and (3, 6) \\ i s y = 2x . \\ (i i) Point A (3, 1) and B (9, 3) are joining each other and a point P (x, y) lies \\ on it, so A, P and B are collinear and area of triangle formed by these \\ points is zero . Then \\ A r e a o f triangle ABP = 0 \\ | \begin{array}{l} 31 1 \\ 9 31 \\ x y 1 \end{array} | = 0 \\ \Rightarrow \frac{1}{2} {3 (3 - y) - 1 (9 - x) + 1 (9 y - 3 x)} = 0 \\ \Rightarrow 9 - 3 y - 9 + x + 9 y - 3 x = 0 \\ \Rightarrow 6 y - 2 x = 0 \\ \Rightarrow 3 y = x \\ Thus, the equation of line joining two points (1, 2) and (3, 6) \\ i s x - 3y = 0 . \end{array}$

(i) (k, 0), (4, 0), (0, 2) (ii) (–2, 0), (0, 4), (0, k)

Ans

$\begin{array}{l} (i) Given : area of triangle = \pm 4 (\begin{array}{l} ∵ Since area of triangle is absolute \\ value of Δ \end{array}) \\ Then the area of triangle formed by given vertices (k, 0), \\ (4, 0), (0, 2) \\ Δ = \frac{1}{2} | \begin{array}{l} k 0 1 \\ 4 0 1 \\ 0 2 1 \end{array} | \\ \pm 4 = \frac{1}{2} {k (0 - 2) - 0 (4 - 0) + 1 (8 - 0)} \\ \pm 4 = \frac{1}{2} (- 2 k - 0 + 8) \\ \pm 8 = - 2 k + 8 \\ \Rightarrow - 2 k + 8 = 8 or - 2 k + 8 = - 8 \\ \Rightarrow k = 0 or k = \frac{- 16}{- 2} = 8 \\ Thus, value of k is either 0 or 8 . \\ (ii) Given : area of triangle = \pm 4 (\begin{array}{l} ∵ Since area of triangle is absolute \\ value of Δ \end{array}) \\ Then the area of triangle formed by given vertices (- 2, 0), \\ (0, 4), (0, k) \\ Δ = \frac{1}{2} | \begin{array}{l} - 2 0 1 \\ 0 4 1 \\ 0 k 1 \end{array} | \\ \pm 4 = \frac{1}{2} {- 2 (4 - k) - 0 (0 - 0) + 1 (0 - 0)} \\ \pm 4 = \frac{1}{2} (- 8 + 2 k - 0 + 0) \\ \pm 8 = 2 k - 8 \\ \Rightarrow 2 k - 8 = 8 or 2 k - 8 = - 8 \\ \Rightarrow k = \frac{16}{2} = 8 or k = \frac{0}{2} = 0 \\ Thus, value of k is either 8 or 0 . \end{array}$

Q.63 Show that points A(a, b + c), B(b, c + a), C(c, a + b) are collinear.

Ans

$\begin{array}{l} Theare a of triangle formed by given vertices A (a, b + c), \\ B (b, c + a), C (c, a + b) \\ Δ = \frac{1}{2} | \begin{matrix} a & b + c & 1 \\ b & c + a & 1 \\ c & a + b & 1 \end{matrix} | \\ = \frac{1}{2} {a (c + a - a - b) - (b + c) (b - c) + 1 (ba + b^{2} - c^{2} - ca)} \\ = \frac{1}{2} {ac - ab - (b^{2} - c^{2}) + (ba + b^{2} - c^{2} - ca)} \\ = \frac{1}{2} (ac - ab - b^{2} + c^{2} + ba + b^{2} - c^{2} - ca) \\ = 0 \\ Since, area of triangle formed by A, B and C is zero, so these \\ points are collinear . \end{array}$

Q.64 Find area of the triangle with vertices at the point given in each of the following:
(i) (1, 0), (6, 0), (4, 3)
(ii) (2, 7), (1, 1), (10, 8)
(iii) (–2, –3), (3, 2), (–1, –8)

Ans

$\begin{array}{l} (i) Theare a of triangle formed by given vertices (1, 0), (6, 0), (4, 3) \\ Δ = \frac{1}{2} | \begin{array}{l} 1 0 1 \\ 6 0 1 \\ 4 3 1 \end{array} | \\ = \frac{1}{2} {1 (0 - 3) - 0 (6 - 4) + 1 (18 - 0)} \\ = \frac{1}{2} (- 3 - 0 + 18) \\ = \frac{1}{2} (15) \\ = \frac{15}{2} sq . units \\ (ii) Theare a of triangle formed by given vertices (2, 7), (1, 1), (10, 8) \\ Δ = \frac{1}{2} | \begin{array}{l} 2 7 1 \\ 1 1 1 \\ 10 8 1 \end{array} | \\ = \frac{1}{2} {2 (1 - 8) - 7 (1 - 10) + 1 (8 - 10)} \\ = \frac{1}{2} (- 14 + 63 - 2) \\ = \frac{1}{2} (47) \\ = \frac{47}{2} sq . units \\ (iii) The area of triangle formed by given vertices (- 2, - 3), \\ (3, 2), (- 1, - 8) \\ Δ = \frac{1}{2} | \begin{array}{l} - 2 - 3 1 \\ 3 2 1 \\ - 1 - 8 1 \end{array} | \\ = \frac{1}{2} {- 2 (2 + 8) - (- 3) (3 + 1) + 1 (- 24 + 2)} \\ = \frac{1}{2} (- 20 + 12 - 22) \\ = \frac{1}{2} (- 30) \\ = - 15 sq . units \\ Since, area of triangle can never be negative . \\ So, area of triangle = 15 square units . \end{array}$

Q.65 Which of the following is correct
(A) Determinant is a square matrix.
(B) Determinant is a number associated to a matrix.
(C) Determinant is a number associated to a square matrix.
(D) None of these.

Ans

To every square matrix A = [a_ij] of order n, we can associate a number (real or complex) called determinant of the square matrix A, where a_ij = (i,j)^th element of A.
Therefore, option (C) is correct

Q.66 Let A b a square matrix of order 3 x 3, then |kA| is equal to

(A) k|A| (B) k²|A| (C) k³|A| (D) 3k|A|

Ans

$\begin{array}{l} Let A be a square matrix of order 3 \times 3, \\ then A = [\begin{array}{l} a_{1} a_{2} a_{3} \\ b_{1} b_{2} b_{3} \\ c_{1} c_{2} c_{3} \end{array}] \\ and \\ kA = [\begin{array}{l} {ka}_{1} {ka}_{2} {ka}_{3} \\ {kb}_{1} {kb}_{2} {kb}_{3} \\ {kc}_{1} {kc}_{2} {kc}_{3} \end{array}] \\ | kA | = | \begin{array}{l} {ka}_{1} {ka}_{2} {ka}_{3} \\ {kb}_{1} {kb}_{2} {kb}_{3} \\ {kc}_{1} {kc}_{2} {kc}_{3} \end{array} | \\ = k^{3} | \begin{array}{l} a_{1} a_{2} a_{3} \\ b_{1} b_{2} b_{3} \\ c_{1} c_{2} c_{3} \end{array} | [Taking k common from R_{1}, R_{2} and R_{3} .] \\ = k^{3} | A | \\ Thus, option (C) is correct . \end{array}$

Q.67 By using properties of determinants, show that:

$|\begin{array}{l} 1 + a^{2} ab ab \\ ab b^{2} + 1 bc \\ ca cb c^{2} + 1 \end{array}| = 1 + a^{2} + b^{2} + c^{2}$

Ans

$\begin{array}{l} L . H . S . = | \begin{array}{l} 1 + a^{2} ab ac \\ ab b^{2} + 1 bc \\ ca cb c^{2} + 1 \end{array} | \\ Applying R_{1} \to \frac{1}{a} R_{1}, R_{2} \to \frac{1}{b} R_{2}, R_{3} \to \frac{1}{c} R_{3} \\ = abc | \begin{array}{l} \frac{1}{a} + a b c \\ a b + \frac{1}{b} c \\ a b c + \frac{1}{c} \end{array} | \\ Applying R_{1} \to R_{1} - R_{2}, R_{2} \to R_{2} - R_{3} \\ = abc | \begin{array}{l} \frac{1}{a} - \frac{1}{b} 0 \\ 0 \frac{1}{b} - \frac{1}{c} \\ a b c + \frac{1}{c} \end{array} | \\ Applying C_{1} \to {aC}_{1}, C_{2} \to {bC}_{2}, C_{3} \to {cC}_{3} \\ = \frac{abc}{abc} | \begin{array}{l} 1 - 1 0 \\ 0 1 - 1 \\ a^{2} b^{2} c^{2} + 1 \end{array} | \\ Applying C_{2} \to C_{2} + C_{1} \\ = | \begin{array}{l} 1 0 0 \\ 0 1 - 1 \\ a^{2} a^{2} + b^{2} c^{2} + 1 \end{array} | \\ Expanding along R_{1}, we get \\ = | \begin{array}{l} 1 - 1 \\ a^{2} + b^{2} c^{2} + 1 \end{array} | \\ = (c^{2} + 1) - (- 1) (a^{2} + b^{2}) \\ = 1 + c^{2} + a^{2} + b^{2} \\ = 1 + a^{2} + b^{2} + c^{2} \\ = R . H . S . \\ Hence, the given result is proved . \end{array}$

Q.68 By using properties of determinants, show that:

$|\begin{array}{l} 1 + a^{2} - b^{2} 2 ab - 2 b \\ 2 ab 1 - a^{2} + b^{2} 2 a \\ 2 b - 2 a 1 - a^{2} - b^{2} \end{array}| = {(1 + a^{2} + b^{2})}^{3}$

Ans

$\begin{array}{l} L . H . S . = | \begin{array}{l} 1 + a^{2} - b^{2} 2 ab - 2 b \\ 2 ab 1 - a^{2} + b^{2} 2 a \\ 2 b - 2 a 1 - a^{2} - b^{2} \end{array} | \\ Apply R_{1} \to R_{1} + {bR}_{3}, R_{2} \to R_{2} - {aR}_{3} \\ = | \begin{matrix} 1 + a^{2} + b^{2} & 0 & - b (1 + a^{2} + b^{2}) \\ 0 & 1 + a^{2} + b^{2} & a (1 + a^{2} + b^{2}) \\ 2 b & - 2 a & 1 - a^{2} - b^{2} \end{matrix} | \\ = {(1 + a^{2} + b^{2})}^{2} | \begin{matrix} 1 & 0 & - b \\ 0 & 1 & a \\ 2 b & - 2 a & 1 - a^{2} - b^{2} \end{matrix} | \\ Expanding along R_{1}, we get \\ = {(1 + a^{2} + b^{2})}^{2} [1 | \begin{array}{l} 1 a \\ - 2 a 1 - a^{2} - b^{2} \end{array} | + (- b) | \begin{array}{l} 01 \\ 2 b - 1 a \end{array} |] \\ = {(1 + a^{2} + b^{2})}^{2} (1 - a^{2} - b^{2} + 2 a^{2} - b \times - 2 b) \\ = {(1 + a^{2} + b^{2})}^{2} (1 - a^{2} - b^{2} + 2 a^{2} + 2 b^{2}) \\ = {(1 + a^{2} + b^{2})}^{3} = R . H . S . \\ Hence, it is proved . \end{array}$

Q.69 By using properties of determinants, show that:

$|\begin{array}{l} 1 x x^{2} \\ x^{2} 1 x \\ x x^{2} 1 \end{array}| = {(1 - x^{3})}^{2}$

Ans

$\begin{array}{l} L . H . S . = | \begin{array}{l} 1 x x^{2} \\ x^{2} 1 x \\ x x^{2} 1 \end{array} | \\ = | \begin{array}{l} 1 + x + x^{2} 1 + x + x^{2} 1 + x + x^{2} \\ x^{2} 1 x \\ x x^{2} 1 \end{array} | [R_{1} \to R_{1} + R_{2} + R_{3}] \\ = (1 + x + x^{2}) | \begin{array}{l} 1 1 1 \\ x^{2} 1 x \\ x x^{2} 1 \end{array} | \\ = (1 + x + x^{2}) | \begin{array}{l} 1 0 0 \\ x^{2} 1 - x^{2} x - x^{2} \\ x x^{2} - x 1 - x \end{array} | [\begin{array}{l} C_{2} \to C_{2} - C_{1}, \\ C_{3} \to C_{3} - C_{1} \end{array}] \\ = (1 + x + x^{2}) | \begin{array}{l} 1 0 0 \\ x^{2} (1 - x) (1 + x) x (1 - x) \\ x x (x - 1) (1 - x) \end{array} | \\ = (1 + x + x^{2}) (1 - x) (1 - x) | \begin{array}{l} 1 0 0 \\ x^{2} (1 + x) x \\ x - x 1 \end{array} | \\ [\begin{array}{l} Taking (1 - x) common from \\ C_{2} and C_{3} . \end{array}] \\ Expanding along R_{1}, we get \\ = (1 + x + x^{2}) (1 - x) (1 - x) (1 + x + x^{2}) \\ = {(1 - x) (1 + x + x^{2})} {(1 - x) (1 + x + x^{2})} \\ = (1 - x^{3}) (1 - x^{3}) \\ = {(1 - x^{3})}^{2} = R . H . S . \\ Hence Proved . \end{array}$

Q.70 By using properties of determinants, show that:

$\begin{array}{l} (i) |\begin{array}{l} a - b - c 2 a 2 a \\ 2 b b - c - a 2 b \\ 2 c 2 c c - a - b \end{array}| = {(a + b + c)}^{3} \\ (ii) |\begin{array}{l} x + y + 2 z x y \\ z y + z + 2 x y \\ z x z + x + 2 y \end{array}| = 2 {(x + y + z)}^{3} \end{array}$

Ans

$\begin{array}{l} (i) | \begin{array}{l} a - b - c 2 a 2 a \\ 2 b b - c - a 2 b \\ 2 c 2 c c - a - b \end{array} | = {(a + b + c)}^{3} \\ L . H . S . = | \begin{array}{l} a - b - c 2 a 2 a \\ 2 b b - c - a 2 b \\ 2 c 2 c c - a - b \end{array} | \\ = | \begin{array}{l} a + b + c a + b + c a + b + c \\ 2 b b - c - a 2 b \\ 2 c 2 c c - a - b \end{array} | [\begin{array}{l} Apply \\ R_{1} \to R_{1} + R_{2} + R_{3} \end{array}] \\ = (a + b + c) | \begin{array}{l} 1 1 1 \\ 2 b b - c - a 2 b \\ 2 c 2 c c - a - b \end{array} | \\ = (a + b + c) | \begin{array}{l} 1 0 0 \\ 2 b - b - c - a 0 \\ 2 c 0 - c - a - b \end{array} | [\begin{array}{l} C_{2} \to C_{2} - C_{1}, \\ C_{3} \to C_{3} - C_{1} \end{array}] \\ = (a + b + c) | \begin{array}{l} - b - c - a 0 \\ 0 - c - a - b \end{array} | \\ = (a + b + c) {(- b - c - a)}^{2} \\ = {(a + b + c)}^{3} \\ = R . H . S . \\ (ii) L . H . S . = | \begin{array}{l} x + y + 2 z x y \\ z y + z + 2 x y \\ z x z + x + 2 y \end{array} | \\ = | \begin{array}{l} 2 x + 2 y + 2 z x y \\ 2 x + 2 y + 2 z y + z + 2 x y \\ 2 x + 2 y + 2 z x z + x + 2 y \end{array} | [C_{1} \to C_{1} + C_{2} + C_{3}] \\ = (2 x + 2 y + 2 z) | \begin{array}{l} 1 x y \\ 1 y + z + 2 x y \\ 1 x z + x + 2 y \end{array} | \\ = (2 x + 2 y + 2 z) | \begin{array}{l} 1 x y \\ 0 y + z + x 0 \\ 0 0 z + x + y \end{array} | [\begin{array}{l} R_{2} \to R_{2} - R_{1} \\ R_{3} \to R_{3} - R_{1} \end{array}] \\ = {(2 x + 2 y + 2 z)}^{3} | \begin{array}{l} 1 x y \\ 0 1 0 \\ 0 0 1 \end{array} | \\ Expanding along C_{1}, we get \\ = {(2 x + 2 y + 2 z)}^{3} | \begin{array}{l} 1 0 \\ 0 1 \end{array} | \\ = {(2 x + 2 y + 2 z)}^{3} (1 - 0) \\ = {(2 x + 2 y + 2 z)}^{3} = R . H . S . \end{array}$

Q.71 By using properties of determinants, show that:

$\begin{array}{l} (i) |\begin{array}{l} x + 4 2 x 2 x \\ 2 x x + 4 2 x \\ 2 x 2 x x + 4 \end{array}| = (5 x + 4) {(4 - x)}^{2} \end{array}$ $\begin{array}{l} (ii) |\begin{array}{l} y + k y y \\ y y + 4 y \\ y y y + 4 \end{array}| = k^{2} (3 y + k) \end{array}$

Ans

$\begin{array}{l} (i) L . H . S . = | \begin{array}{l} x + 4 2 x 2 x \\ 2 x x + 4 2 x \\ 2 x 2 x x + 4 \end{array} | \\ = | \begin{array}{l} 5 x + 4 2 x 2 x \\ 5 x + 5 x + 4 2 x \\ 5 x + 5 2 x x + 4 \end{array} | [Apply C_{1} \to C_{1} + C_{2} + C_{3}] \\ = (5 x + 4) | \begin{array}{l} 1 2 x 2 x \\ 1 x + 4 2 x \\ 1 2 x x + 4 \end{array} | \\ = (5 x + 4) | \begin{array}{l} 1 2 x 2 x \\ 0 4 - x 0 \\ 0 0 4 - x \end{array} | [\begin{array}{l} Apply \\ R_{2} \to R_{2} - R_{1}, R_{3} \to R_{3} - R_{1} \end{array}] \\ Expanding along C_{1}, we get \\ = (5 x + 4) | \begin{array}{l} 4 - x 0 \\ 0 4 - x \end{array} | \\ = (5 x + 4) {(4 - x)}^{2} \\ = R . H . S . \\ (ii) L . H . S . = | \begin{array}{l} y + k y y \\ y y + k y \\ y y y + k \end{array} | \\ = | \begin{array}{l} 3 y + k y y \\ 3 y + k y + k y \\ 3 y + k y y + k \end{array} | [Apply C_{1} \to C_{1} + C_{2} + C_{3}] \\ = (3 y + k) | \begin{array}{l} 1 y y \\ 1 y + k y \\ 1 y y + k \end{array} | \\ [Apply R_{2} \to R_{2} - R_{1}, R_{3} \to R_{3} - R_{1}] \\ = (3 y + k) | \begin{array}{l} 1 y y \\ 0 k 0 \\ 0 0 k \end{array} | \\ Expanding along C_{1}, we get \\ = (3 y + k) | \begin{array}{l} k 0 \\ 0 k \end{array} | \\ = (3 y + k) \times k^{2} \\ = k^{2} (3 y + k) \end{array}$

Q.72 By using properties of determinants, show that:

$|\begin{array}{l} x x^{2} yz \\ y y^{2} zx \\ z z^{2} xy \end{array}| = (x - y) (y - z) (z - x) (xy + yz + zx)$

Ans

$\begin{array}{l} L . H . S . = | \begin{array}{l} x x^{2} yz \\ y y^{2} zx \\ z z^{2} xy \end{array} | \\ = | \begin{array}{l} x - y x^{2} - y^{2} yz - zx \\ y - z y^{2} - z^{2} zx - xy \\ z z^{2} xy \end{array} | \\ [Applying R_{1} \to R_{1} - R_{2}, R_{2} \to R_{2} - R_{3}] \\ = | \begin{array}{l} (x - y) (x - y) (x + y) - z (x - y) \\ (y - z) (y - z) (y + z) - x (y - z) \\ z z^{2} xy \end{array} | \\ = (x - y) (y - z) | \begin{array}{l} 1 (x + y) - z \\ 1 (y + z) - x \\ z z^{2} xy \end{array} | [\begin{array}{l} Taking common (x - y) \\ and (y - z) from R_{1} and R_{2} \\ respectively \end{array}] \\ = (x - y) (y - z) | \begin{array}{l} 0 (x - z) (x - z) \\ 1 (y + z) - x \\ z z^{2} xy \end{array} | \\ [Applying R_{1} \to R_{1} - R_{2}] \\ = (x - y) (y - z) (z - x) | \begin{array}{l} 0 - 1 - 1 \\ 1 (y + z) - x \\ z z^{2} xy \end{array} | [\begin{array}{l} Taking common \\ (z - x) from R_{1} \end{array}] \\ = (x - y) (y - z) (z - x) | \begin{array}{l} 0 0 - 1 \\ 1 (x + y + z) - x \\ z z^{2} - xy xy \end{array} | [\begin{array}{l} Applying \\ C_{2} \to C_{2} - C_{3} \end{array}] \\ Expanding along R_{1}, we get \\ = (x - y) (y - z) (z - x) {(- 1) | \begin{array}{l} 1 (x + y + z) \\ z z^{2} - xy \end{array} |} \\ = (x - y) (y - z) (z - x) \times (- 1) {z^{2} - xy - (z) (x + y + z)} \\ = (x - y) (y - z) (z - x) \times (- 1) {z^{2} - xy - zx - zy - z^{2}} \\ = (x - y) (y - z) (z - x) (xy + yz + zx) \\ = R . H . S . \end{array} a$

Q.73 By using properties of determinants, show that:

$\begin{array}{l} (i) |\begin{array}{l} 1 a a^{2} \\ 1 b b^{2} \\ 1 c c^{2} \end{array}| = (a - b) (b - c) (c - a) \\ (ii) |\begin{array}{l} 1 1 1 \\ a b c \\ a^{3} b^{3} c^{3} \end{array}| = (a - b) (b - c) (c - a) (a + b + c) \end{array}$

Ans

$\begin{array}{l} (i) L . H . S . = | \begin{array}{l} 1 a a^{2} \\ 1 b b^{2} \\ 1 c c^{2} \end{array} | \\ = | \begin{array}{l} 0 a - b a^{2} - b^{2} \\ 0 b - c b^{2} - c^{2} \\ 1 c c^{2} \end{array} | [Apply R_{1} \to R_{1} - R_{2} and R_{2} \to R_{2} - R_{3}] \\ = (a - b) (b - c) | \begin{array}{l} 0 1 a + b \\ 0 1 b + c \\ 1 c c^{2} \end{array} | \\ Expanding along C_{3}, we get \\ = (a - b) (b - c) | \begin{array}{l} 1 a + b \\ 1 b + c \end{array} | \\ = (a - b) (b - c) (b + c - a - b) \\ = (a - b) (b - c) (c - a) = R . H . S . \\ (ii) L . H . S . = | \begin{array}{l} 1 1 1 \\ a b c \\ a^{3} b^{3} c^{3} \end{array} | \\ = | \begin{array}{l} 0 0 1 \\ a - b b - c c \\ a^{3} - b^{3} b^{3} - c^{3} c^{3} \end{array} | \\ [Applying C_{1} \to C_{1} - C_{2} and C_{2} \to C_{2} - C_{3}] \\ = | \begin{array}{l} 0 0 1 \\ a - b b - c c \\ (a - b) (a^{2} + ab + b^{2}) (b - c) (b^{2} + bc + c^{2}) c^{3} \end{array} | \\ = (a - b) (b - c) | \begin{array}{l} 0 0 1 \\ 1 1 c \\ (a^{2} + ab + b^{2}) (b^{2} + bc + c^{2}) c^{3} \end{array} | \\ = (a - b) (b - c) | \begin{array}{l} 1 1 \\ (a^{2} + ab + b^{2}) (b^{2} + bc + c^{2}) \end{array} | [Expending along R_{1}] \\ = (a - b) (b - c) {(b^{2} + bc + c^{2}) - (a^{2} + ab + b^{2})} \\ = (a - b) (b - c) (b^{2} + bc + c^{2} - a^{2} - ab - b^{2}) \\ = (a - b) (b - c) (bc + c^{2} - a^{2} - ab) \\ = (a - b) (b - c) {c^{2} - a^{2} + bc - ab} \\ = (a - b) (b - c) {(c - a) (c + a) - b (c - a)} \\ = (a - b) (b - c) (c - a) (a + b + c) \end{array}$

Q.74 By using properties of determinants, show that:

$|\begin{array}{l} - a^{2} ab ac \\ b a - b^{2} b c \\ ca cb - c^{2} \end{array}| = 4 a^{2} b^{2} c^{2}$

Ans

$\begin{matrix} L . H . S . = Δ \\ = |\begin{array}{l} - a^{2} abac \\ ba - b^{2} bc \\ cacb - c^{2} \end{array}| \\ = abc |\begin{array}{l} - a a a \\ b - b b \\ c c - c \end{array}| [\begin{array}{l} Applying C_{1} \to \frac{1}{a} C_{1}, C_{2} \to \frac{1}{b} C_{2} \\ and C_{3} \to \frac{1}{c} C_{3} \end{array}] \\ = abc |\begin{array}{l} - a a a \\ b - b b \\ c c - c \end{array}| \\ = abc |\begin{array}{l} 0 a a \\ 2 b - b b \\ 0 c - c \end{array}| [Applying C_{1} \to C_{1} + C_{3}] \\ = abc \times - 2 b |\begin{array}{l} a a \\ c - c \end{array}| \\ = abc \times - 2 b (- ac - ac) \\ = abc \times - 2 b (- 2 ac) \\ = 4 a^{2} b^{2} c^{2} = R . H . S . \end{matrix}$

Q.75 By using properties of determinants, show that:

$|\begin{array}{l} 0 a - b \\ - a 0 - c \\ b c 0 \end{array}| = 0$

Ans

$\begin{array}{l} Δ = |\begin{array}{l} 0 a - b \\ - a 0 - c \\ bc 0 \end{array}| \\ = \frac{1}{c} |\begin{array}{l} 0 ac - bc \\ - a 0 - c \\ bc 0 \end{array}| [R_{1} \to {cR}_{1}] \\ = \frac{1}{c} |\begin{array}{l} abac 0 \\ - a 0 - c \\ bc 0 \end{array}| [R_{1} \to R_{1} - {bR}_{2}] \\ = \frac{a}{c} |\begin{array}{l} bc 0 \\ - a 0 - c \\ bc 0 \end{array}| [Taking a common from R_{1} .] \\ = \frac{a}{c} \times 0 [Two rows are identical, so Δ = 0.] \\ = 0 \end{array}$

Q.76

$|\begin{array}{l} b + cq + ry + z \\ c + ar + pz + x \\ a + bp + qx + y \end{array}| = 2 |\begin{array}{l} \end{array}|$

Ans

$\begin{array}{l} L . H . S . = | \begin{array}{l} b + c q + r y + z \\ c + a r + p z + x \\ a + b p + q x + y \end{array} | \\ = | \begin{array}{l} b + c q + r y + z \\ c + a r + p z + x \\ a p x \end{array} | + | \begin{array}{l} b + c q + r y + z \\ c + a r + p z + x \\ b q y \end{array} | \\ = Δ_{1} + Δ_{2} [Let] \\ Δ_{1} = | \begin{array}{l} b + c q + r y + z \\ c + a r + p z + x \\ a p x \end{array} | \\ = | \begin{array}{l} b + c q + r y + z \\ c r z \\ a p x \end{array} | [R_{2} \to R_{2} - R_{3}] \\ = | \begin{array}{l} b q y \\ c r z \\ a p x \end{array} | [R_{1} \to R_{1} - R_{2}] \\ = (- 1) | \begin{array}{l} a p x \\ c r z \\ b q y \end{array} | [R_{1} \leftrightarrow R_{3}] \\ = = | \begin{array}{l} a p x \\ b q y \\ c r z \end{array} | \\ Δ_{2} = | \begin{array}{l} b + c q + r y + z \\ c + a r + p z + x \\ b q y \end{array} | \\ = | \begin{array}{l} c r z \\ c + a r + p z + x \\ b q y \end{array} | [R_{1} \to R_{1} - R_{3}] \\ = | \begin{array}{l} c r z \\ a p x \\ b q y \end{array} | [R_{2} \to R_{2} - R_{1}] {(- 1)}^{2} | \begin{array}{l} a p x \\ b q y \\ c r z \end{array} | [R_{1} \leftrightarrow R_{3}] \\ = | \begin{array}{l} a p x \\ c r z \\ b q y \end{array} | [R_{1} \leftrightarrow R_{2}] \\ = | \begin{array}{l} a p x \\ b q y \\ c r z \end{array} | [R_{2} \leftrightarrow R_{3}] \\ ∴ Δ = Δ_{1} + Δ_{2} \\ = 2 | \begin{array}{l} a p x \\ b q y \\ c r z \end{array} | \end{array}$

Q.77 Using the property of determinants and without expanding in, prove that:

$|\begin{array}{l} 1 bc a (b + c) \\ 1 ca b (c + a) \\ 1 ab c (a + b) \end{array}| = 0$

Ans

$\begin{array}{l} Δ = | \begin{array}{l} 1 bc a (b + c) \\ 1 ca b (c + a) \\ 1 ab c (a + b) \end{array} | \\ = | \begin{array}{l} 1 bc ab + ac \\ 1 ca bc + ba \\ 1 ab ca + cb \end{array} | \\ = | \begin{array}{l} 1 bc ab + ac + bc \\ 1 ca bc + ba + ca \\ 1 ab ca + cb + ab \end{array} | [Apply C_{3} \to C_{3} + C_{2}] \\ = (ab + bc + ca) | \begin{array}{l} 1 bc 1 \\ 1 ca 1 \\ 1 ab 1 \end{array} | [Taking common (ab + bc + ca) from C_{3} .] \\ = (ab + bc + ca) \times 0 [Δ = 0 because two columns are identical .] \\ = 0 \end{array}$

Q.78 Using the property of determinants and without expanding in, prove that:

$|\begin{array}{l} 2 7 65 \\ 3 8 75 \\ 5 9 86 \end{array}| = 0$

Ans

$\begin{array}{l} Δ = |\begin{array}{l} 2765 \\ 3875 \\ 5986 \end{array}| \\ Apply C_{3} \to C_{3} + C_{1} \\ = |\begin{array}{l} 2763 + 2 \\ 3872 + 3 \\ 5981 + 5 \end{array}| \\ = |\begin{array}{l} 2763 \\ 3872 \\ 5981 \end{array}| + |\begin{array}{l} 272 \\ 383 \\ 595 \end{array}| \\ = |\begin{array}{l} 2763 \\ 3872 \\ 5981 \end{array}| + 0 [∵ Δ = 0 because two columns are identical .] \\ Apply C_{3} \to \frac{1}{9} C_{3} \\ = 9 |\begin{array}{l} 277 \\ 388 \\ 599 \end{array}| \\ = 9 \times 0 [∵ Δ = 0 because two columns are identical .] \\ = 0 \end{array}$

Q.79 Using the property of determinants and without expanding in, prove that:

$|\begin{array}{l} a - b b - c c - a \\ b - c c - a a - b \\ c - a a - b b - c \end{array}| = 0$

Ans

$\begin{array}{l} | \begin{array}{l} a - b b - c c - a \\ b - c c - a a - b \\ c - a a - b b - c \end{array} | = | \begin{array}{l} a - b + b - c + c - a b - c c - a \\ b - c + c - a + a - b c - a a - b \\ c - a + a - b + b - c a - b b - c \end{array} | \\ [Apply C_{1} \to C_{1} + C_{2} + C_{3}] \\ = | \begin{array}{l} 0 b - c c - a \\ 0 c - a a - b \\ 0 a - b b - c \end{array} | \\ = 0 \\ [Since, all the elements of a column are zero .] \end{array}$

Q.80 Using the property of determinants and without expanding in, prove that:

$|\begin{array}{l} x a x + a \\ y b y + b \\ z c z + c \end{array}| = 0$

Ans

$\begin{array}{l} | \begin{array}{l} x a x + a \\ y b y + b \\ z c z + c \end{array} | = | \begin{array}{l} x a x \\ y b y \\ z c z \end{array} | + | \begin{array}{l} x a a \\ y b b \\ z c c \end{array} | \\ = 0 + 0 [\begin{array}{l} If two columns of a determinant is \\ identical, then their value is zero . \end{array}] \\ = 0 \end{array}$

Q.81

$\begin{array}{l} If | \begin{array}{l} x 2 \\ 18 x \end{array} | = | \begin{array}{l} 6 2 \\ 18 6 \end{array} |, then x is equal to \\ (A) 6 (B) \pm 6 (C) - 6 (D) 0 . \end{array}$

Ans

$\begin{array}{l} ∵ | \begin{array}{l} x 2 \\ 18 x \end{array} | = | \begin{array}{l} 6 2 \\ 18 6 \end{array} | \\ ∴ x^{2} - 36 = 36 - 36 \\ \Rightarrow x^{2} = 36 \\ \Rightarrow x = \pm 6 \\ Therefore, option (B) is correct . \end{array}$

Q.82

$\begin{array}{l} Find value so fx, if \\ (i) | \begin{array}{l} 2 4 \\ 5 1 \end{array} | = | \begin{array}{l} 2 x 4 \\ 6 x \end{array} | (ii) | \begin{array}{l} 2 3 \\ 4 5 \end{array} | = | \begin{array}{l} x 3 \\ 2 x 5 \end{array} | \end{array}$

Ans

$\begin{array}{l} (i) | \begin{array}{l} 2 4 \\ 5 1 \end{array} | = | \begin{array}{l} 2 x 4 \\ 6 x \end{array} | \\ \Rightarrow 2 - 20 = 2 x^{2} - 24 \\ \Rightarrow - 18 = 2 x^{2} - 24 \\ \Rightarrow 24 - 18 = 2 x^{2} \\ \Rightarrow x^{2} = \frac{6}{2} \\ \Rightarrow x = \pm \sqrt{3} \\ (ii) | \begin{array}{l} 2 3 \\ 4 5 \end{array} | = | \begin{array}{l} x 3 \\ 2 x 5 \end{array} | \\ \Rightarrow 10 - 12 = 5 x - 6 x \\ \Rightarrow - 2 = - x \\ \Rightarrow x = 2 . \end{array}$

Q.83

$If A= [\begin{array}{l} 1 1 - 2 \\ 2 1 - 3 \\ 5 4 - 9 \end{array}], find | A | .$

Ans

$\begin{array}{l} A = [\begin{array}{l} 1 1 - 2 \\ 2 1 - 3 \\ 5 4 - 9 \end{array}], \\ | A | = | \begin{array}{l} 1 1 - 2 \\ 2 1 - 3 \\ 5 4 - 9 \end{array} | [Expending along R_{1}] \\ = 1 | \begin{array}{l} 1 - 3 \\ 4 - 9 \end{array} | - 1 | \begin{array}{l} 2 - 3 \\ 5 - 9 \end{array} | + (- 2) | \begin{array}{l} 2 1 \\ 5 4 \end{array} | \\ = (- 9 + 12) - 1 (- 18 + 15) - 2 (8 - 5) \\ = 3 + 3 - 6 \\ = 0 \end{array}$

Q.84

$\begin{array}{l} Evaluate the determinants \\ (i) | \begin{array}{l} 3 - 1 - 2 \\ 0 0 - 1 \\ 3 - 5 0 \end{array} | (ii) | \begin{array}{l} 3 - 4 5 \\ 1 1 - 2 \\ 2 3 1 \end{array} | \\ (iii) | \begin{array}{l} 0 1 2 \\ 1 0 - 3 \\ - 2 3 0 \end{array} | (iv) | \begin{array}{l} 2 - 1 - 2 \\ 0 2 - 1 \\ 3 - 5 0 \end{array} | \end{array}$

Ans

$\begin{matrix} (i) | \begin{array}{l} 3 - 1 - 2 \\ 0 0 - 1 \\ 3 - 5 0 \end{array} | = 3 | \begin{array}{l} 0 - 1 \\ - 5 0 \end{array} | - (- 1) | \begin{array}{l} 0 - 1 \\ 3 0 \end{array} | + (- 2) | \begin{array}{l} 0 0 \\ 3 - 5 \end{array} | \\ [Expanding along R_{1}] \\ = 3 (0 - 5) + 1 (0 + 3) - 2 (0 - 0) \\ = - 15 + 3 - 0 \\ = - 12 \\ (ii) | \begin{array}{l} 3 - 4 5 \\ 1 1 - 2 \\ 2 3 1 \end{array} | = 3 | \begin{array}{l} 1 - 2 \\ 3 1 \end{array} | - (- 4) | \begin{array}{l} 1 - 2 \\ 2 1 \end{array} | + 5 | \begin{array}{l} 1 1 \\ 2 3 \end{array} | \\ = 3 (1 + 6) + 4 (1 + 4) + 5 (3 - 2) \\ = 21 + 20 + 5 \\ = 46 \end{matrix}$ $\begin{matrix} (iii) | \begin{array}{l} 0 1 2 \\ - 1 0 - 3 \\ - 2 3 0 \end{array} | = 0 | \begin{array}{l} 0 - 3 \\ 3 0 \end{array} | - 1 | \begin{array}{l} - 1 - 3 \\ - 2 0 \end{array} | + 2 | \begin{array}{l} - 1 0 \\ - 2 3 \end{array} | \\ [E x p a n d i n g {along R}_{1}] \\ = 0 (0 + 9) - 1 (0 - 6) + 2 (- 3 - 0) \\ = 0 + 6 - 6 \\ = 0 \\ (i v) | \begin{array}{l} 2 - 1 - 2 \\ 0 2 - 1 \\ 3 - 5 0 \end{array} | = 2 | \begin{array}{l} 2 - 1 \\ - 5 0 \end{array} | - (- 1) | \begin{array}{l} 0 - 1 \\ 3 0 \end{array} | - 2 | \begin{array}{l} 0 2 \\ 3 - 5 \end{array} | \\ [E x p a n d i n g {along R}_{1}] \\ = 2 (0 - 5) + 1 (0 + 3) - 2 (0 - 6) \\ = - 10 + 3 + 12 \\ = 5 \end{matrix}$

Q.85

$If A= [\begin{array}{l} 1 0 1 \\ 0 1 2 \\ 0 0 4 \end{array}], then show that | 3 A | = 27 | A | .$

Ans

$\begin{array}{l} Given \\ A = [\begin{array}{l} 1 0 1 \\ 0 1 2 \\ 0 0 4 \end{array}] \\ 3 A = [\begin{array}{l} 3 0 3 \\ 0 3 6 \\ 0 0 12 \end{array}] \\ | A | = | \begin{array}{l} 3 0 3 \\ 0 3 6 \\ 0 0 12 \end{array} | \\ = 3 (36 - 0) - 0 (0 - 0) + 3 (0 - 0) \\ = 108 + 0 + 0 \\ = 108 \\ | 3 A | = | \begin{array}{l} 9 0 9 \\ 0 9 18 \\ 0 0 36 \end{array} | \\ = 9 (324 - 0) - 0 (0 - 0) + 9 (0 - 0) \\ = 2916 + 0 + 0 \\ = 27 \times 108 \\ = 27 \times | A | \\ | 3 A | = 27 | A | \\ Hence Proved . \end{array}$

Q.86

$If A= |\begin{array}{l} 1 2 \\ 4 2 \end{array}|, then show that | 2 A | = 4 | A | .$

Ans

$\begin{array}{l} Given : \\ [A] = [\begin{array}{l} 1 2 \\ 4 2 \end{array}] \\ | A | = | \begin{array}{l} 1 2 \\ 4 2 \end{array} | \\ = 2 - 8 \\ = - 6 \\ 2 A = [\begin{array}{l} 2 4 \\ 8 4 \end{array}] \\ L . H . S . = \\ | 2 A | = | \begin{array}{l} 2 4 \\ 8 4 \end{array} | \\ = 8 - 32 \\ = - 24 \\ = 4 \times (- 6) \\ = 4 | A | = R . H . S . \end{array}$

Q.87 Evaluate the determinants

$| \begin{array}{l} (x^{2} - x + 1) (x - 1) \\ (x + 1) (x + 1) \end{array} |$

Ans

$\begin{array}{l} | \begin{array}{l} x^{2} - x + 1 x - 1 \\ x + 1 x + 1 \end{array} | = (x + 1) (x^{2} - x + 1) - (x - 1) (x + 1) \\ = (x^{3} - 1) - (x^{2} - 1) = x^{3} - 1 - x^{2} + 1 \\ = x^{3} - x^{2} \end{array}$

Q.88 Evaluate the determinants

$|\begin{array}{l} 2 4 \\ - 5 - 1 \end{array}|$

Ans

$\begin{array}{l} | \begin{array}{l} 2 4 \\ - 5 - 1 \end{array} | = 2 \times (- 1) - 4 (- 5) \\ = - 2 + 20 \\ = 18 \end{array}$

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Related Chapters

Chapter 1 - Relations and Functions

Chapter 2 - Inverse Trigonometric Functions

Chapter 3 - Matrices

Chapter 5 - Continuity and Differentiability

Chapter 6 - Application of Derivatives

Chapter 7 - Integrals

Chapter 8 - Application of Integrals

Chapter 9 - Differential Equations

Chapter 10 - Vector Algebra

Chapter 11 - Three Dimensional Geometry

Chapter 12 - Linear Programming

Chapter 13 - Probability

FAQs (Frequently Asked Questions)

Determinants are one of the main topics that appear in the exam, and students can understand the essences of integrals and values of square matrices under this chapter. In addition, practising from NCERT Solutions Class 12 Mathematics Chapter 4 helps students understand the concept better as the examples are explained well, with step-by-step solutions for all the exercise problems.

There are six exercises in NCERT Solutions Class 12 Mathematics Chapter 4 Determinants, one of the essential topics. However, one shall also focus on the area of a triangle, minors, cofactors, and inverse of a matrix. Besides, Class 12 Mathematics Chapter 4 includes short answer questions and long answer question types that help students prepare for their examination.

Linear equations, problems on the theory of determinants, and multiple-choice questions are the most common forms of questions. Besides, the solutions also cover problems in the area of a triangle, minors, and cofactors. Short answer types of questions and long answer types are also present. The NCERT Solutions will help evaluate the maximum number and properties of determinants.

NCERT Solutions Class 12 Mathematics Chapter 4 – Determinants

Key Topics Covered In NCERT Solutions Class 12 Mathematics Chapter 4

4.1 Introduction

4.2 Determinant

4.3 Properties of Determinants

4.4 Area of Triangle

4.5 Minors and Cofactors

4.6 The Adjoint and Inverse of a Matrix

4.7 Applications of Determinants and Matrices

NCERT Solutions Class 12 Mathematics Chapter 4 Exercise & Answer Solutions

NCERT Exemplar Class 12 Mathematics

Key Features of NCERT Solutions Class 12 Mathematics Chapter 4

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Related Chapters

Chapter 1 - Relations and Functions

Chapter 2 - Inverse Trigonometric Functions

Chapter 3 - Matrices

Chapter 5 - Continuity and Differentiability

Chapter 6 - Application of Derivatives

Chapter 7 - Integrals

Chapter 8 - Application of Integrals

Chapter 9 - Differential Equations

Chapter 10 - Vector Algebra

Chapter 11 - Three Dimensional Geometry

Chapter 12 - Linear Programming

Chapter 13 - Probability

FAQs (Frequently Asked Questions)

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NCERT Solutions Class 12 Mathematics Chapter 4 – Determinants

Key Topics Covered In NCERT Solutions Class 12 Mathematics Chapter 4

4.1 Introduction

4.2 Determinant

4.3 Properties of Determinants

4.4 Area of Triangle

4.5 Minors and Cofactors

4.6 The Adjoint and Inverse of a Matrix

4.7 Applications of Determinants and Matrices

NCERT Solutions Class 12 Mathematics Chapter 4 Exercise & Answer Solutions

NCERT Exemplar Class 12 Mathematics

Key Features of NCERT Solutions Class 12 Mathematics Chapter 4

Please register to view this section

Related Chapters

Chapter 1 - Relations and Functions

Chapter 2 - Inverse Trigonometric Functions

Chapter 3 - Matrices

Chapter 5 - Continuity and Differentiability

Chapter 6 - Application of Derivatives

Chapter 7 - Integrals

Chapter 8 - Application of Integrals

Chapter 9 - Differential Equations

Chapter 10 - Vector Algebra

Chapter 11 - Three Dimensional Geometry

Chapter 12 - Linear Programming

Chapter 13 - Probability

FAQs (Frequently Asked Questions)

1. What are the benefits of studying from NCERT Solutions Class 12 Mathematics Chapter 4?

2. Which are the main exercises in NCERT Solutions Class 12 Mathematics Chapter 4?

3. Which forms of question are present in NCERT Solutions Class 12 Mathematics Chapter 4?

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