# CBSE Class 12 Maths Revision Notes Chapter 4

## Class 12 Mathematics Chapter 4 Notes

In Class 12, students deal with high pressure and stress, mainly in Mathematics. The marks scored in the final board examination will be required to appear for future studies in their field of interest. Mathematics is considered to be a tricky subject. A strong understanding of all basic concepts and regular practice is very important to understanding complex topics. Students are advised to use the best study material, such as the Class 12 Mathematics chapter 4 notes, to help them get a clear understanding of determinants and concepts related to it.

The concepts explained in this chapter are very useful to analyse and find the solution. In Class 12 Mathematics chapter 4 notes, students will gain knowledge of determinants up to order 3. With the help of determinants, we can find out the uniqueness of a solution. The chapter has a wide range of applications in the Engineering, Science, etc., sectors. Also, concepts based on the properties of the matrix, cofactors, inverse, adjoint, and area of a triangle are covered in the Class 12 Mathematics chapter 4 notes.

Extramarks, an online learning platform, provides professional help and gives students the best academic notes to ace their exams. With the help of the class 12 chapter 4 mathematics notes, students can prepare for both the board exams and the competitive exams. Our notes are provided with detailed information, graphical representation, and easy language for students’ overall development.

## Key Topics Covered In Class 12 Mathematics Chapter 4 Notes

The main topics covered in the Class 12 Mathematics chapter 4 notes are:

• Introduction
• Determinant Definition
• Properties
• Area of a Triangle
• Minors and Cofactors
• The inverse of a Matrix
• Applications
• Solution of a system of linear equations

A brief of the key topics covered in class 12 Mathematics chapter 4 notes is as under.

### INTRODUCTION:

The Class 12 Mathematics chapter 4 notes include various complex concepts related to determinants and their applications. To ace this chapter, students must learn concepts introduced in Class 11 as well as chapter 3 of the NCERT books.

RECALL:

If the system of equations is given as

a1x + b1y=c1 and a2x + b2y=c2, then the matric representation will be

 a1 b1 x = c1 a2 b2 y c2

The solution of this system is all the values of variables x and y, which satisfy the linear equations in the system.

### DEFINITION OF DETERMINANT:

The determinant of any square matrix is a scalar value that is calculated from all elements of the matrix, which implies, a for every square matrix A of order n is associated with a number called the determinant of that matric A and is denoted by A (mod A), det (A) or Δ .

Consider A =

 a11 b12 a21 b22

The scalar value for matrix A is a1b2b1a2

A= a11b22b12a21

 a11 b12 a21 b22

NOTE: Only square matrices (number of rows and columns are equal) have determinants.

### TYPES OF DETERMINANTS:

1. First Order Determinant: The determinant of a matrix, say A of order one, is known as the First-order determinant. The value of the determinant is said to be the element of matrix A.

Consider A = 

Therefore determinant = 2= 2

1. Second Order Determinant – The determinant of a matrix, say A of order two, is known as the Second-order determinant.

Consider A =

 4 2 5 3

Therefore, A

 4 2 5 3

(4 x 3) – (5 x 2) = 12 – 10 = 2

1. Third Order Determinant – The determinant of a matrix, say A of order three, is known as the third-order determinant.

Consider A =

 a1 b1 c1 a2 b2 c2 a3 b3 c3

Then A

 a1 b1 c1 a2 b2 c2 a3 b3 c3

as a1(b2c3b3c2) – b1(a2c3a3c2) + c1(a2b3a3b2)

NOTE

Consider two square matrices A and B having order n and A=kB, then |A|=kn|B|, n N

### PROPERTIES OF DETERMINANTS:

Let us glance through some properties of the Determinants, which are discussed in Class 12 Mathematics Chapter 4 Notes.

Property 1– the value of the determinant remains the same even after interchanging the rows and columns. Symbolic representation Ci Ri

Consider  A

 4 2 5 3

(4 x 3) – (5 x 2) = 12 – 10 = 2

And  A

 4 5 2 3

(4 x 3) – (2 x 5) = 12 – 10 = 2

This implies that:

If A is a square matrix, then det (A) = det (A’), where A’ is the transpose of matrix A.

Property 2– From the matrix, if the elements of any two columns (or rows) of the determinants are interchanged, then the sign of determinants will change. Symbolic representation Ci Cj or Ri Rj

Property 3– If the matrix has any two rows or columns equal or identical,  then we can say that the value of the determinant is equal to 0.

Property 4– If every element of a row or a column in a matrix is multiplied by k, then to obtain the value of the determinant, the original determinant is multiplied by k, where k is a constant.

Property 5– If the elements of any row or a column of a determinant are expressed as the sum of two or more terms, then we can say that the original determinant can also be expressed as the sum of two or more determinants.

Property 6– If for each element of a row or a column of the given determinant, the equimultiples of the analogous elements of other row or column respectively are added, then we can say that the value of the original determinant remains the same.

The symbolic representation of the operation performed is Ri Ri+ kRj or Ci Ci+ kCj

Property 7– If every element in a row or column of the given determinant is zero, then the value of the determinant is zero.

Property 8- Consider an upper triangular matrix or low triangular matrix, i.e., all elements on one side of the diagonal of the matrix are zeroes. The value of the determinant of such a matrix can be calculated by multiplying all diagonal elements.

In simple words, the determinant of an upper (lower) triangular matrix is equal to the product of all the elements in the diagonal.

Visit the Extramarks platform to access the class 12 mathematics chapter 4 notes. Gain in-depth information and understand each property in detail.

AREA OF A TRIANGLE:

Let the vertices of a triangle be (x1, y1),(x2, y2) and (x3, y3). Then the area of the triangle is defined as

A=12[x1(y2y3) + x3(y3y1) + x3(y1y3)].

Representation of area in the form of determinant = = 12 x determinant

Determinant is given as

 x1 y1 1 x2 y2 1 x3 y3 1

Remember:

1. The area is a positive quantity. Therefore determinant will be positive
2. In case the area is already given, then use both the positive and negative values of the determinant.
3. The area of a triangle with three collinear points is equal to zero.

The class 12 mathematics notes chapter 4 includes several problems based on the area of the triangle for students to practice and master this concept.

Minors of the Determinant:

Representation: Mij

Minor of any element is calculated by deleting the ith row and jth column in which the element lies. For a21 we delete the 2nd row and first column.

Minor of any element of the given determinant of order n, where n ≥ 2 is a determinant of order n – 1.

Cofactor of the Determinant:

Representation: Aij

Cofactor of an element aij is calculated by multiplying the minor of the element with (-1)i+j

Cofactor of element aij (Aij) = (-1)i+jMij

The adjoint of a matrix is the matrix obtained by the transpose of a cofactor matrix.

Singular Matrices

A singular matrix is defined as the square matrix whose determinant is zero.

Non-Singular Matrices

A Non-singular matrix is defined as the square matrix whose determinant is a non-zero value.

### THEOREMS:

Theorem 1 – Let A be any square matrix of order n, then A (adj A) = (adj A) A =|A| I, where I is the identity matrix of the same order.

Theorem 2 – Let A and B be the non-singular matrices of order n, then their product, i.e., AB and BA, will also be non-singular matrices of the same order n.

Theorem 3 – Let A and B be two square matrices of order n. The determinant of the product of matrix AB is given as the product of the respective determinants. It is written as AB=|A||B|.

Theorem 4 – A square matrix is invertible (inverse exists)  iff the matrix is non-singular. So, for any matrix A, the inverse of the matrix is A-1= 1A(adj A).

Refer to the class 12 chapter 4 mathematics notes to access unlimited problems for extra practice.

APPLICATIONS:

• Matrices and determinants are used to solve systems of linear equations with respect to two or three variables. They are also used to check the consistency of the given system of linear equations.
• A Consistent system is said to be a system of equations whose solution can be found or exists.
• An Inconsistent system is said to be a system of equations whose solution cannot be found or does not exist.
• The value of the determinant is a number that is capable of determining the uniqueness of the solution of the given system of linear equations.

SOLUTION TO THE SYSTEM OF LINEAR EQUATIONS:

Consider a system,

a1x + b1y=c1 and a2x + b2y=c2, then the matric representation will be

 a1 b1 x = c1 a2 b2 y c2

It is represented as AX = B ….. (1)

Case 1: A  is a non-singular matrix

In this case, we premultiply A-1to both sides of equation 1

Therefore, we get (A-1) AX = (A-1).B

Using associativity, (A-1A) X = (A-1).B

I X = (A-1).B

We get, X = (A-1).B

The value of X provides a unique solution. This method is commonly known as the matrix method.

Case 2: A  is a singular matrix

In this case, firstly, we calculate (adj A) B

Depending on the value of (adj A) B, we get our result.

(adj A)B is a non-zero matrix: Solution does not exist. The system of equations is said to be inconsistent with no solution.

(adj A)B is a zero matrix: Solution exists. The system of equations is said to be either consistent with many solutions or inconsistent.

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## Chapter 4 Mathematics Class 12 Notes: Exercises & Answer Solutions

Based on the latest guidelines and norms of CBSE, our class 12 mathematics chapter 4 notes provide an engaging and fun learning experience. Detailed answers, stepwise solutions, formulas, derivations, and important questions are all included in the notes. Students will gain a deeper knowledge of determinants and also learn how to apply the theoretical knowledge to solve the system of a linear equation to find its solution.

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Q.1 Find the area of a triangle whose vertices are given by (x1, y1), (x2, y2) and (x3, y3).

Ans

$\mathrm{Area}\mathrm{of}\mathrm{\Delta } =\frac{1}{2}\mathrm{ }\left|\begin{array}{l}{\mathrm{x}}_{1}{\mathrm{y}}_{1}1\\ {\mathrm{x}}_{2}{\mathrm{y}}_{2}1\\ {\mathrm{x}}_{3}{\mathrm{y}}_{3}1\end{array}\right|$

Q.2

$\mathrm{Evaluate}\left|\begin{array}{l}1 2 4 \\ –1 3 0 \\ 4 1 0\end{array}\right|$

Ans

$\begin{array}{l}\left|\begin{array}{l}1 2 4 \\ –1 3 0 \\ 4 1 0\end{array}\right|\mathrm{ }=\mathrm{ }1\mathrm{ }\left|\begin{array}{l}3 0\\ 1 0\end{array}\right|\mathrm{ }-2\mathrm{ }\left|\begin{array}{l}-1 0\\ 4 0\end{array}\right|+4\left|\begin{array}{l}-1 3\\ 4 1\end{array}\right|\\ =\mathrm{ }1\left(0\right)-2\left(0\right)+4\left(-1-12\right)\\ \mathrm{ }= 4\left(-13\right)\mathrm{ }=\mathrm{ }-\mathrm{ }52\end{array}$

Q.3

$\begin{array}{l}\mathrm{Show}\mathrm{that}\left|\begin{array}{l}\mathrm{a}\mathrm{b}\mathrm{c}\\ \mathrm{a}+2\mathrm{x}\mathrm{b}+2\mathrm{y} \mathrm{c}+2\mathrm{z}\\ \mathrm{x} \mathrm{y} \mathrm{z}\end{array}\right|\mathrm{ }=\mathrm{ }0 \mathrm{by}\\ \mathrm{using}\mathrm{properties}\mathrm{of}\mathrm{determinants}.\end{array}$

Ans

$\begin{array}{l}\left|\begin{array}{l}\mathrm{a}\mathrm{b}\mathrm{c}\\ \mathrm{a}+2\mathrm{x}\mathrm{b}+2\mathrm{y} \mathrm{c}+2\mathrm{z}\\ \mathrm{x} \mathrm{y} \mathrm{z}\end{array}\right|\mathrm{ }=\mathrm{ }\left|\begin{array}{l}\mathrm{a}\mathrm{b}\mathrm{c}\\ \mathrm{a}\mathrm{b}\mathrm{c}\\ \mathrm{x} \mathrm{y} \mathrm{z}\end{array}\right|+\left|\begin{array}{l}\mathrm{a}\mathrm{b}\mathrm{c}\\ 2\mathrm{x}2\mathrm{y} 2\mathrm{z}\\ \mathrm{x} \mathrm{y} \mathrm{z}\end{array}\right|\\ \left(\begin{array}{l}\mathrm{All}\mathrm{elements}\mathrm{of}{\mathrm{R}}_{2} \mathrm{are}\mathrm{expressed}\mathrm{as}\mathrm{sum}\mathrm{of}\\ \mathrm{two}\mathrm{terms}.\mathrm{Therefore},\mathrm{the}\mathrm{determinant}\mathrm{can}\\ \mathrm{be}\mathrm{expressed}\mathrm{as}\mathrm{sum}\mathrm{of}\mathrm{two}\mathrm{determinant}.\end{array}\right)\\ =\mathrm{ }0\mathrm{ }+2\mathrm{ }\left|\begin{array}{l}\mathrm{a}\mathrm{b}\mathrm{c}\\ \mathrm{a}\mathrm{b}\mathrm{c}\\ \mathrm{x} \mathrm{y} \mathrm{z}\end{array}\right|\mathrm{ }\left(2\mathrm{is}\mathrm{taken}\mathrm{out}\mathrm{common}\mathrm{from}{\mathrm{R}}_{2}.\right)\\ \left(\begin{array}{l}\mathrm{If}\mathrm{any}\mathrm{two}\mathrm{rows}\mathrm{or}\mathrm{columns}\mathrm{of}\mathrm{a}\mathrm{determinant}\mathrm{are}\\ \mathrm{identical}\mathrm{then}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{determinant}\mathrm{is}\mathrm{zero}.\end{array}\right)\\ = 0\mathrm{ }+\mathrm{ }0\mathrm{ }=\mathrm{ }0\end{array}$

Q.4 If P, Q, R are three non-null square matrices of the same order, write the condition on P such that PQ = PR ⇒ Q = R.

Ans

P must be invertible or |P| ≠ 0.

Q.5

$\mathrm{If}\mathrm{A}=\left|\begin{array}{l}12\\ 42\end{array}\right|,\mathrm{then}\mathrm{show}\mathrm{that}\left|2\mathrm{A}\right| = 4\left|\mathrm{A}\right|$

Ans

$\begin{array}{l} \mathrm{A}=\left|\begin{array}{l}12\\ 42\end{array}\right|\\ ⇒\mathrm{ }2\mathrm{A}\mathrm{ }= \mathrm{A}=\left|\begin{array}{l}24\\ 84\end{array}\right|\\ \mathrm{ }\therefore \mathrm{ }\left|\mathrm{A}\right|=\left|\begin{array}{l}12\\ 42\end{array}\right|\mathrm{ }=\mathrm{ }2-8\mathrm{ }=-6\\ \left|2\mathrm{A}\right|=\left|\begin{array}{l}24\\ 84\end{array}\right|\mathrm{ }=\mathrm{ }8-32\mathrm{ }=-24\\ ⇒\mathrm{ }\left|2\mathrm{A}\right|= -24 = 4×–6= 4\left|\mathrm{A}\right|\mathrm{}\end{array}$

Q.6

$\begin{array}{l}\mathrm{Examine}\mathrm{the}\mathrm{consistency}\mathrm{of}\mathrm{the}\mathrm{system}\mathrm{of}\mathrm{equations}:\\ 5\mathrm{x}–\mathrm{y}+4\mathrm{z}= 5\\ 2\mathrm{x}+3\mathrm{y}+5\mathrm{z}= 2\\ 5\mathrm{x}-2\mathrm{y}+6\mathrm{z}= -1\end{array}$

Ans

$\begin{array}{l}\mathrm{The}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{as}\mathrm{AX}=\mathrm{B}\\ \mathrm{where}\mathrm{A}=\left|\begin{array}{l}5-1 4\\ 23 5\\ 5-2 6\end{array}\right|\mathrm{ },\mathrm{ }\mathrm{X}\mathrm{ }=\mathrm{ }\left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right] \mathrm{and}\mathrm{B}=\mathrm{ }\left[\begin{array}{l}5\\ 2\\ -1\end{array}\right]\\ \\ \mathrm{Here},\left|\mathrm{A}\right|\mathrm{ }=\mathrm{ }5\left(18+10\right)\mathrm{ }+\left(12-25\right)\mathrm{ }+4\left(-4-15\right)\\ =\mathrm{ }5\left(28\right)+\left(-13\right)+4\left(-19\right)\\ =\mathrm{ }140-13-76\\ =\mathrm{ }51\ne 0\\ \mathrm{Since}\left|\mathrm{A}\right|\mathrm{ }\ne \mathrm{ }0,\mathrm{the}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{is}\mathrm{consistent}.\end{array}$

Q.7

$\begin{array}{l}\mathrm{If}\mathrm{A}=\left|\begin{array}{l}1-1 2\\ 02 -3\\ 3-2 4\end{array}\right| \mathrm{and}\mathrm{B}\left|\begin{array}{l}–20 1\\ 92 -3\\ 61 -2\end{array}\right| \mathrm{then}\mathrm{find}\mathrm{AB}.\\ \mathrm{Using}\mathrm{AB},\mathrm{solve}\mathrm{the}\mathrm{folowing}\mathrm{system}\mathrm{of}\mathrm{equations}:\\ \mathrm{x}-\mathrm{y}+2\mathrm{z}=1\\ 2\mathrm{y}-3\mathrm{z}=1\\ 3\mathrm{x}-2\mathrm{y}+4\mathrm{z}= 2\end{array}$

Ans

$\begin{array}{l}\mathrm{Here},\mathrm{AB}=\left|\begin{array}{l}1-1 2\\ 02 -3\\ 3-2 4\end{array}\right| \left|\begin{array}{l}–20 1\\ 92 -3\\ 61 -2\end{array}\right|\\ =\mathrm{ }\left|\begin{array}{l}-\mathrm{2}-9+1 2 0-2+21+3-\mathrm{4}\\ 0+18-18 0+4-30-6+6\\ -6-18+240-4+4 3 +6-8\end{array}\right|\\ =\left|\begin{array}{l}10 0\\ 01 0\\ 00 1\end{array}\right|\\ \mathrm{Then}\mathrm{AB}=\mathrm{I}\\ \mathrm{Thus}{\mathrm{A}}^{-1}\mathrm{ }=\mathrm{ }\mathrm{B}\\ {\left|\begin{array}{l}1-1 2\\ 02 -3\\ 3-2 4\end{array}\right|}^{-1}= \left|\begin{array}{l}–20 1\\ 92 -3\\ 61 -2\end{array}\right|\\ \mathrm{Now}\mathrm{given}\mathrm{equations}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{as}{\mathrm{A}}_{1}\mathrm{X}\mathrm{ }=\mathrm{ }{\mathrm{B}}_{1}\\ \mathrm{where},{\mathrm{A}}_{1}\mathrm{ }=\mathrm{ }\left|\begin{array}{l}1-1 2\\ 02 -3\\ 3-2 4\end{array}\right|\mathrm{ }\mathrm{X}\mathrm{ }=\mathrm{ }\left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right] \mathrm{and}{\mathrm{B}}_{1}\mathrm{}=\mathrm{ }\left[\begin{array}{l}1\\ 1\\ 2\end{array}\right]\\ \mathrm{Now} {\mathrm{A}}_{1\mathrm{ }}=\mathrm{ }\mathrm{A}\\ \therefore \mathrm{ }{{\mathrm{A}}_{1}}^{-1}\mathrm{ }=\mathrm{ }{\mathrm{A}}^{-1}\\ \mathrm{ }⇒\mathrm{ }{{\mathrm{A}}_{1}}^{-1}\mathrm{ }=\mathrm{ }\left|\begin{array}{l}–20 1\\ 92 -3\\ 61 -2\end{array}\right|\\ \mathrm{as}\mathrm{X}= {{\mathrm{A}}_{1}}^{-1}\mathrm{ }{\mathrm{B}}_{1}\\ \mathrm{ }\therefore \mathrm{ }\mathrm{X}\mathrm{ }=\mathrm{ }\left|\begin{array}{l}–20 1\\ 92 -3\\ 61 -2\end{array}\right|\left[\begin{array}{l}1\\ 1\\ 2\end{array}\right]\\ \mathrm{X}\mathrm{ }=\mathrm{ }\left|\begin{array}{l}–2+ 0 + 1\\ 9+2 -6\\ 6+1 -4\end{array}\right|\left[\begin{array}{l}0\\ 5\\ 3\end{array}\right]\\ \mathrm{Hence},\mathrm{x}= 0,\mathrm{y}= 5\mathrm{and}\mathrm{z}= 3\end{array}$

Q.8 The sum of three numbers is 6. If we multiply third number by 3 and add second number to it we get 11. By adding first and third numbers we get double of the second number. Represent it algebraically and find the numbers using matrix method.

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{the}\mathrm{first},\mathrm{second}\mathrm{and}\mathrm{third}\mathrm{numbers}\mathrm{be}\mathrm{x},\mathrm{y}\mathrm{and}\mathrm{z}.\\ \therefore \mathrm{x}+\mathrm{y}+\mathrm{z}= 6\\ \mathrm{y}+3\mathrm{z} = 11\\ \mathrm{x}+\mathrm{z}= 2\mathrm{y}\mathrm{or}\mathrm{x}-2\mathrm{y}+\mathrm{z}= 0\\ \mathrm{This}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{as}\mathrm{AX}=\mathrm{B},\mathrm{where}\\ \\ \mathrm{A} = \left[\begin{array}{l}11 1\\ 01 3\\ 1-2 1\end{array}\right],\mathrm{ }\mathrm{X}\mathrm{ }=\mathrm{ }\left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right] \mathrm{and}\mathrm{B}=\left[\begin{array}{l}6\\ 11\\ 0\end{array}\right]\\ \mathrm{Here},\mathrm{A}\mathrm{is}\mathrm{non}\mathrm{singular}\mathrm{matrix}\mathrm{and}\mathrm{so}\mathrm{its}\mathrm{inverse}\mathrm{exists}.\\ \mathrm{Now}\mathrm{we}\mathrm{find}\mathrm{adjA}\\ {\mathrm{c}}_{11}\mathrm{ }=\mathrm{ }7 {\mathrm{c}}_{12}\mathrm{ }=\mathrm{ }3 {\mathrm{c}}_{13}\mathrm{ }=\mathrm{ }-1\\ {\mathrm{c}}_{21}\mathrm{ }=\mathrm{ }-3 {\mathrm{c}}_{22}\mathrm{ }=\mathrm{ }0 {\mathrm{c}}_{23}\mathrm{ }=\mathrm{ }3\\ {\mathrm{c}}_{31}\mathrm{ }=\mathrm{ }2 {\mathrm{c}}_{32}\mathrm{ }=\mathrm{ }-3 {\mathrm{c}}_{33}\mathrm{ }=\mathrm{ }1\\ \mathrm{Hence},\mathrm{adj}\mathrm{A}=\mathrm{A} = \left[\begin{array}{l}7-3 2\\ 30 -3\\ –13 1\end{array}\right]\\ \mathrm{Thus}{\mathrm{A}}^{-1}=\mathrm{}\frac{\mathrm{adjA}}{\left|\mathrm{A}\right|}=\mathrm{ }\frac{1}{9}\mathrm{ }\left[\begin{array}{l}7-3 2\\ 30 -3\\ –13 1\end{array}\right]\\ \mathrm{Since} \mathrm{X}={\mathrm{A}}^{-1}\mathrm{B}\\ \mathrm{Therefore} \mathrm{X}=\frac{1}{9}\mathrm{ }\left[\begin{array}{l}7-3 2\\ 30 -3\\ –13 1\end{array}\right]\left[\begin{array}{l}6\\ 11\\ 0\end{array}\right]\\ \left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]\mathrm{ }=\mathrm{ }\frac{1}{9}\mathrm{ }\left[\begin{array}{l}42-33 0\\ 18+0 -0\\ –6+ 3 +0\end{array}\right]\mathrm{ }=\mathrm{ }\frac{1}{9}\mathrm{ }\left[\begin{array}{l}9\\ 18\\ 3\end{array}\right]=\mathrm{ }\left[\begin{array}{l}1\\ 2\\ 3\end{array}\right]\\ \mathrm{Thus} \mathrm{x} =1, \mathrm{y}= 2\mathrm{and}\mathrm{z}= 3\end{array}$

Q.9

$\begin{array}{l}\mathrm{Solve}\mathrm{the}\mathrm{following}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{by}\mathrm{matrix}\mathrm{method}.\\ \left|\begin{array}{l}3\mathrm{x}-2\mathrm{y}+3\mathrm{z} = 8\\ 2\mathrm{x} + \mathrm{y}–\mathrm{z} = 1\\ 4\mathrm{x}-3\mathrm{y}+2\mathrm{z} =4\end{array}\right|\end{array}$

Ans

$\begin{array}{l}\mathrm{The}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{as}\mathrm{AX}=\mathrm{B},\mathrm{where}\\ \mathrm{A}= \left[\begin{array}{l}3 -2 3\\ 2 1 -1\\ 4 -3 2\end{array}\right]\mathrm{ }\mathrm{x}\mathrm{ }=\mathrm{ }\left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]\mathrm{ }\mathrm{and}\mathrm{B}=\left[\begin{array}{l}8\\ 1\\ 4\end{array}\right]\\ \mathrm{Here},\left|\mathrm{A}\right|\mathrm{ }=3\left(2-3\right)+2\left(4+4\right)+3\left(-6-4\right)\\ =\mathrm{ }-17\ne \mathrm{ }0\\ \mathrm{Hence},\mathrm{A}\mathrm{is}\mathrm{nonsingular}\mathrm{matrix}\mathrm{and}\mathrm{so}\mathrm{its}\mathrm{inverse}\mathrm{exists}.\\ \mathrm{Now}\mathrm{cofactors}\mathrm{of}\mathrm{A}\mathrm{are}\\ {\mathrm{c}}_{11\mathrm{ }}=\mathrm{ }-1 {\mathrm{c}}_{12} =\mathrm{ }-8 {\mathrm{c}}_{13}\mathrm{ }=\mathrm{ }-10\\ {\mathrm{c}}_{12\mathrm{ }}=\mathrm{ }-1 {\mathrm{c}}_{22} =\mathrm{ }-8 {\mathrm{c}}_{23}\mathrm{ }=\mathrm{ }-1\\ {\mathrm{c}}_{31\mathrm{ }}=\mathrm{ }-1 {\mathrm{c}}_{32} =\mathrm{ }-8 {\mathrm{c}}_{33}\mathrm{ }=\mathrm{ }-7\\ \mathrm{adjoint}\mathrm{of}\mathrm{A}={\mathrm{C}}^{\mathrm{T}}=\mathrm{ }\left[\begin{array}{l}-1 -5 -1\\ -8 -6 9\\ -10 1 7\end{array}\right]\\ \therefore {\mathrm{A}}^{-1}\mathrm{ }=\mathrm{ }\frac{\mathrm{adjA}}{\left|\mathrm{A}\right|}\mathrm{ }=\mathrm{ }\frac{-1}{7}= \left[\begin{array}{l}-1 -5 -1\\ -8 -6 9\\ -10 1 7\end{array}\right]\\ \mathrm{So}\mathrm{X}={\mathrm{A}}^{-1}\mathrm{B}\frac{-1}{17} = \left[\begin{array}{l}-1 -5 -1\\ -8 -6 9\\ -10 1 7\end{array}\right]\mathrm{ }\left[\begin{array}{l}8\\ 1\\ 4\end{array}\right]\\ \left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]\mathrm{ }=\mathrm{ }\frac{-1}{17}\left[\begin{array}{l}-17\\ -34\\ -51\end{array}\right]\mathrm{ }=\mathrm{ }\left[\begin{array}{l}1\\ 2\\ 3\end{array}\right]\\ \mathrm{Hence},\mathrm{x}= 1,\mathrm{y}= 2\mathrm{and}\mathrm{z}= 3.\end{array}$

Q.10

$\begin{array}{l}\mathrm{If} \mathrm{a}, \mathrm{b}, \mathrm{care}\mathrm{in}\mathrm{A}.\mathrm{P}\mathrm{then}\mathrm{without}\mathrm{expanding}\mathrm{show}\mathrm{that}\\ \left|\begin{array}{l}\mathrm{x}+2\mathrm{x}+3\mathrm{x}+2\mathrm{a}\\ \mathrm{x}+3\mathrm{x}+4\mathrm{x}+2\mathrm{b}\\ \mathrm{x}+4\mathrm{x}+5\mathrm{x}+2\mathrm{c}\end{array}\right|\mathrm{ }=\mathrm{ }0\end{array}$

Ans

$\begin{array}{l}\mathrm{Here},\mathrm{\Delta } = \left|\begin{array}{l}\mathrm{x}+2\mathrm{x}+3\mathrm{x}+2\mathrm{a}\\ \mathrm{x}+3\mathrm{x}+4\mathrm{x}+2\mathrm{b}\\ \mathrm{x}+4\mathrm{x}+5\mathrm{x}+2\mathrm{c}\end{array}\right|\mathrm{ }\\ =\mathrm{ }\left|\begin{array}{l}\mathrm{x}+2\mathrm{x}+3\mathrm{x}+2\mathrm{a}\\ \mathrm{x}+3\mathrm{x}+4\mathrm{x}+2\mathrm{b}\\ 1 1 2\left(\mathrm{c}-\mathrm{b}\right)\end{array}\right|\mathrm{ }\left(\mathrm{by}\mathrm{applying}{\mathrm{R}}_{3}\to {\mathrm{R}}_{3}-{\mathrm{R}}_{2}\right)\\ =\mathrm{ }\left|\begin{array}{l}\mathrm{x}+2\mathrm{x}+3\mathrm{x}+2\mathrm{a}\\ 1 1 2\left(\mathrm{b}-\mathrm{a}\right)\\ 1 1 2\left(\mathrm{c}-\mathrm{b}\right)\end{array}\right|\mathrm{ }\left(\mathrm{by}\mathrm{applying}{\mathrm{R}}_{2}\to {\mathrm{R}}_{2}-{\mathrm{R}}_{1}\right)\mathrm{ }\\ \mathrm{Now}\mathrm{because}\mathrm{a},\mathrm{b},\mathrm{c}\mathrm{are}\mathrm{in}\mathrm{A}.\mathrm{P}.,\mathrm{we}\mathrm{have}\\ \mathrm{b}-\mathrm{a}\mathrm{ }=\mathrm{ }\mathrm{c}\mathrm{ }-\mathrm{b}\\ \mathrm{Hence}\mathrm{in}\mathrm{\Delta } {\mathrm{R}}_{2}=\mathrm{ }{\mathrm{R}}_{3}\\ \mathrm{If}\mathrm{any}\mathrm{two}\mathrm{rows}\mathrm{or}\mathrm{columns}\mathrm{of}\mathrm{a}\mathrm{determinant}\mathrm{are}\mathrm{identical}\\ \mathrm{then}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{determinant}\mathrm{is}\mathrm{zero}.\\ ⇒\mathrm{ }\mathrm{\Delta } = 0\end{array}$

Q.11

$\begin{array}{l}\mathrm{Without}\mathrm{expanding},\mathrm{prove}\mathrm{that}\\ \left|\begin{array}{l}\mathrm{b}+\mathrm{c}\mathrm{q}+\mathrm{r} \mathrm{y}+\mathrm{z}\\ \mathrm{c}+\mathrm{a} \mathrm{r}+\mathrm{p} \mathrm{z}+\mathrm{x}\\ \mathrm{a}+\mathrm{b} \mathrm{p}+\mathrm{q} \mathrm{x}+\mathrm{y}\end{array}\right|\mathrm{ }=\mathrm{ }2\mathrm{ }\left|\begin{array}{l}\mathrm{a}\mathrm{q} \mathrm{x}\\ \mathrm{b} \mathrm{q}\mathrm{y}\\ \mathrm{c}\mathrm{r}\mathrm{z}\end{array}\right|\end{array}$

Ans

$\begin{array}{l}\mathrm{Here},\mathrm{LHS}=\left|\begin{array}{l}\mathrm{b}+\mathrm{c}\mathrm{q}+\mathrm{r} \mathrm{y}+\mathrm{z}\\ \mathrm{c}+\mathrm{a} \mathrm{r}+\mathrm{p} \mathrm{z}+\mathrm{x}\\ \mathrm{a}+\mathrm{b} \mathrm{p}+\mathrm{q} \mathrm{x}+\mathrm{y}\end{array}\right|\mathrm{ }\\ \mathrm{By}\mathrm{applying}{\mathrm{R}}_{1}\to {\mathrm{R}}_{1}-\left({\mathrm{R}}_{2}+{\mathrm{R}}_{3}\right), \mathrm{we}\mathrm{get}\\ \mathrm{LHS}=\left|\begin{array}{l}-2\mathrm{a}-2\mathrm{p} -2\mathrm{x}\\ \mathrm{c}+\mathrm{a} \mathrm{r}+\mathrm{p} \mathrm{z}+\mathrm{x}\\ \mathrm{a}+\mathrm{b} \mathrm{p}+\mathrm{q} \mathrm{x}+\mathrm{y}\end{array}\right|\mathrm{ }\\ =\mathrm{ }2\mathrm{ }\left|\begin{array}{l}-\mathrm{a}-\mathrm{p} -\mathrm{x}\\ \mathrm{c}+\mathrm{a} \mathrm{r}+\mathrm{p} \mathrm{z}+\mathrm{x}\\ \mathrm{a}+\mathrm{b} \mathrm{p}+\mathrm{q} \mathrm{x}+\mathrm{y}\end{array}\right|\mathrm{ }\left[\mathrm{Taking}2\mathrm{common}\mathrm{from}{\mathrm{R}}_{1}\right]\\ =\mathrm{ }2\mathrm{ }\left|\begin{array}{l}-\mathrm{a}-\mathrm{p} -\mathrm{x}\\ \mathrm{c} \mathrm{r} \mathrm{z}\\ \mathrm{b} \mathrm{q} \mathrm{y}\end{array}\right|\mathrm{ }\left[\mathrm{By}\mathrm{applying}{\mathrm{R}}_{2}\to {\mathrm{R}}_{2}+{\mathrm{R}}_{1}\mathrm{ }\mathrm{and}\mathrm{ }{\mathrm{R}}_{3}\to {\mathrm{R}}_{3}+{\mathrm{R}}_{1}\right]\\ =\mathrm{ }2\mathrm{ }\left|\begin{array}{l}\mathrm{a} \mathrm{p} \mathrm{x}\\ \mathrm{c} \mathrm{r} \mathrm{z}\\ \mathrm{b} \mathrm{q} \mathrm{y}\end{array}\right| \left[\mathrm{Taking}– 1\mathrm{common}\mathrm{from}{\mathrm{R}}_{1}\right]\\ =\mathrm{ }2\mathrm{ }\left|\begin{array}{l}\mathrm{a} \mathrm{p} \mathrm{x}\\ \mathrm{b} \mathrm{q} \mathrm{y}\\ \mathrm{c} \mathrm{r} \mathrm{z}\end{array}\right| \left[\mathrm{By}\mathrm{applying}{\mathrm{R}}_{2}↔{\mathrm{R}}_{3}\right]\\ =\mathrm{ }\mathrm{RHS}\end{array}$

Q.12

$\begin{array}{l}\mathrm{Using}\mathrm{properties}\mathrm{of}\mathrm{determinants},\mathrm{show}\mathrm{that}\\ \mathrm{ }\left|\begin{array}{l}1+\mathrm{a} 1 1\\ 1 1+\mathrm{b} 1\\ 1 1 1+\mathrm{c}\end{array}\right|\mathrm{ }=\mathrm{ }\mathrm{abc}\mathrm{ }\left(1+\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{b}}+\frac{1}{\mathrm{c}}\right)\mathrm{ }=\mathrm{ }\mathrm{abc}+\mathrm{bc}+\mathrm{ca}+\mathrm{ab}.\end{array}$

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{\Delta } =\left|\begin{array}{l}1+\mathrm{a} 1 1\\ 1 1+\mathrm{b} 1\\ 1 1 1+\mathrm{c}\end{array}\right|\\ \mathrm{By}\mathrm{applying}{\mathrm{c}}_{1}\to {\mathrm{c}}_{2}\mathrm{ }\mathrm{and}\mathrm{ }{\mathrm{c}}_{3}\to {\mathrm{c}}_{3}-{\mathrm{c}}_{2},\mathrm{ }\mathrm{we}\mathrm{get}\\ \left|\begin{array}{l}\mathrm{a} 1 0\\ -\mathrm{b} 1+\mathrm{b} -\mathrm{b}\\ 0 1 \mathrm{c}\end{array}\right|\\ \mathrm{Expanding}\mathrm{along}{\mathrm{R}}_{1}, \mathrm{we}\mathrm{obtain}\\ \mathrm{ }\mathrm{\Delta }\mathrm{ }=\mathrm{ }\mathrm{a}\mathrm{ }\left|\begin{array}{l}1+\mathrm{b} -\mathrm{b}\\ 1 \mathrm{c}\end{array}\right|-1\mathrm{ }\left|\begin{array}{l}-\mathrm{b} -\mathrm{b}\\ 0 \mathrm{c}\end{array}\right|+\mathrm{ }0\\ =\mathrm{ }\mathrm{a}\mathrm{ }\left(\mathrm{c}+\mathrm{bc}+\mathrm{b}\right)-\left(-\mathrm{bc}\right)\\ =\mathrm{ }\mathrm{ac}+\mathrm{abc}+\mathrm{ab}+\mathrm{bc}\mathrm{ }\\ =\mathrm{ }\mathrm{abc}+\mathrm{bc}+\mathrm{ca}+\mathrm{ab}\mathrm{ }\\ =\mathrm{ }\mathrm{abc}\mathrm{ }\left[\mathrm{a}+\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{b}}+\frac{1}{\mathrm{c}}\right]\end{array}$

Q.13

$\begin{array}{l}\mathrm{If}\mathrm{x},\mathrm{y},\mathrm{z}\mathrm{are}\mathrm{different}\mathrm{and}\mathrm{Let}\mathrm{\Delta }\mathrm{ = }\left|\begin{array}{l}\mathrm{x}\mathrm{ }{\mathrm{x}}^{2}\mathrm{ }1+{\mathrm{x}}^{3}\\ \mathrm{y}\mathrm{ }{\mathrm{y}}^{2}\mathrm{ }1+{\mathrm{y}}^{3}\\ \mathrm{z}\mathrm{ }{\mathrm{z}}^{2}\mathrm{ }1+{z}^{3}\end{array}\right|\mathrm{ }=\mathrm{ }0,\\ \mathrm{then}\mathrm{show}\mathrm{that}1+\mathrm{xyz}\mathrm{is}\mathrm{a}\mathrm{factor}\mathrm{of}\mathrm{\Delta }\mathrm{.}\end{array}$

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{\Delta } = \left|\begin{array}{l}\mathrm{x} {\mathrm{x}}^{2} 1+{\mathrm{x}}^{3}\\ \mathrm{y} {\mathrm{y}}^{2} 1+{\mathrm{y}}^{3}\\ \mathrm{z} {\mathrm{z}}^{2} 1+{\mathrm{z}}^{3}\end{array}\right|\mathrm{ }=\mathrm{ }0,\\ \mathrm{All}\mathrm{elements}\mathrm{of}{\mathrm{c}}_{3}\mathrm{ }\mathrm{are}\mathrm{expressed}\mathrm{as}\mathrm{sum}\mathrm{of}\mathrm{two}\mathrm{terms}.\\ \mathrm{Therefore},\mathrm{the}\mathrm{determinant}\mathrm{can}\mathrm{be}\mathrm{expressed}\mathrm{as}\mathrm{sum}\mathrm{of}\\ \mathrm{two}\mathrm{determinants}.\\ =\mathrm{ }\left|\begin{array}{l}\mathrm{x} {\mathrm{x}}^{2} 1\\ \mathrm{y} {\mathrm{y}}^{2} 1\\ \mathrm{z} {\mathrm{z}}^{2} 1\end{array}\right|\mathrm{ }+\mathrm{ }\left|\begin{array}{l}\mathrm{x} {\mathrm{x}}^{2} {\mathrm{x}}^{3}\\ \mathrm{y} {\mathrm{y}}^{2} {\mathrm{y}}^{3}\\ \mathrm{z} {\mathrm{z}}^{2} {\mathrm{z}}^{3}\end{array}\right|\\ \mathrm{Using}{\mathrm{c}}_{3}↔\mathrm{ }{\mathrm{c}}_{2}\mathrm{and}\mathrm{then}{\mathrm{c}}_{1}↔\mathrm{ }{\mathrm{c}}_{2} \mathrm{in}\mathrm{first}\mathrm{determinant}\mathrm{and}\\ \mathrm{taking}\mathrm{out}\mathrm{x},\mathrm{y}\mathrm{and}\mathrm{z}\mathrm{common}\mathrm{from}{\mathrm{R}}_{1},\mathrm{ }{\mathrm{R}}_{2} \mathrm{and}{\mathrm{R}}_{3} \mathrm{respectively}\\ \mathrm{in}\mathrm{second}\mathrm{determinant},\mathrm{we}\mathrm{get}\\ = {\left(-1\right)}^{2}\mathrm{ }\left|\begin{array}{l}1 \mathrm{x} {\mathrm{x}}^{2}\\ 1 \mathrm{y} {\mathrm{y}}^{2}\\ 1 \mathrm{z} {\mathrm{z}}^{2}\end{array}\right|\mathrm{ }+\mathrm{xyz}\mathrm{ }\left|\begin{array}{l}1 \mathrm{x} {\mathrm{x}}^{2}\\ 1 \mathrm{y} {\mathrm{y}}^{2}\\ 1 \mathrm{z} {\mathrm{z}}^{2}\end{array}\right|\\ \mathrm{\Delta }\mathrm{ }=\mathrm{ }\left|\begin{array}{l}1 \mathrm{x} {\mathrm{x}}^{2}\\ 1 \mathrm{y} {\mathrm{y}}^{2}\\ 1 \mathrm{z} {\mathrm{z}}^{2}\end{array}\right|\mathrm{ }\left(1+\mathrm{xyz}\right)\\ \mathrm{Hence},\left(1+\mathrm{xyz}\right)\mathrm{is}\mathrm{factor}\mathrm{of}\mathrm{\Delta }\mathrm{.}\end{array}$

Q.14

$\begin{array}{l}\mathrm{Using}\mathrm{properties}\mathrm{of}\mathrm{determinants},\mathrm{prove}\mathrm{that}\\ \left|\begin{array}{l}\mathrm{a} \mathrm{a}+\mathrm{b} \mathrm{a}+\mathrm{b}+\mathrm{c} \\ 2\mathrm{a} 3\mathrm{a}+2\mathrm{b} 4\mathrm{a}+3\mathrm{b}+2\mathrm{c}\\ 3\mathrm{a} 6\mathrm{a}+3\mathrm{b} 10\mathrm{a}+6\mathrm{b}+3\mathrm{c}\end{array}\right|\mathrm{ }=\mathrm{ }{\mathrm{a}}^{3}\end{array}$

Ans

$\begin{array}{l}\mathrm{By}\mathrm{applying}{\mathrm{R}}_{2}\to {\mathrm{R}}_{2}-2{\mathrm{R}}_{1}\mathrm{ }\mathrm{and}\mathrm{ }{\mathrm{R}}_{3}\to {\mathrm{R}}_{3}-3{\mathrm{R}}_{1}\\ \mathrm{\Delta }\mathrm{ }=\mathrm{ }\left|\begin{array}{l}\mathrm{a}\mathrm{a}+\mathrm{b} \mathrm{a}+\mathrm{b}+\mathrm{c} \\ 0 \mathrm{a} 2\mathrm{a}+\mathrm{b}\\ 0 3\mathrm{a} 7\mathrm{a}+3\mathrm{b}\end{array}\right|\\ \mathrm{Now}\mathrm{applying}{\mathrm{R}}_{3}\to {\mathrm{R}}_{3}-{\mathrm{R}}_{2},\mathrm{ }\mathrm{we}\mathrm{ }\mathrm{get}\\ \mathrm{\Delta }\mathrm{ }=\mathrm{ }\left|\begin{array}{l}\mathrm{a}\mathrm{a}+\mathrm{b} \mathrm{a}+\mathrm{b}+\mathrm{c} \\ 0 \mathrm{a} 2\mathrm{a}+\mathrm{b}\\ 0 0 \mathrm{a}\end{array}\right|\\ \mathrm{Expanding}\mathrm{along}{\mathrm{c}}_{1}\mathrm{ }\mathrm{we}\mathrm{obtain}\\ \mathrm{ }\mathrm{\Delta } = \mathrm{a} \left|\begin{array}{l}\mathrm{a}2\mathrm{a}+\mathrm{b}\\ 0 \mathrm{a}\mathrm{ }\end{array}\right|\mathrm{ }+\mathrm{ }0+0\\ =\mathrm{ }\mathrm{a}\left({\mathrm{a}}^{2}-0\right)\mathrm{ }=\mathrm{ }\mathrm{a}\left({\mathrm{a}}^{2}\right)\mathrm{ }=\mathrm{ }{\mathrm{a}}^{3}\end{array}$

Q.15

$\mathrm{If}\mathrm{A} = \left[\begin{array}{l}2 5\\ 3 6\end{array}\right], \mathrm{verify}\mathrm{that}\mathrm{A}\left(\mathrm{adjA}\right) = \left|\mathrm{A}\right| \mathrm{I}.$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have} \left|\mathrm{A}\right|\mathrm{ }=4-6=-\mathrm{2}\\ \left|\mathrm{A}\right|\mathrm{ }\mathrm{I}=-2 \left[\begin{array}{l}1 0\\ 0 1\end{array}\right]\\ =\mathrm{ }\left[\begin{array}{l}-2 0\\ 0 -2\end{array}\right] .\dots ..\left(1\right)\\ \mathrm{Cofactors}\mathrm{of}\mathrm{A}\\ {\mathrm{c}}_{11}\mathrm{ }=\mathrm{ }4 {\mathrm{c}}_{12}\mathrm{ }=\mathrm{ }-2\\ {\mathrm{c}}_{21}\mathrm{ }=\mathrm{ }-3 {\mathrm{c}}_{22}\mathrm{ }=\mathrm{ }1\\ ⇒\mathrm{ }\mathrm{adjA}\mathrm{ }=\mathrm{ }{\mathrm{c}}^{\mathrm{T}}\\ =\mathrm{ }\left[\begin{array}{l}4 -2\\ -3 1\end{array}\right]\\ \mathrm{Now}\mathrm{A}\left(\mathrm{adjA}\right)\mathrm{ }=\mathrm{ }\left[\begin{array}{l}1 2\\ 3 4\end{array}\right]\left[\begin{array}{l}4 -2\\ -3 1\end{array}\right]\\ =\mathrm{ }\left[\begin{array}{l}1×4+2×\left(-3\right) 1×\left(-2\right)+2×1\\ 3×4+4×\left(-3\right) 3×\left(-2\right)+4×1\end{array}\right]\\ =\mathrm{ }\left[\begin{array}{l}4-6 -2+2\\ 12-12 -6+4\end{array}\right]\\ =\mathrm{ }\left[\begin{array}{l}-2 2\\ 0 -2\end{array}\right] .\dots \mathrm{ }\left(2\right)\\ \mathrm{by}\left(1\right) \mathrm{and}\left(2\right)\\ \mathrm{A} \left(\mathrm{adjA}\right)\mathrm{ }=\mathrm{ }\left|\mathrm{A}\right|\mathrm{I}\end{array}$

Q.16 Find the area of the triangle whose vertices are (3, 8), (-4, 2) and (5, 1).

Ans

$\begin{array}{l} \mathrm{The}\mathrm{area}\mathrm{of}\mathrm{triangle}\mathrm{is}\mathrm{given}\mathrm{by}\\ \mathrm{}\mathrm{\Delta } = \frac{1}{2}\mathrm{ }\left|\begin{array}{l}3 8 1\mathrm{ }\\ -4 2 1\mathrm{ }\\ 5 1 1\end{array}\right|\\ =\mathrm{ }\frac{1}{2}\mathrm{ }\left[3\left(2-1\right)-8\left(-4-5\right)+1\left(-4-10\right)\right]\\ =\mathrm{ }\frac{1}{2}\mathrm{ }\left(3+72-14\right)\\ =\mathrm{ }\frac{61}{2} \mathrm{Sq}.\mathrm{units}\end{array}$

Q.17

$\mathrm{Without}\mathrm{expanding},\mathrm{evaluate} \mathrm{\Delta } = \left|\begin{array}{l}1 \mathrm{a} \mathrm{bc} \\ 2 \mathrm{b} \mathrm{ca}\\ 5 \mathrm{c} \mathrm{ab}\end{array}\right|\mathrm{.}$

Ans

$\begin{array}{l}\mathrm{Applying}{\mathrm{R}}_{2}\to \mathrm{ }{\mathrm{R}}_{2}-{\mathrm{R}}_{1}\mathrm{ }\mathrm{and}\mathrm{ }{\mathrm{R}}_{3}\to \mathrm{ }{\mathrm{R}}_{3}\mathrm{ }-{\mathrm{R}}_{1}, \mathrm{We}\mathrm{ }\mathrm{get}\\ \mathrm{ }\mathrm{\Delta }\mathrm{ }= \left|\begin{array}{l}1 \mathrm{a} \mathrm{bc} \\ 0 \mathrm{b}-\mathrm{c} \mathrm{c}\left(\mathrm{a}-\mathrm{b}\right)\\ 0 \mathrm{c}-\mathrm{a} \mathrm{b}\left(\mathrm{a}-\mathrm{c}\right)\end{array}\right|\\ \mathrm{Taking}\mathrm{factors}\left(\mathrm{b}-\mathrm{a}\right)\mathrm{ }\mathrm{and}\mathrm{}\left(\mathrm{c}-\mathrm{a}\right)\mathrm{common}\mathrm{from}{\mathrm{R}}_{2}\mathrm{ }\mathrm{and}\mathrm{ }{\mathrm{R}}_{3},\mathrm{ }\\ \mathrm{respectively}\mathrm{we}\mathrm{get}\\ \mathrm{ }\mathrm{\Delta } = \left(\mathrm{b}-\mathrm{a}\right)\left(\mathrm{c}-\mathrm{a}\right)\mathrm{ }\left|\begin{array}{l}1 \mathrm{a} \mathrm{bc} \\ 0 1 -\mathrm{c}\\ 0 1 -\mathrm{b}\end{array}\right|\\ =\mathrm{ }\left(\mathrm{b}-\mathrm{a}\right)\left(\mathrm{c}-\mathrm{a}\right)\left(-\mathrm{b}+\mathrm{c}\right)\\ =\mathrm{ }\left(\mathrm{a}-\mathrm{b}\right)\left(\mathrm{b}-\mathrm{c}\right)\left(\mathrm{c}-\mathrm{a}\right)\end{array}$

Q.18

$\text{If A = }\left|\begin{array}{l}1 1 -2\mathrm{ }\\ 2 1 -3\\ 5 4 -\mathrm{9}\end{array}\right|, \mathrm{find}\left|\mathrm{A}\right|.$

Ans

$\begin{array}{l}\mathrm{Expanding}\mathrm{with}{\mathrm{R}}_{1}\\ \left|\mathrm{A}\right|\mathrm{ }=\mathrm{ }1\mathrm{ }\left|\begin{array}{l}1 -3\\ 4 -9\end{array}\right|\mathrm{ }-1\mathrm{ }\left|\begin{array}{l}2 -3\\ 5 -9\end{array}\right|\mathrm{ }-2\left|\begin{array}{l}2 1\\ 5 4\end{array}\right|\\ =\mathrm{ }\left(-9+12\right)-1\left(-18+15\right)-2\left(8-5\right)\\ =\mathrm{ }3-\left(-3\right)-2\left(3\right)\\ =\mathrm{ }3+3-6\\ =\mathrm{ }0\end{array}$

Q.19

$\text{Find values of x for which}\left|\begin{array}{l}3 \mathrm{x}\\ \mathrm{x} 1\end{array}\right|\mathrm{ }=\mathrm{ }\left|\begin{array}{l}3 2\\ 4 1\end{array}\right|.$

Ans

$\begin{array}{l} \mathrm{We}\mathrm{have},\left|\begin{array}{l}3 \mathrm{x}\\ \mathrm{x} 1\end{array}\right|\mathrm{ }=\mathrm{ }\left|\begin{array}{l}3 2\\ 4 1\end{array}\right|.\\ ⇒ 3-{\mathrm{x}}^{2}\mathrm{ }=\mathrm{ }3-8\\ ⇒ {\mathrm{x}}^{2}\mathrm{ }=\mathrm{ }8\\ ⇒ \mathrm{x}\mathrm{ }=\mathrm{ }±2\sqrt{2}\end{array}$

Q.20

$\mathrm{Find}\mathrm{values}\mathrm{of}\mathrm{x}\mathrm{for}\mathrm{which}\left|\begin{array}{l}1 \mathrm{x}\\ \mathrm{x} 1\end{array}\right|\mathrm{ }=\mathrm{ }\left|\begin{array}{l}4 1\\ 4 1\end{array}\right|$

Ans

$\begin{array}{l} \mathrm{We}\mathrm{have},\left|\begin{array}{l}1 \mathrm{x}\\ \mathrm{x} 1\end{array}\right|\mathrm{ }=\mathrm{ }\left|\begin{array}{l}4 1\\ 4 1\end{array}\right|\\ ⇒ 1-{\mathrm{x}}^{2}\mathrm{ }=\mathrm{ }4-4\\ ⇒ 1-{\mathrm{x}}^{2}\mathrm{ }=\mathrm{ }0\\ ⇒ {\mathrm{x}}^{2}\mathrm{ }=\mathrm{ }1\\ ⇒ \mathrm{x}\mathrm{ }=\mathrm{ }±1\end{array}$

Q.21

$\text{Evaluate: }\left|\begin{array}{l}\mathrm{x} \mathrm{x}+1\mathrm{ }\\ \mathrm{x}-1 \mathrm{x}\end{array}\right|.$

Ans

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

Q.22 The value of determinant will _______ if its rows and columns are interchanged.

Ans

remain unchanged

Q.23 The value of determinant will be _____, if its two rows or two columns are interchanged with each other.

Ans

negative

Q.24 If A is an invertivle square matrix, then write the matrix adj (AT) – (adj A)T.

Ans

If A is an invertible square matrix, then adj (AT) = (adj A)T

Q.25

$\mathrm{Without}\mathrm{expanding},\mathrm{prove}\mathrm{that} \left|\begin{array}{l}\mathrm{x} \mathrm{y} \mathrm{z} \\ \mathrm{y}+\mathrm{z} \mathrm{z}+\mathrm{x} \mathrm{x}+\mathrm{y}\\ 1 1 1\end{array}\right| = 0$

Ans

$\begin{array}{l}\mathrm{By}\mathrm{applying}{\mathrm{R}}_{2}\to {\mathrm{R}}_{2}+{\mathrm{R}}_{1},\mathrm{ }\mathrm{we}\mathrm{ }\mathrm{get}\\ \left|\begin{array}{l}\mathrm{x} \mathrm{y} \mathrm{z} \\ \mathrm{y}+\mathrm{z} \mathrm{z}+\mathrm{x} \mathrm{x}+\mathrm{y}\\ 1 1 1\end{array}\right|\mathrm{ }=\mathrm{ }\left|\begin{array}{l}\mathrm{x} \mathrm{y} \mathrm{z} \\ \mathrm{x}+\mathrm{y} \mathrm{x}+\mathrm{y} \mathrm{x}+\mathrm{y} +\mathrm{z}\\ 1 1 1\end{array}\right|\\ =\mathrm{ }\left|\begin{array}{l}\mathrm{x} \mathrm{y} \mathrm{z} \\ 1 1 1\\ 1 1 1\end{array}\right|\mathrm{ }\left(\mathrm{Taking}\mathrm{x}+\mathrm{y}+\mathrm{z}\mathrm{common}\mathrm{from}{\mathrm{R}}_{2}.\right)\\ =\mathrm{ }\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)\mathrm{ }×\mathrm{ }0\\ =\mathrm{ }0\end{array}$

Q.26 Let A be a 3 × 3 square matrix such that A(adj A) = 2I, where I is the identity matrix. Write the value of |adj A|.

Ans

∵ A (adj (A)) = |A|I

∴ A (adj A) = 2I ⇒ |A| = 2

Now, |adj A| = |A|n-1 ⇒ |adj A| = 23-1 = 4

Q.27 If A is a square matrix of order 3 such that adj (2A) = kadj (A), then write the value of k.

Ans

Since, adj (kA) = kn – 1adj(A), where n is the order
⇒ k = 4

Q.28

$\begin{array}{l}\mathrm{Without}\mathrm{expanding}\mathrm{evaluate}\mathrm{the}\mathrm{determinant}\\ \left|\begin{array}{l}41 1 5\\ 79 7 9\\ 29 5 3\end{array}\right|\end{array}$

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{\Delta } = \left|\begin{array}{l}41 1 5\\ 79 7 9\\ 29 5 3\end{array}\right|.\\ \mathrm{Then},\mathrm{applying}{\mathrm{c}}_{1}\to {\mathrm{c}}_{2}+\left(-8\right)\mathrm{ }{\mathrm{c}}_{3}, \mathrm{we}\mathrm{get}\\ \mathrm{ }\mathrm{\Delta } = \left|\begin{array}{l}1 1 5\\ 7 7 9\\ 5 5 3\end{array}\right|\mathrm{ }=\mathrm{ }0\mathrm{ }\left[\because \mathrm{ }{\mathrm{c}}_{1}\mathrm{ }\mathrm{and}\mathrm{ }{\mathrm{c}}_{2}\mathrm{ }\mathrm{are}\mathrm{identical}\right]\end{array}$

Q.29

$\text{If A = }\left[\begin{array}{l}1 2\\ 3 4\end{array}\right], \mathrm{verify}\mathrm{A}\left(\mathrm{adjA}\right)\mathrm{ }=\mathrm{ }\left|\mathrm{A}\right|\mathrm{I}.$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have}\left|\mathrm{A}\right|\mathrm{I}\mathrm{ }=\mathrm{ }-2 \left[\begin{array}{l}1 0\\ 0 1\end{array}\right]\\ =\mathrm{ }\left[\begin{array}{l}-2 0\\ 0 -2\end{array}\right] .\dots ..\mathrm{ }\left(1\right)\\ \mathrm{Cofactors}\mathrm{of}\mathrm{A}\\ {\mathrm{c}}_{11}\mathrm{ }=\mathrm{ }4 {\mathrm{c}}_{12}\mathrm{ }=\mathrm{ }-3\\ {\mathrm{c}}_{21}\mathrm{ }=\mathrm{ }-2 {\mathrm{c}}_{22}\mathrm{ }=\mathrm{ }1\\ ⇒\mathrm{ }\mathrm{adj}\mathrm{ }\mathrm{A}\mathrm{ }=\mathrm{ }{\mathrm{C}}^{\mathrm{T}} \\ =\mathrm{ }\left[\begin{array}{l}4 -2\\ -3 1\end{array}\right]\mathrm{ }\\ \mathrm{Now}\mathrm{A}\left(\mathrm{adjA}\right)\mathrm{ }=\mathrm{ }\left[\begin{array}{l}1 2\\ 3 4\end{array}\right]\mathrm{ }\left[\begin{array}{l}4 -2\\ -3 1\end{array}\right]\\ =\mathrm{ }\left[\begin{array}{l}1×4+2×\left(-3\right) 1×\left(-2\right)+2×1\mathrm{ }\\ 3×4+4×\left(-3\right) 3×\left(-2\right)+4×1\mathrm{ }\end{array}\right]\\ =\mathrm{ }\left[\begin{array}{l}4-6 -2+2\mathrm{ }\\ 12-12 -6+4\end{array}\right]\\ =\mathrm{ }\left[\begin{array}{l}-2 0\mathrm{ }\\ 0 -2\end{array}\right] .\dots \dots \left(2\right)\\ \mathrm{by}\left(1\right)\mathrm{and}\left(2\right)\\ \mathrm{A} \left(\mathrm{adjA}\right)\mathrm{ }=\mathrm{ }\left|\mathrm{A}\right|\mathrm{I}\end{array}$

Q.30

$\begin{array}{l}\mathrm{Using}\mathrm{properties}\mathrm{of}\mathrm{determinants},\mathrm{prove}\mathrm{that}\\ \left|\begin{array}{l}{\mathrm{b}}^{2+}{\mathrm{c}}^{2} \mathrm{ab} \mathrm{ac}\\ \mathrm{ba} {\mathrm{c}}^{2}+{\mathrm{a}}^{2} \mathrm{bc}\\ \mathrm{ca} \mathrm{cb} {\mathrm{a}}^{2}+{\mathrm{b}}^{2}\end{array}\right|\mathrm{ }=\mathrm{ }4{\mathrm{a}}^{2}{\mathrm{b}}^{2}{\mathrm{c}}^{2}.\end{array}$

Ans

$\begin{array}{l}\mathrm{Let} \mathrm{\Delta }\mathrm{ }\left|\begin{array}{l}{\mathrm{b}}^{2+}{\mathrm{c}}^{2} \mathrm{ab} \mathrm{ac}\\ \mathrm{ba} {\mathrm{c}}^{2}+{\mathrm{a}}^{2} \mathrm{bc}\\ \mathrm{ca} \mathrm{cb} {\mathrm{a}}^{2}+{\mathrm{b}}^{2}\end{array}\right|\mathrm{ }\\ \mathrm{Multiplying}{\mathrm{R}}_{1},{\mathrm{R}}_{2} \mathrm{and}{\mathrm{R}}_{3} \mathrm{by}\mathrm{a},\mathrm{b}\mathrm{and}\mathrm{c}\mathrm{respectively},\mathrm{we}\mathrm{get}\\ \mathrm{\Delta } = \frac{1}{\mathrm{abc}}\mathrm{ }\left|\begin{array}{l}\mathrm{a}\left({\mathrm{b}}^{2+}{\mathrm{c}}^{2}\right) {\mathrm{a}}^{2}\mathrm{b} {\mathrm{a}}^{2}\mathrm{c}\\ \mathrm{ }{\mathrm{b}}^{2}\mathrm{a} \mathrm{b}\left({\mathrm{c}}^{2}+{\mathrm{a}}^{2}\right) {\mathrm{b}}^{2}\mathrm{c}\\ {\mathrm{c}}^{2}\mathrm{a} {\mathrm{c}}^{2}\mathrm{b} \mathrm{c}\left({\mathrm{a}}^{2}+{\mathrm{b}}^{2}\right)\end{array}\right|\\ ⇒\mathrm{ }\mathrm{\Delta } = \frac{\mathrm{abc}}{\mathrm{abc}}\mathrm{ }\left|\begin{array}{l}{\mathrm{b}}^{2+}{\mathrm{c}}^{2} {\mathrm{a}}^{2} {\mathrm{a}}^{2}\\ {\mathrm{b}}^{2} {\mathrm{c}}^{2}+{\mathrm{a}}^{2} {\mathrm{b}}^{2}\\ {\mathrm{c}}^{2} {\mathrm{c}}^{2} {\mathrm{a}}^{2}+{\mathrm{b}}^{2}\end{array}\right|\mathrm{ }\left[\begin{array}{l}\mathrm{Taking}\mathrm{a},\mathrm{b}\mathrm{anc}\mathrm{c}\mathrm{common}\mathrm{from}\\ {\mathrm{c}}_{1},\mathrm{ }{\mathrm{c}}_{2}\mathrm{ }\mathrm{and}\mathrm{ }{\mathrm{c}}_{3} \mathrm{respectively}.\end{array}\right]\\ ⇒ \mathrm{\Delta } = \left|\begin{array}{l}2\left({\mathrm{b}}^{2+}{\mathrm{c}}^{2}\right) 2\left({\mathrm{a}}^{2}+{\mathrm{c}}^{2}\right) 2\mathrm{ }\left({\mathrm{a}}^{2}+{\mathrm{b}}^{2}\right)\\ {\mathrm{b}}^{2} {\mathrm{c}}^{2}+{\mathrm{a}}^{2} {\mathrm{b}}^{2}\\ {\mathrm{c}}^{2} {\mathrm{c}}^{2} {\mathrm{a}}^{2}+{\mathrm{b}}^{2}\end{array}\right|\mathrm{ }\left[\mathrm{Applying}{\mathrm{R}}_{1},\mathrm{ }{\mathrm{R}}_{2}\mathrm{ }\mathrm{and}\mathrm{ }{\mathrm{R}}_{3}\right]\\ ⇒ \mathrm{\Delta } = 2 \left|\begin{array}{l}\left({\mathrm{b}}^{2+}{\mathrm{c}}^{2}\right) \left({\mathrm{a}}^{2}+{\mathrm{c}}^{2}\right) \left({\mathrm{a}}^{2}+{\mathrm{b}}^{2}\right)\\ {\mathrm{b}}^{2} {\mathrm{c}}^{2}+{\mathrm{a}}^{2} {\mathrm{b}}^{2}\\ {\mathrm{c}}^{2} {\mathrm{c}}^{2} {\mathrm{a}}^{2}+{\mathrm{b}}^{2}\end{array}\right|\mathrm{ }\left[\mathrm{Taking}2\mathrm{common}\mathrm{from}{\mathrm{R}}_{1}\right]\\ ⇒ \mathrm{\Delta } = 2 \left|\begin{array}{l}\left({\mathrm{b}}^{2+}{\mathrm{c}}^{2}\right) \left({\mathrm{a}}^{2}+{\mathrm{c}}^{2}\right) \left({\mathrm{a}}^{2}+{\mathrm{b}}^{2}\right)\\ -{\mathrm{c}}^{2} 0 -{\mathrm{a}}^{2}\\ -{\mathrm{b}}^{2} -{\mathrm{a}}^{2} 0\end{array}\right|\mathrm{ }\left[\mathrm{Applying}{\mathrm{R}}_{2}\to {\mathrm{R}}_{2}-{\mathrm{R}}_{1}\mathrm{ }\mathrm{and} {\mathrm{R}}_{3}\to {\mathrm{R}}_{3}-{\mathrm{R}}_{1}\right]\\ ⇒ \mathrm{\Delta } = 2 \left|\begin{array}{l}0 {\mathrm{c}}^{2} {\mathrm{b}}^{2}\\ -{\mathrm{c}}^{2} 0 -{\mathrm{a}}^{2}\\ -{\mathrm{b}}^{2} -{\mathrm{a}}^{2} 0\end{array}\right|\mathrm{ }\left[\mathrm{Applying}{\mathrm{R}}_{1}\to {\mathrm{R}}_{1}+{\mathrm{R}}_{2}+{\mathrm{R}}_{3}\right]\\ ⇒ \mathrm{\Delta } = 2\left\{-{\mathrm{c}}^{2} \left|\begin{array}{l}-{\mathrm{c}}^{2} -{\mathrm{a}}^{2}\\ -{\mathrm{b}}^{2} 0\end{array}\right|+{\mathrm{b}}^{2\mathrm{ }}\left|\begin{array}{l}-{\mathrm{c}}^{2} 0\\ -{\mathrm{b}}^{2} -{\mathrm{a}}^{2}\end{array}\right|}\right\\mathrm{ }\\ ⇒ \mathrm{\Delta } = 2\left\{{\mathrm{a}}^{2}{\mathrm{b}}^{2}{\mathrm{c}}^{2}+{\mathrm{a}}^{2}{\mathrm{b}}^{2}{\mathrm{c}}^{2}}\right\\mathrm{ }=\mathrm{ }4{\mathrm{a}}^{2}{\mathrm{b}}^{2}{\mathrm{c}}^{2}\end{array}$

Q.31

$\begin{array}{l}\mathrm{If}\mathrm{f}\left(\mathrm{x}\right)\mathrm{ }=\mathrm{ }{\mathrm{ax}}^{2}+\mathrm{bx}+\mathrm{c}\mathrm{ }\mathrm{isaquadratic}\mathrm{function}\mathrm{such}\mathrm{that}\mathrm{f}\left(1\right)\mathrm{ }=\mathrm{ }8,\\ \mathrm{f}\left(2\right)\mathrm{ }=\mathrm{ }11\mathrm{and}\mathrm{f}\left(-3\right)\mathrm{ }=\mathrm{ }6, \mathrm{find}\mathrm{f}\left(\mathrm{x}\right)\mathrm{ }\mathrm{by}\mathrm{using}\mathrm{determinants}.\\ \mathrm{Also},\mathrm{find}\mathrm{f}\left(0\right).\end{array}$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have}\mathrm{f}\left(\mathrm{x}\right)\mathrm{ }={\mathrm{ax}}^{2}+\mathrm{bx}+\mathrm{c}\\ \mathrm{ }\therefore \mathrm{f}\left(1\right)\mathrm{ }=\mathrm{ }8\mathrm{ }⇒\mathrm{ }\mathrm{a}+\mathrm{b}+\mathrm{c}\mathrm{ }= 8\\ \mathrm{f}\left(2\right)\mathrm{ }=\mathrm{ }11\mathrm{ }⇒\mathrm{ }4\mathrm{a}+2\mathrm{b}+\mathrm{c}\mathrm{ }= 11\\ \mathrm{and} \mathrm{f}\left(-3\right)\mathrm{ }=\mathrm{ }6\mathrm{ }⇒\mathrm{ }9\mathrm{a}-3\mathrm{b}+\mathrm{c}\mathrm{ }= 6\mathrm{ }\\ \mathrm{Thus},\mathrm{we}\mathrm{obtain}\mathrm{the}\mathrm{following}\mathrm{system}\mathrm{of}\mathrm{equations}\\ \mathrm{a}+\mathrm{b}+\mathrm{c} = 8\\ 4\mathrm{a}+2\mathrm{b}+\mathrm{c} = 11\\ 9\mathrm{a} -3\mathrm{b}+\mathrm{c}= 6\\ \mathrm{Here},\\ \mathrm{D} = \left|\begin{array}{l}11 1\\ 42 1\\ 9-3 1\end{array}\right|\mathrm{ }=\mathrm{ }1\left(2+3\right)-\left(4-9\right)+1\left(12-18\right)\\ =\mathrm{ }5+5-30\mathrm{ }=\mathrm{ }-20\\ {\mathrm{D}}_{1} = \left|\begin{array}{l}81 1\\ 112 1\\ 6-3 1\end{array}\right|\mathrm{ }=\mathrm{ }8\left(2+3\right)-1\left(11-6\right)+1\left(33-12\right)\\ =\mathrm{ }40-5-45\mathrm{ }=\mathrm{ }-10\\ {\mathrm{D}}_{2} = \left|\begin{array}{l}18 1\\ 411 1\\ 96 1\end{array}\right|\mathrm{ }=\mathrm{ }1\left(11-6\right)-8\left(4-9\right)+1\left(24-99\right)\\ =\mathrm{ }5+40-75\mathrm{ }=\mathrm{ }-30\\ {\mathrm{D}}_{3} = \left|\begin{array}{l}11 8\\ 42 11\\ 9-3 6\end{array}\right|\mathrm{ }=\mathrm{ }1\left(12+33\right)-1\left(24-99\right)+8\left(-12-18\right)\\ =\mathrm{ }45+75-240\mathrm{ }=\mathrm{ }-120\\ \mathrm{a}\mathrm{ }=\mathrm{ }\frac{{\mathrm{D}}_{1}}{\mathrm{D}}\mathrm{ }=\mathrm{ }\frac{-10}{-20}\mathrm{ }=\mathrm{ }\frac{1}{2}\\ \mathrm{b}\mathrm{ }=\mathrm{ }\frac{{\mathrm{D}}_{2}}{\mathrm{D}}\mathrm{ }=\mathrm{ }\frac{-30}{-20}\mathrm{ }=\mathrm{ }\frac{3}{2}\\ \mathrm{c}\mathrm{ }=\mathrm{ }\frac{{\mathrm{D}}_{3}}{\mathrm{D}}\mathrm{ }=\mathrm{ }\frac{-120}{-20}\mathrm{ }=\mathrm{ }6\\ \mathrm{Hence},\mathrm{f}\left(\mathrm{x}\right)\mathrm{ }=\mathrm{ }\frac{1}{2}\mathrm{ }{\mathrm{x}}^{2}+\frac{3}{2}\mathrm{x}+6\\ ⇒ \mathrm{f}\left(0\right)\mathrm{ }=\mathrm{ }6.\end{array}$

Q.32

$\begin{array}{l}\mathrm{If} \mathrm{m}\in \mathrm{N}\mathrm{and}\mathrm{m}\ge 2,\mathrm{without}\mathrm{expanding},\mathrm{prove}\mathrm{that}\\ \left|\begin{array}{l}11 1\\ {\mathrm{m}}_{{\mathrm{c}}_{1}} \mathrm{m}{+1}_{{\mathrm{c}}_{1}} \mathrm{m}{+2}_{{\mathrm{c}}_{1}}\\ {\mathrm{m}}_{{\mathrm{c}}_{2}} \mathrm{m}{+1}_{{\mathrm{c}}_{2}} \mathrm{m}{+2}_{{\mathrm{c}}_{2}}\end{array}\right| =\mathrm{ }1\end{array}$

Ans

$\begin{array}{l}\mathrm{Let} \mathrm{\Delta }=\left|\begin{array}{l}11 1\\ {\mathrm{m}}_{{\mathrm{c}}_{1}} \mathrm{m}{+1}_{{\mathrm{c}}_{1}} \mathrm{m}{+2}_{{\mathrm{c}}_{1}}\\ {\mathrm{m}}_{{\mathrm{c}}_{2}} \mathrm{m}{+1}_{{\mathrm{c}}_{2}} \mathrm{m}{+2}_{{\mathrm{c}}_{2}}\end{array}\right|\mathrm{ }. \mathrm{Then} \\ \mathrm{\Delta }=\left|\begin{array}{l}11 1\\ {\mathrm{m}}_{{\mathrm{c}}_{1}} \mathrm{m}{+1}_{{\mathrm{c}}_{1}} \mathrm{m}{+1}_{{\mathrm{c}}_{0}}+\mathrm{m}+{1}_{{\mathrm{c}}_{1}}\\ {\mathrm{m}}_{{\mathrm{c}}_{2}} \mathrm{m}{+1}_{{\mathrm{c}}_{2}} \mathrm{m}{+1}_{{\mathrm{c}}_{1}}+\mathrm{m}{+1}_{{\mathrm{c}}_{2}}\end{array}\right| \left[\because \mathrm{ }{\mathrm{n}}_{{\mathrm{c}}_{\mathrm{t}}}+{\mathrm{n}}_{{\mathrm{c}}_{\mathrm{t}-1}}=\mathrm{ }\mathrm{n}+{1}_{{\mathrm{c}}_{\mathrm{r}}}\right]\\ \mathrm{\Delta }=\left|\begin{array}{l}11 1\\ {\mathrm{m}}_{{\mathrm{c}}_{1}} \mathrm{m}{+1}_{{\mathrm{c}}_{1}} \mathrm{m}{+1}_{{\mathrm{c}}_{0}} \\ {\mathrm{m}}_{{\mathrm{c}}_{2}} \mathrm{m}{+1}_{{\mathrm{c}}_{2}} \mathrm{m}{+1}_{{\mathrm{c}}_{1}}\end{array}\right| \left[\mathrm{Applying}{\mathrm{c}}_{3}\to \mathrm{ }{\mathrm{c}}_{3}-{\mathrm{c}}_{2}\right]\\ ⇒ \mathrm{\Delta }=\left|\begin{array}{l}11 1\\ {\mathrm{m}}_{{\mathrm{c}}_{1}} \mathrm{m}{+1}_{{\mathrm{c}}_{1}}\mathrm{ }+{\mathrm{m}}_{{\mathrm{c}}_{0}}+\mathrm{m}+{1}_{{\mathrm{c}}_{0}}\\ {\mathrm{m}}_{{\mathrm{c}}_{2}} {\mathrm{m}}_{{\mathrm{c}}_{1}}+ {\mathrm{m}}_{{\mathrm{c}}_{2}} \mathrm{m}{+1}_{{\mathrm{c}}_{1}}\end{array}\right| \left[\because \mathrm{ }{\mathrm{n}}_{{\mathrm{c}}_{\mathrm{r}}}+{\mathrm{n}}_{{\mathrm{c}}_{\mathrm{r}-1}}=\mathrm{ }\mathrm{n}+{1}_{{\mathrm{c}}_{\mathrm{r}}}\right]\\ ⇒ \mathrm{\Delta }=\left|\begin{array}{l}10 0\\ {\mathrm{m}}_{{\mathrm{c}}_{1}} \mathrm{m}{+1}_{{\mathrm{c}}_{0}}\mathrm{ }\mathrm{m}+{1}_{{\mathrm{c}}_{0}}\\ {\mathrm{m}}_{{\mathrm{c}}_{2}} {\mathrm{m}}_{{\mathrm{c}}_{1}} \mathrm{m}{+1}_{{\mathrm{c}}_{1}}\end{array}\right| \left[\mathrm{Applying}{\mathrm{c}}_{2}\to \mathrm{ }{\mathrm{c}}_{2}-{\mathrm{c}}_{1}\right]\\ ⇒ \mathrm{\Delta }=\left|\begin{array}{l}{\mathrm{m}}_{{\mathrm{c}}_{0}} \mathrm{m}{+1}_{{\mathrm{c}}_{0}}\mathrm{ }\\ {\mathrm{m}}_{{\mathrm{c}}_{1}} \mathrm{m}{+1}_{{\mathrm{c}}_{1}}\end{array}\right| \left[\mathrm{Expanding}\mathrm{along}{\mathrm{R}}_{1}\right]\\ ⇒ \mathrm{\Delta }=\left|\begin{array}{l}1 1\mathrm{ }\\ \mathrm{m} \mathrm{m}+1\end{array}\right|\mathrm{ }=\mathrm{ }\left(\mathrm{m}+1-\mathrm{m}\right)\mathrm{ }=\mathrm{ }1\end{array}$

Q.33

$\begin{array}{l}\mathrm{Solve}\mathrm{the}\mathrm{following}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{by}\mathrm{cramer}‘\mathrm{s}\mathrm{rule}:\\ \mathrm{ }\frac{\mathrm{2}}{\mathrm{x}}\mathrm{+}\frac{\mathrm{3}}{\mathrm{y}}\mathrm{+}\frac{\mathrm{10}}{\mathrm{z}} = 4, \frac{\mathrm{4}}{\mathrm{x}}\mathrm{+}\frac{\mathrm{6}}{\mathrm{y}}\mathrm{+}\frac{\mathrm{5}}{\mathrm{z}} = 1, \mathrm{and} \frac{\mathrm{6}}{\mathrm{x}}\mathrm{+}\frac{\mathrm{9}}{\mathrm{y}}\mathrm{+}\frac{\mathrm{20}}{\mathrm{z}} = 2.\end{array}$

Ans

$\begin{array}{l}\mathrm{Let}\frac{1}{\mathrm{x}}\mathrm{ }=\mathrm{ }\mathrm{\mu },\mathrm{ }\frac{1}{\mathrm{y}}\mathrm{ }=\mathrm{ }\mathrm{\nu }\mathrm{ }\mathrm{and}\frac{1}{\mathrm{z}}\mathrm{ }=\mathrm{ }\mathrm{\omega }. \mathrm{Then},\mathrm{the}\mathrm{above}\mathrm{system}\mathrm{of}\mathrm{equation}\mathrm{can}\mathrm{be}\\ \mathrm{written}\mathrm{as}\\ 2\mathrm{\mu }+3\mathrm{\nu }+10\mathrm{\omega }\mathrm{ }=\mathrm{ }4\\ 4\mathrm{\mu }-6\mathrm{\nu }+5\mathrm{\omega }\mathrm{ }=\mathrm{ }1\\ 6\mathrm{\mu }+9\mathrm{\nu }-20\mathrm{\omega }\mathrm{ }=\mathrm{ }2\\ \mathrm{Here},\mathrm{D}= \left|\begin{array}{l}23 10\\ 4-6 5\\ 69 -20\end{array}\right|\\ =\mathrm{ }2\left(120-45\right)-3\left(-80-30\right)+10\left(36+36\right)\\ ⇒ \mathrm{D}\mathrm{ }=\mathrm{ }150+330+720\mathrm{ }=\mathrm{ }1200\\ {\mathrm{D}}_{1}= \left|\begin{array}{l}43 10\\ 1-6 5\\ 29 -20\end{array}\right|\\ =\mathrm{ }4\left(120-45\right)-3\left(-20-10\right)+10\left(9+12\right)\\ \mathrm{ }⇒ {\mathrm{D}}_{1}\mathrm{ }=\mathrm{ }300+90+210\mathrm{ }=\mathrm{ }600\\ {\mathrm{D}}_{2}= \left|\begin{array}{l}24 10\\ 41 5\\ 62 -20\end{array}\right|\\ =\mathrm{ }2\left(-20-10\right)-4\left(-80-30\right)+10\left(8+6\right)\\ ⇒ {\mathrm{D}}_{2}\mathrm{ }=\mathrm{ }-60+440+20\mathrm{ }=\mathrm{ }400\\ {\mathrm{D}}_{3}= \left|\begin{array}{l}23 4\\ 4-6 1\\ 69 2\end{array}\right|\\ =\mathrm{ }2\left(-12-9\right)-3\left(8-6\right)+4\left(36+36\right)\\ ⇒ {\mathrm{D}}_{3}\mathrm{ }=\mathrm{ }-42-6+288=\mathrm{ }240\\ \therefore \mathrm{ }\mathrm{\mu }\mathrm{ }=\mathrm{ }\frac{{\mathrm{D}}_{1}}{\mathrm{D}}\mathrm{ }=\mathrm{ }\frac{600}{1200}\mathrm{ }=\mathrm{ }\frac{1}{2}\\ ⇒\mathrm{ }\frac{1}{\mathrm{x}}\mathrm{ }=\mathrm{ }\frac{1}{2}\\ ⇒\mathrm{ }\mathrm{x}\mathrm{ }=\mathrm{ }2\\ \mathrm{\nu }\mathrm{ }=\mathrm{ }\frac{{\mathrm{D}}_{2}}{\mathrm{D}}=\mathrm{ }\frac{400}{1200} =\mathrm{ }\frac{1}{3}\\ ⇒\mathrm{ }\frac{1}{\mathrm{y}}\mathrm{ }=\mathrm{ }\frac{1}{3}\\ ⇒\mathrm{ }\mathrm{y}=\mathrm{ }3,\\ \mathrm{ }\mathrm{\omega } =\mathrm{ }\frac{{\mathrm{D}}_{3}}{\mathrm{D}}\mathrm{ }=\mathrm{ }\frac{240}{1200}\mathrm{ }=\mathrm{ }\frac{1}{5}\\ ⇒ \frac{1}{\mathrm{z}}\mathrm{ }=\mathrm{ }\frac{1}{5}\\ \mathrm{Hence},\mathrm{x}= 2,\mathrm{y}= 3\mathrm{and}\mathrm{z}= 5.\end{array}$

Q.34

$\begin{array}{l}\mathrm{A}\mathrm{triangle}\mathrm{has}\mathrm{its}\mathrm{theree}\mathrm{sides}\mathrm{equal}\mathrm{to}\mathrm{a},\mathrm{b}\mathrm{and}\mathrm{c},\mathrm{If}\mathrm{the}\mathrm{coordinates}\\ \mathrm{of}\mathrm{its}\mathrm{vertices}\mathrm{are}\mathrm{A}\left({\mathrm{x}}_{1},\mathrm{ }{\mathrm{y}}_{1}\right),\mathrm{ }\mathrm{B}\left({\mathrm{x}}_{2},\mathrm{ }{\mathrm{y}}_{2}\right)\mathrm{ }\mathrm{and}\mathrm{ }\mathrm{C}\mathrm{ }\left({\mathrm{x}}_{3},{\mathrm{y}}_{3}\right), \mathrm{show}\mathrm{that}\\ {\left|\begin{array}{l}{\mathrm{x}}_{1}{\mathrm{y}}_{1}2\\ {\mathrm{x}}_{2}{\mathrm{y}}_{2}2\\ {\mathrm{x}}_{3}{\mathrm{y}}_{3}2\end{array}\right|}^{2}\mathrm{ }=\mathrm{ }\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)\right)\left(\mathrm{b}+\mathrm{c}-\mathrm{a}\right)\mathrm{ }\left(\mathrm{a}+\mathrm{b}-\mathrm{c}\right)\\ \mathrm{by}\mathrm{using}\mathrm{determinants}. \end{array}$

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{\Delta } \mathrm{be}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{triangle}\mathrm{ABC}.\mathrm{Then},\\ \mathrm{\Delta } = \frac{1}{2}\left|\begin{array}{l}{\mathrm{x}}_{1}{\mathrm{y}}_{1}1\\ {\mathrm{x}}_{2}{\mathrm{y}}_{2}1\\ {\mathrm{x}}_{3}{\mathrm{y}}_{3}1\end{array}\right|\\ ⇒ 2\mathrm{\Delta } = \left|\begin{array}{l}{\mathrm{x}}_{1}{\mathrm{y}}_{1}1\\ {\mathrm{x}}_{2}{\mathrm{y}}_{2}1\\ {\mathrm{x}}_{3}{\mathrm{y}}_{3}1\end{array}\right|\\ ⇒ 4\mathrm{\Delta } =2 \left|\begin{array}{l}{\mathrm{x}}_{1}{\mathrm{y}}_{1}1\\ {\mathrm{x}}_{2}{\mathrm{y}}_{2}1\\ {\mathrm{x}}_{3}{\mathrm{y}}_{3}1\end{array}\right|\mathrm{ }\left(\mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{by}2.\right)\\ ⇒ 4\mathrm{\Delta } = \left|\begin{array}{l}{\mathrm{x}}_{1}{\mathrm{y}}_{1}2\\ {\mathrm{x}}_{2}{\mathrm{y}}_{2}2\\ {\mathrm{x}}_{3}{\mathrm{y}}_{3}2\end{array}\right|\mathrm{ }\left(\mathrm{All}\mathrm{the}\mathrm{elements}\mathrm{of}{\mathrm{c}}_{3}\mathrm{are}\mathrm{multiplied}\mathrm{by}2.\right)\\ ⇒ 16{\mathrm{\Delta }}^{2} = {\left|\begin{array}{l}{\mathrm{x}}_{1}{\mathrm{y}}_{1}2\\ {\mathrm{x}}_{2}{\mathrm{y}}_{2}2\\ {\mathrm{x}}_{3}{\mathrm{y}}_{3}2\end{array}\right|}^{2}\mathrm{ }\left(\mathrm{On}\mathrm{squaring}\mathrm{both}\mathrm{sides}.\right) .\dots .\left(\mathrm{i}\right)\\ \mathrm{We}\mathrm{also}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{triangle}\mathrm{ABC}\mathrm{is}\mathrm{given}\mathrm{by}\\ \mathrm{\Delta }\mathrm{ }=\mathrm{ }\sqrt{\mathrm{s}\left(\mathrm{s}-\mathrm{a}\right)\left(\mathrm{s}-\mathrm{b}\right)\left(\mathrm{s}-\mathrm{c}\right)}, \mathrm{where} \mathrm{s}=\frac{1}{2}\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)\\ \mathrm{But},\mathrm{s} =\frac{1}{2}\mathrm{ }\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)\\ ⇒\mathrm{ }\mathrm{s}-\mathrm{a} =\frac{1}{2}\mathrm{ }\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)-\mathrm{c}\mathrm{ }\\ =\frac{1}{2}\mathrm{ }\left(\mathrm{b}+\mathrm{c}-\mathrm{a}\right)\\ \mathrm{Similarly},\mathrm{s}–\mathrm{c}=\frac{1}{2}\left(\mathrm{a}+\mathrm{b}-\mathrm{c}\right).\\ \therefore \mathrm{ }{\mathrm{\Delta }}^{2}\mathrm{ }=\mathrm{ }\frac{1}{2}\mathrm{ }\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right).\frac{1}{2}\mathrm{ }\left(\mathrm{b}+\mathrm{c}-\mathrm{a}\right).\frac{1}{2}\left(\mathrm{c}+\mathrm{a}-\mathrm{b}\right).\frac{1}{2}\left(\mathrm{a}+\mathrm{b}-\mathrm{c}\right).\\ ⇒\mathrm{ }16{\mathrm{\Delta }}^{2}\mathrm{ }=\mathrm{ }\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)\left(\mathrm{b}+\mathrm{c}-\mathrm{a}\right)\left(\mathrm{c}+\mathrm{a}-\mathrm{b}\right)\left(\mathrm{a}+\mathrm{b}-\mathrm{c}\right) .\dots .\mathrm{ }\left(\mathrm{ii}\right)\\ \mathrm{From}\mathrm{}\left(\mathrm{i}\right)\mathrm{ }\mathrm{and}\left(\mathrm{ii}\right)\mathrm{ }\mathrm{we}\mathrm{get}\\ \left|\begin{array}{l}{\mathrm{x}}_{1}{\mathrm{y}}_{1}2\\ {\mathrm{x}}_{2}{\mathrm{y}}_{2}2\\ {\mathrm{x}}_{3}{\mathrm{y}}_{3}2\end{array}\right| =\mathrm{ }\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)\left(\mathrm{b}+\mathrm{c}-\mathrm{a}\right)\left(\mathrm{c}+\mathrm{a}-\mathrm{b}\right)\left(\mathrm{a}+\mathrm{b}-\mathrm{c}\right)\end{array}$

Q.35

$\begin{array}{l}\mathrm{Using}\mathrm{cramer}‘\mathrm{s}\mathrm{rule},\mathrm{find}\mathrm{values}\mathrm{of}\mathrm{a}\mathrm{and}\mathrm{b}\mathrm{for}\mathrm{which}\\ \mathrm{the}\mathrm{system}\mathrm{of}\mathrm{equations}\\ 2\mathrm{x}+\mathrm{ay}+6\mathrm{z}= 8\\ \mathrm{a}+2\mathrm{y}+\mathrm{bz}\mathrm{ }=\mathrm{ }5\\ \mathrm{x}+\mathrm{ }\mathrm{y}+\mathrm{ }3\mathrm{z} =\mathrm{ }4\\ \mathrm{has}\\ \left(\mathrm{i}\right) \mathrm{a}\mathrm{unique}\mathrm{solution}\\ \left(\mathrm{ii}\right)\mathrm{infinitely}\mathrm{many}\mathrm{soutions}\\ \left(\mathrm{iii}\right)\mathrm{no}\mathrm{solutions}.\end{array}$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have},\\ \mathrm{D} = \left|\begin{array}{l}26 6\\ 12\mathrm{b}\\ 11 3\end{array}\right|\\ ⇒ \mathrm{D}= 2\left(6-\mathrm{b}\right)-\mathrm{a}\left(3-\mathrm{b}\right)+6\left(1-2\right)\\ ⇒ \mathrm{D}= 12- 2\mathrm{b}– 3\mathrm{b}+\mathrm{ab}– 6\\ ⇒ \mathrm{D}= 6 – 3\mathrm{a}– 2\mathrm{b}+\mathrm{ab}=\left(\mathrm{b}-3\right)\left(\mathrm{a}-2\right)\\ \mathrm{D}=\left|\begin{array}{l}8\mathrm{a}6\\ 52\mathrm{b}\\ 41 3\end{array}\right|\\ {\mathrm{D}}_{1}\mathrm{ }=\mathrm{ }8\left(6-\mathrm{b}\right)-\mathrm{a}\left(15-4\mathrm{b}\right)+6\left(5-8\right)\\ ⇒{\mathrm{D}}_{1}\mathrm{ }=\mathrm{ }48-8\mathrm{b}-15\mathrm{a}+4\mathrm{ab}-18\\ ⇒{\mathrm{D}}_{1}\mathrm{ }=\mathrm{ }30-15\mathrm{a}-8\mathrm{b}+4\mathrm{ab}=\mathrm{ }\left(\mathrm{a}-2\right)\left(4\mathrm{b}-15\right)\\ {\mathrm{D}}_{2} \left|\begin{array}{l}28 6\\ 15\mathrm{b}\\ 14 3\end{array}\right|\mathrm{ }\\ {\mathrm{D}}_{2}\mathrm{ }=\mathrm{ }2\left(15-4\mathrm{b}\right)-8\left(3-\mathrm{b}\right)+6\left(4-5\right)\\ ⇒{\mathrm{D}}_{2}\mathrm{ }=30-8\mathrm{b}-24+8\mathrm{b}-6=0\\ \mathrm{and}\\ {\mathrm{D}}_{3} \left|\begin{array}{l}2\mathrm{a}8\\ 12 5\\ 11 4\end{array}\right|\mathrm{ }\\ {\mathrm{D}}_{3}\mathrm{ }=2\left(8-5\right)-\mathrm{a}\left(4-50+8\right)\left(1-2\right)\\ ⇒\mathrm{ }{\mathrm{D}}_{3}\mathrm{ }=\mathrm{ }6+\mathrm{a}-8=\mathrm{a}-2\\ \left(\mathrm{i}\right)\mathrm{ }\mathrm{For}\mathrm{ }\mathrm{unique} \mathrm{solution},\mathrm{we}\mathrm{must}\mathrm{have}\\ \mathrm{D} \ne 0 ⇒\mathrm{ }\left(\mathrm{a}-2\right)\mathrm{ }\left(\mathrm{b}-3\right)\mathrm{ }\ne \mathrm{ }0\\ ⇒\mathrm{ }\mathrm{a}\ne 2,\mathrm{ }\mathrm{or}\mathrm{ }\mathrm{b}\mathrm{ }\ne \mathrm{ }3.\\ \left(\mathrm{ii}\right) \mathrm{for}\mathrm{infinitely}\mathrm{many}\mathrm{solutions},\mathrm{we}\mathrm{must}\mathrm{have}\\ \mathrm{D}\mathrm{ }=\mathrm{ }{\mathrm{D}}_{1}\mathrm{ }=\mathrm{ }{\mathrm{D}}_{2}\mathrm{ }=\mathrm{ }{\mathrm{D}}_{3}\mathrm{ }=\mathrm{ }0\\ ⇒\mathrm{ }\left(\mathrm{a}-2\right)\mathrm{ }\left(\mathrm{b}-3\right)\mathrm{ }=\mathrm{ }0,\mathrm{ }\left(\mathrm{a}-2\right)\left(4\mathrm{b}-15\right)\mathrm{ }=\mathrm{ }0\mathrm{ }\mathrm{and}\mathrm{ }\mathrm{a}\mathrm{ }-2\mathrm{ }=0.\\ ⇒ \mathrm{a}= 2\\ \mathrm{Putting}\mathrm{a}= 2\mathrm{in}\mathrm{hte}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations},\mathrm{we}\mathrm{get}\\ 2\mathrm{x}+ 2\mathrm{y}+ 6\mathrm{z} =8 \\ \mathrm{x}+ 2\mathrm{y}+ 6\mathrm{z} =5\\ \mathrm{x}+\mathrm{y}+ 3\mathrm{z} =4\\ \mathrm{x}+\mathrm{y}+\mathrm{z} =4\\ \mathrm{x}+ 2\mathrm{y}+\mathrm{bz} =5\\ \mathrm{Putting}\mathrm{z}=\mathrm{k},\mathrm{we}\mathrm{get}\\ \mathrm{x}+\mathrm{y} = 4 –\mathrm{k}\\ \mathrm{x}+ 2\mathrm{y} = 5 –\mathrm{bk}\\ \mathrm{Solving}\mathrm{these}\mathrm{two}\mathrm{equations},\mathrm{we}\mathrm{get}\\ \mathrm{x} = 3 - 3\mathrm{k} + \mathrm{bk},\mathrm{ }\mathrm{y}\mathrm{ }=\mathrm{ }1-\mathrm{bk}+\mathrm{k}\\ \mathrm{Thus},\mathrm{the}\mathrm{given}\mathrm{system}\mathrm{has}\mathrm{infinitely}\mathrm{many}\mathrm{solutions}\mathrm{given}\mathrm{by}\\ \mathrm{x}= 3-\mathrm{ }2\mathrm{k}+\mathrm{bk},\mathrm{ }\mathrm{y}\mathrm{ }=\mathrm{ }1-\mathrm{bk}+\mathrm{k},\mathrm{z}=\mathrm{ }\mathrm{k},\mathrm{ }\mathrm{where}\mathrm{ }\mathrm{k}\mathrm{ }\in \mathrm{ }\mathrm{R}.\\ \mathrm{Hence},\mathrm{the}\mathrm{system}\mathrm{has}\mathrm{infinitely}\mathrm{many}\mathrm{solutions}\mathrm{for}\mathrm{a}= 2.\\ \mathrm{ }\left(\mathrm{iii}\right) \mathrm{For}\mathrm{no}\mathrm{solution},\mathrm{we}\mathrm{must}\mathrm{have}\\ \mathrm{D} = 0\mathrm{and}\mathrm{at}\mathrm{least}\mathrm{one}\mathrm{of}{\mathrm{D}}_{1},\mathrm{ }{\mathrm{D}}_{2}\mathrm{ }\mathrm{and}\mathrm{ }\mathrm{is}\mathrm{non}\mathrm{zero}.\\ \mathrm{ }\mathrm{Clearly},\mathrm{for}\mathrm{b}= 3,\mathrm{we}\mathrm{have}\\ \mathrm{D}= 0\mathrm{and}{\mathrm{D}}_{3}\mathrm{ }\ne \mathrm{ }0.\\ \mathrm{Hence},\mathrm{the}\mathrm{system}\mathrm{has}\mathrm{no}\mathrm{solution}\mathrm{for}\mathrm{b}= 3.\end{array}$

## 1. Should I practice all the questions included in the Class 12 Mathematics chapter 4 notes?

The NCERT Solutions includes unlimited practice questions along with their solutions. These stepwise solutions teach students the shortcut techniques and easy way out of complex problems. Every question introduces different concepts of the chapter. By practising all the questions included in the Class 12 Mathematics chapter 4 notes, students will learn how to approach a specific question. They will gain a deeper understanding and insightful knowledge of all the topics covered in this chapter.

## 2. How do I prepare for the Class 12 Mathematics board exams?

Few preparation tips for the upcoming Class 12 board examination are mentioned below-

1.Know the exam details:

It is very important for students to keep themselves updated about the latest norms of the CBSE Syllabus, exam patterns, marking system, etc.