# CBSE Class 12 Maths Revision Notes Chapter 3

## Class 12 Mathematics Chapter 3 Notes

Mathematics is a compulsory and essential subject in the CBSE Curriculum. Most students find it difficult to tackle sums and problems in Mathematics. In this subject, strong basic knowledge is essential to easily understand the higher level and complex topics. The Class 12 Mathematics notes chapter 3 – matrices give a clear understanding of dimensional vector spaces. In this chapter, students will learn to represent quadratic equations to find their solutions.

Extramarks make boring textbook concepts interesting with the chapter-wise solutions. The Class 12 Mathematics chapter 3 notes help to practice and gain a deep understanding of the concepts. Students should also benefit from various practice tests, mock tests, and CBSE revision notes available for free on the Extramarks web portal.

## Key Topics Covered In Class 12 Mathematics Chapter 3 Notes

The main topics covered under class 12 chapter 3 Mathematics notes are as under

### Matrix:

It is an ordered array of functions. These functions or numbers are called elements or entries of a matrix.

Rows and Columns of a matrix:

The horizontal entries from the row of the matrix and the vertical entries from the column of the matrix.  For Eg. The matrix A has two rows and columns. R1 has elements {1, 2} and R2 has elements {0, 1}. Similarly, we can say that C1 has elements {1, 0} and C2 has elements {2, 1}.

A=

 1 2 0 1

Order of a matrix:

If A is a matrix with m rows and n columns, the order of matrix A is given as the product of a number of rows and columns, i.e., the order of a matrix is= m x n.

The matrix of order m x n has mn elements.

 a11 a12….. a1n a21 a22….. a2n am1 am2… amn

Where, 1 i m, i j n; i, j N.

### Types of Matrices:

1. Row Matrix:

The matrix A with only one row is called a Row Matrix. In general, it is given as A = aij1 X is a row matrix. The order of A is 1 x n.

E.g. [ 1     3   0]

2. Column Matrix:

A matrix having only one column is known as a Column Matrix. In general, it is given as A = aijm X 1is a column matrix. The order of A is m x 1

Eg.

 1 2

3. Square Matrix:

If the number of rows (m) and numbers of columns(n) are equal, then the matrix is known as a square matrix. So, m = n = a. Therefore, the order of the matrix is a x a, where a N

Eg.

 1 2 0 1

4. Diagonal Matrix:

A diagonal matrix has non-zero elements in the diagonal, and all other elements are zero.

For E.g.

 1 0 0 0 8 0 0 0 –7

5. Scalar Matrix:

It is an expansion of a diagonal matrix. The only difference is that all diagonal elements are equal.

For E.g.

 -7 0 0 0 -7 0 0 0 –7

6. Identity Matrix:

A matrix with all diagonal elements as 1 is known as an Identity matrix, for E.g.

 1 0 0 0 1 0 0 0 1

7. Zero Matrix:

A zero matrix, also called a null matrix, is a matrix where the elements are 0.

For E.g.

 0 0 0 0 0 0 0 0 0

### Equality of Matrices:

If A and B are two matrices, then,

(i) The order of the matrix A and B will be the same.

(ii) The corresponding elements are the same, i.e., aij= bij

Operations on Matrices:

Addition and Subtraction of two matrices are carried out only if the order of both matrices A and B are the same. 1 ≤ i ≤ m and 1 ≤ j ≤ n, we have,

If A= xijm X n and B= yijm X n then A + B = xij+ yijm X n

Subtraction:

If A= xijm X n and B= yijm X n then A – B = xijyijm X n

Properties:

a) Commutative Property:

If A = xij and B = yij are matrices of order m x n then A + B = B + A.

(b) Associative Property:

For A = xij, B = yij, C = zij are matrices of the same order m x n then, A + (B + C) = (A + B) + C.

If A = xij is a m x n matrix and O is zero matrices of the same order, then A + O = O + A = A. where O is the additive identity of the matrix.

Let A = xijm X n be a matrix, then there exists -A = -xijm X n such that A + (-A) = (-A) + A = O. The matrix (-A) is known as the additive inverse of A.

Note: A + B is only defined if A and B are of the same order.

### Negative of a Matrix:

When a matrix A is multiplied by -1, it gives the negative of matrix A.

(-1) A = – A

Multiplication of Matrices:

If A and B are two matrices, then their product is defined when the number of columns (m) of matrix A is equal to the number of rows (n) of B, i.e., m = n.

The row entries are multiplied by corresponding column entries. Therefore, the first-row entry is multiplied by the first column entry.

Properties of Matrix Multiplication:

1. Non-Commutative Law:

If A = xij and B = yij are matrices of order m x n, then A B B A.

1. Associative Law:

For A = xij, B = yij, C = zij are matrices of the same order m x n then, A (BC)=(AB) C.

1. Distributive Law:

For A = xij, B = yij, C = zij are matrices of same order m x n then,

A(B+C)=AB+AC

(A+B)C=AC+BC

Existence of Multiplicative Identity:

If A = xij is a m x n matrix and I is a matrix of the same order, then A I= I A = A. where I is the multiplicative identity of the matrix.

Multiplication of a Matrix by a scalar:

Multiplying a matrix A by a scalar k will result in each element of the matrix being multiplied by the scalar quantity to obtain a new matrix.

If k = 2 and P =

 1 2 3 0

Then k.P =

 2 8 6 0

Properties:

1.  If k is a scalar quantity, then k(A + B) = kA + kB.
2. If k and l are scalars, then (k + l)A = kA + lA.

Transpose of a Matrix:

The transpose of a matrix, denoted by P’ or PT, is obtained by changing the rows with columns.

For Eg. P=

 1 2 3 0

Then P’ or PT is given as =

 1 0 2 3

Properties:

1. (PT )T = P
2. (kP)T =kPT
3. (A + B)T =AT + BT
4. ABT =BT AT

### Special Types of Matrices:

1. Symmetric Matrices:

When the original square matrix is equal to its transpose, it is known as a symmetric matrix, i.e., P = PT

2. Skew-symmetric Matrices:

When the original square matrix is equal to the negative of its transpose, it is known as a skew-symmetric matrix, i.e., -P = PT.

#### Determinant of a Matrix:

A determinant of a square matrix A is the given as the Subtraction of the products of the elements within a matrix.

If A =

 p q r s

Then A = ps – qr

The class 12 Mathematics chapter 3 notes provided by Extramarks ensure that students are well prepared for their examination as all the concepts are clearly defined. In addition to the notes, students may access other study materials for free.

## Chapter 3 Mathematics class 12 notes: Exercises & Answer Solutions

The Class 12 Mathematics chapter 3 notes give detailed information about all concepts related to Matrices. Topics such as Algebra of matrices, multiplication, determinants, and applications of matrices are explained in detail in the Class 12 Mathematics chapter 3 notes.  Extramarks provide important definitions, formulas, properties, and theorems in a detailed and well-structured manner. With the help of the Extramarks Class 12 Mathematics chapter 3 notes, students will also gain knowledge about matric operations, rank and types of matrices.

Students may refer to various academic notes such as the Class 12 Mathematics chapter 3 notes, CBSE revision notes and important questions prepared by the experts at Extramarks. Access the Extramarks Exercises & Answer Solutions of this chapter by clicking on the links given below:

## NCERT Exemplar Class 12 Mathematics

The platform of Extramarks provides the NCERT Exemplar, which includes an extensive set of CBSE extra questions for students to practice. With the help of these notes, students can gain deeper knowledge and master the basic techniques of the subject. Solving CBSE sample papers will also help students to assess themselves based on their performance. Study materials such as NCERT Exemplar, Class 12 Mathematics chapter 3 notes and other NCERT books enable quick learning and teach several short-cut methods to tackle difficult questions.

With the help of the NCERT Exemplar and Class 12 Mathematics chapter 3 notes, students can perform well in the Class 12 CBSE board exams. Get access to the best academic materials from Extramarks.

### Key Features of Class 12 Mathematics chapter 3 notes

Key features of Extramarks class 12 Mathematics chapter 3 notes are as under.

• It is focused on providing a basic understanding of all concepts.
• It follows the CBSE Syllabus
• It is prepared by the experts at Extramarks.
• It provides authentic knowledge in an easy language.
• It aims to clarify all the doubts.
• It enhances time management skills.

Q.1

$\begin{array}{l}\mathrm{Obtain}\mathrm{the}\mathrm{inverse}\mathrm{of}\mathrm{the}\mathrm{matrix}\mathrm{A}=\left[\begin{array}{l}\mathrm{}104\\ \mathrm{}116\\ -30-10\end{array}\right]\\ \mathrm{using}\mathrm{row}\mathrm{operatins}.\end{array}$

Ans

$\begin{array}{l}\mathrm{Write}\mathrm{A}=\mathrm{IA},\mathrm{ie}.,\\ \mathrm{}\left[\begin{array}{l}\mathrm{}104\\ \mathrm{}116\\ -30-10\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{l}100\\ 010\\ 001\end{array}\right]\mathrm{A}\\ \mathrm{One}\mathrm{applying}{\mathrm{R}}_{2}\to {\mathrm{R}}_{2}-{\mathrm{R}}_{1}\mathrm{and}{\mathrm{R}}_{3}\to {\mathrm{R}}_{3}-{\mathrm{R}}_{1},\mathrm{we}\mathrm{get}\\ \mathrm{}\left[\begin{array}{l}104\\ 012\\ 002\end{array}\right]=\left[\begin{array}{l}100\\ -110\\ 301\end{array}\right]\mathrm{A}\\ \mathrm{On}\mathrm{applying}{\mathrm{R}}_{3}\to \mathrm{}\frac{1}{2}{\mathrm{R}}_{3},\\ \mathrm{}\left[\begin{array}{l}104\\ 012\\ 001\end{array}\right]=\left[\begin{array}{l}100\\ -110\\ \frac{3}{2}0\frac{1}{2}\end{array}\right]\mathrm{A}\\ \\ \mathrm{On}\mathrm{applying}{\mathrm{R}}_{1}\to {\mathrm{R}}_{1}-4{\mathrm{R}}_{3}\mathrm{and}{\mathrm{R}}_{2}\to {\mathrm{R}}_{2}-2{\mathrm{R}}_{3},\mathrm{we}\mathrm{get}\\ \left[\begin{array}{l}100\\ 010\\ 001\end{array}\right]=\left[\begin{array}{l}-50-2\\ -41-1\\ \frac{3}{2}0\frac{1}{2}\end{array}\right]\mathrm{A}\\ \mathrm{Hence},{\mathrm{A}}^{-1}=\mathrm{}\left[\begin{array}{l}-50-2\\ -41-1\\ \frac{3}{2}0\frac{1}{2}\end{array}\right]\end{array}$

Q.2 Show that for any square matrix A, A + A’ is a symmetric matrix and A – A’ is a skew symmetric matrix.

Ans

Let B = A + A’
B’ = (A + A’)’
= A’ + (A’)’ [as (A + B)’ = A’ + B’]
= A’ + A [ (A’)’ = A]
= B
∴ B = A + A’ is a symmetric matrix.

Now let C = A – A’
C’ = (A – A’)’
= A’ – (A’)’
= A’ – A
= -(A – A’)
∴ C = A – A’ is a skew symmetric matrix.

Q.3

$\text{Find the matrix x so that x}\left[\begin{array}{l}123\\ 456\end{array}\right]=\left[\begin{array}{l}-7-8-9\\ 246\end{array}\right]$

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{x}=\left[\begin{array}{l}\mathrm{a}\mathrm{b}\\ \mathrm{c}\mathrm{d}\end{array}\right]\\ \mathrm{}\therefore \mathrm{}\left[\begin{array}{l}\mathrm{a}\mathrm{b}\\ \mathrm{c}\mathrm{d}\end{array}\right]\left[\begin{array}{l}123\\ 456\end{array}\right]=\mathrm{}\left[\begin{array}{l}-7-8-9\\ 246\end{array}\right]\\ ⇒\mathrm{}\left[\begin{array}{l}\mathrm{a}×1+\mathrm{b}×4\mathrm{a}×2+\mathrm{}\mathrm{b}×5\mathrm{a}×3+\mathrm{b}×6\\ \mathrm{c}×1+\mathrm{d}×4\mathrm{c}×2+\mathrm{}\mathrm{d}×5\mathrm{c}×3+\mathrm{d}×6\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{l}-7-8-9\\ 246\end{array}\right]\\ ⇒\left[\begin{array}{l}\mathrm{a}+4\mathrm{b}2\mathrm{a}+\mathrm{}5\mathrm{b}3\mathrm{a}+6\mathrm{b}\\ \mathrm{c}+4\mathrm{b}2\mathrm{c}+\mathrm{}5\mathrm{b}3\mathrm{c}+6\mathrm{b}\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{l}-7-8-9\\ 246\end{array}\right]\\ \mathrm{By}\mathrm{using}\mathrm{equality}\mathrm{and}\mathrm{solving}\\ \begin{array}{l}\mathrm{a}+ 4\mathrm{b}= – 7\\ 2\mathrm{a}+ 5\mathrm{b}= -8\end{array}}.\dots \dots \left(1\right)\begin{array}{l}\mathrm{c}+ 4\mathrm{d}= 2\\ 2\mathrm{c}+ 5\mathrm{d}= 4\end{array}}.\dots \dots \left(2\right)\\ \mathrm{Now},\mathrm{by}\left(1\right)\mathrm{}\mathrm{a}=1\mathrm{b}=-2\\ \mathrm{by}\mathrm{}\left(2\right)\mathrm{}\mathrm{c}\mathrm{}=\mathrm{}2\mathrm{d}\mathrm{}=\mathrm{}0\\ \mathrm{Hence}\mathrm{x}=\left[\begin{array}{l}1-2\\ 20\end{array}\right]\end{array}$

Q.4

$\text{If A =}\left[\begin{array}{l}3-4\\ 1-1\end{array}\right],\mathrm{}\mathrm{then}\mathrm{prove}\mathrm{that}{\mathrm{A}}^{\mathrm{n}}\mathrm{}=\mathrm{}\left[\begin{array}{l}1+2\mathrm{n}-4\mathrm{n}\\ \mathrm{n}1-2\mathrm{n}\end{array}\right].$

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{P}\left(\mathrm{n}\right)\mathrm{be}\mathrm{the}\mathrm{statement}.\\ {\mathrm{A}}^{\mathrm{n}}=\mathrm{}\left[\begin{array}{l}1+2\mathrm{n}-4\mathrm{n}\\ \mathrm{n}1-2\mathrm{n}\end{array}\right]\\ \mathrm{For}\mathrm{n}= 1,{\mathrm{A}}^{1}\mathrm{}=\mathrm{A}\mathrm{}=\mathrm{}\left[\begin{array}{l}3-4\\ 1-1\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{l}1+2.1-4.1\\ 11-2.1\end{array}\right]\\ ⇒\mathrm{P}\left(1\right)\mathrm{is}\mathrm{true}.\\ \mathrm{Let}\mathrm{P}\left(\mathrm{m}\right)\mathrm{be}\mathrm{true},\mathrm{i}.\mathrm{e}.,\\ {\mathrm{A}}^{\mathrm{m}}=\mathrm{}\left[\begin{array}{l}1+2\mathrm{m}-4\mathrm{m}\\ \mathrm{m}1-2\mathrm{m}\end{array}\right]\\ \mathrm{Now},\mathrm{}{\mathrm{A}}^{\mathrm{m}+1}\mathrm{}=\mathrm{}{\mathrm{AA}}^{\mathrm{m}}=\mathrm{}\left[\begin{array}{l}3-4\\ 1-1\end{array}\right]\left[\begin{array}{l}1+2\mathrm{m}-4\mathrm{m}\\ \mathrm{m}1-2\mathrm{m}\end{array}\right]\\ =\mathrm{}\left[\begin{array}{l}3+2\mathrm{m}-4\mathrm{m}-4\\ 1+\mathrm{m}-2\mathrm{m}-1\end{array}\right]\\ =\mathrm{}\left[\begin{array}{l}1+2\left(\mathrm{m}+1\right)-4\left(\mathrm{m}+1\right)\\ \mathrm{m}+11\mathrm{}-2\mathrm{m}-1\end{array}\right]\\ ⇒\mathrm{}\mathrm{P}\mathrm{}\left(\mathrm{m}+1\right)\mathrm{}\mathrm{is}\mathrm{ture}.\\ \mathrm{Hence}\mathrm{by}\mathrm{mathematical}\mathrm{induction},\mathrm{P}\left(\mathrm{n}\right)\mathrm{}\mathrm{is}\mathrm{}\mathrm{true}\mathrm{}\mathrm{for}\mathrm{}\mathrm{all}\end{array}$

Q.5 If A, B are symmetric matrices of same order, then show that AB – BA is a skew- symmetric matrix.

Ans

Here, (AB – BA)’ = (AB)’ – (BA)’
= B’A’ – A’B’
= –(A’B’ – B’A’)
= –(AB – BA) [ A’ = A, B’ = B]
⇒ (AB – BA)’ = –(AB – BA)
Hence, AB – BA is a skew-symmetric matrix.

Q.6

$\begin{array}{l}\mathrm{Express}\mathrm{A}=\left[\begin{array}{l}33-1\\ -2-21\\ -4-52\end{array}\right]\mathrm{as}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{a}\mathrm{symmetric}\\ \mathrm{and}\mathrm{a}\mathrm{skew}–\mathrm{symmetric}\mathrm{matrix}.\end{array}$

Ans

$\begin{array}{l}\mathrm{Here}\mathrm{A}‘ =\left[\begin{array}{l}3-2-4\\ 3-2-5\\ -112\end{array}\right]\\ \mathrm{Let}\mathrm{P}=\frac{1}{2}\left(\mathrm{A}+\mathrm{A}‘\right)\\ \therefore \mathrm{}\frac{1}{2}\mathrm{}\left(\mathrm{A}+\mathrm{A}‘\right)\mathrm{}=\mathrm{}\frac{1}{2}\mathrm{}\left[\begin{array}{l}33-1\\ -2-21\\ -4-52\end{array}\right]+\left[\begin{array}{l}3-2-4\\ 3-2-5\\ -112\end{array}\right]\\ =\mathrm{}\frac{1}{2}\mathrm{}\left[\begin{array}{l}61-5\\ 1-4-\mathrm{}4\\ -5-44\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{l}3\frac{1}{2}\frac{-5}{2}\\ \frac{1}{2}-2-2\\ \frac{-5}{2}-22\end{array}\right]\\ \mathrm{Now}\mathrm{P}‘ =\left[\begin{array}{l}3\frac{1}{2}\frac{-5}{2}\\ \frac{1}{2}-2-2\\ \frac{-5}{2}-22\end{array}\right]\mathrm{}=\mathrm{}\mathrm{P}\\ \mathrm{Thus}\mathrm{P}=\frac{1}{2}\mathrm{}\left[\mathrm{A}-\mathrm{A}‘\right]\\ \frac{1}{2}\left[\mathrm{A}-\mathrm{A}‘\right]\mathrm{}=\mathrm{}\frac{1}{2}\mathrm{}\left[\left[\begin{array}{l}33-1\\ -2-21\\ -4-52\end{array}\right]-\left[\begin{array}{l}\mathrm{}3-2-4\\ \mathrm{}3-2-5\\ -112\end{array}\right]\right]\\ =\mathrm{}\frac{1}{2}\mathrm{}\left[\left[\begin{array}{l}053\\ -206\\ -3-60\end{array}\right]=\left[\begin{array}{l}\mathrm{}0\frac{5}{2}\frac{3}{2}\\ -\frac{5}{2}-30\\ -\frac{3}{2}-\mathrm{}30\end{array}\right]\right]\\ \mathrm{Then}\mathrm{Q}‘ =\left[\begin{array}{l}0-\frac{5}{2}-\frac{3}{2}\\ \frac{5}{2}0-3\\ \frac{3}{2}30\end{array}\right]\mathrm{}=\mathrm{}-\mathrm{Q}\\ \mathrm{Thus}\mathrm{Q}=\frac{1}{2}\mathrm{}\left(\mathrm{A}-\mathrm{A}‘\right)\mathrm{is}\mathrm{a}\mathrm{skew}\mathrm{symmetric}\mathrm{matrix}.\\ \mathrm{Now},\mathrm{P}+\mathrm{Q}=\left[\begin{array}{l}3\frac{1}{2}\frac{-5}{2}\\ \frac{1}{2}-2-2\\ \frac{-5}{2}-22\end{array}\right]+\left[\begin{array}{l}\mathrm{}0\frac{5}{2}\frac{3}{2}\\ -\frac{5}{2}03\\ -\frac{3}{2}-\mathrm{}30\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{l}33-1\\ -2-21\\ -4-52\end{array}\right]\mathrm{}=\mathrm{}\mathrm{A}\end{array}$

Q.7

$\text{If A =}\left[\begin{array}{l}-2\\ 4\\ 5\end{array}\right],\mathrm{}\mathrm{B}=\left[13-6\right],\mathrm{}\mathrm{verify}\mathrm{}\mathrm{that}\mathrm{}\left(\mathrm{AB}\right)‘\mathrm{}=\mathrm{}\mathrm{B}‘\mathrm{A}‘.$

Ans

$\begin{array}{l}\mathrm{Here}\mathrm{A}=\left[\begin{array}{l}-2\\ 4\\ 5\end{array}\right],\mathrm{}\mathrm{B}=\left[13-6\right]\\ \mathrm{AB}=\left[\begin{array}{l}-2\\ 4\\ 5\end{array}\right]\left[13-6\right]\mathrm{}=\mathrm{}\left[\begin{array}{l}-2-612\\ \mathrm{}412-24\\ \mathrm{}515-30\end{array}\right]\\ \left(\mathrm{AB}‘\right)\mathrm{}=\mathrm{}\left[\begin{array}{l}-245\\ -61215\\ 12-24-30\end{array}\right]\\ \mathrm{Now},\mathrm{A}‘\mathrm{}=\left[-245\right],\mathrm{B}‘\mathrm{}=\mathrm{}\left[\begin{array}{l}1\\ 3\\ -6\end{array}\right]\\ \mathrm{B}‘\mathrm{A}‘\mathrm{}=\mathrm{}\left[\begin{array}{l}1\\ 3\\ -6\end{array}\right]\mathrm{}\left[-245\right]\mathrm{}=\mathrm{}\left[\begin{array}{l}-245\\ -61215\\ 12-24-30\end{array}\right]=\mathrm{}\left(\mathrm{AB}\right)‘\\ \mathrm{Hence}\left(\mathrm{AB}\right)‘\mathrm{}=\mathrm{}\mathrm{B}‘\mathrm{A}‘\end{array}$

Q.8

$\text{If A =}\left[\begin{array}{l}10\\ -17\end{array}\right]\mathrm{and}\mathrm{I}=\left[\begin{array}{l}10\\ 01\end{array}\right],\mathrm{then}\mathrm{find}\mathrm{k}\mathrm{so}\mathrm{that}{\mathrm{A}}^{2}=8\mathrm{A}+\mathrm{kI}.$

Ans

$\begin{array}{l}{\mathrm{A}}^{2}=\mathrm{AA}=\left[\begin{array}{l}10\\ -17\end{array}\right]\mathrm{}\left[\begin{array}{l}10\\ -17\end{array}\right]\\ =\mathrm{}\left[\begin{array}{l}\mathrm{}1+00\mathrm{}+\mathrm{}0\\ -1-70+49\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{l}10\\ -849\end{array}\right]\\ 8\mathrm{A}+\mathrm{KI}= 8\left[\begin{array}{l}10\\ -17\end{array}\right]\mathrm{}+\mathrm{K}\left[\begin{array}{l}10\\ 01\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{l}8+\mathrm{k}0\\ -856+\mathrm{k}\end{array}\right]\mathrm{}\\ \mathrm{Given}{\mathrm{A}}^{2}=8\mathrm{A}+\mathrm{kI}\\ ⇒\mathrm{}\left[\begin{array}{l}10\\ -849\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{l}8+\mathrm{k}0\\ -856+\mathrm{k}\end{array}\right]\\ ⇒\mathrm{}8+\mathrm{k}=1\mathrm{and}56+\mathrm{k}=49\\ ⇒\mathrm{k}=-7\end{array}$

Q.9

$\mathrm{Find}\mathrm{x}\mathrm{and}\mathrm{y}\mathrm{if}\mathrm{x}+\mathrm{y}=\left[\begin{array}{l}52\\ 09\end{array}\right]\mathrm{and}\mathrm{x}–\mathrm{y}=\left[\begin{array}{l}36\\ 0\mathrm{}-1\end{array}\right]$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have}\\ \left(\mathrm{x}+\mathrm{y}\right)+\left(\mathrm{x}-\mathrm{y}\right)\mathrm{}=\mathrm{}\left[\begin{array}{l}52\\ 09\end{array}\right]\mathrm{}+\mathrm{}\left[\begin{array}{l}36\\ 0\mathrm{}-1\end{array}\right]\\ 2\mathrm{x}\mathrm{}=\mathrm{}\left[\begin{array}{l}5+36+2\\ 0\mathrm{}+09-1\end{array}\right]=\mathrm{}\left[\begin{array}{l}88\\ 08\end{array}\right]\\ ⇒\mathrm{x}\mathrm{}=\mathrm{}\frac{1}{2}\left[\begin{array}{l}88\\ 08\end{array}\right]\\ ⇒\mathrm{x}\mathrm{}=\left[\begin{array}{l}44\\ 04\end{array}\right]\\ \mathrm{Now}\left(\mathrm{x}+\mathrm{y}\right)\left(\mathrm{x}-\mathrm{y}\right)\mathrm{}=\mathrm{}\left[\begin{array}{l}52\\ 09\end{array}\right]-\left[\begin{array}{l}36\\ 01\end{array}\right]\\ 2\mathrm{y}\mathrm{}=\mathrm{}\left[\begin{array}{l}5\mathrm{}-32-6\\ 0\mathrm{}-09-1\end{array}\right]\\ ⇒\mathrm{y}\mathrm{}=\frac{1}{2}\mathrm{}\left[\begin{array}{l}2-4\\ 08\end{array}\right]\\ ⇒\mathrm{y}\mathrm{}=\left[\begin{array}{l}1-2\\ 04\end{array}\right]\\ \mathrm{Hence}\mathrm{x}=\left[\begin{array}{l}44\\ 04\end{array}\right]\mathrm{y}\mathrm{}=\frac{1}{2}\mathrm{}\left[\begin{array}{l}1-2\\ 04\end{array}\right]\end{array}$

Q.10

$\begin{array}{l}\text{If}\mathrm{A}=\left[\begin{array}{cc}\begin{array}{c}8\\ 4\\ 3\end{array}& \begin{array}{c}0\\ -2\\ 6\end{array}\end{array}\right]\text{and}\mathrm{B}=\left[\begin{array}{cc}\begin{array}{c}2\\ 4\\ -5\end{array}& \begin{array}{c}-2\\ 2\\ 1\end{array}\end{array}\right],\text{}\mathrm{X}\\ \text{such that}\mathrm{A}+\mathrm{X}=\mathrm{B}.\end{array}$

Ans

$\begin{array}{l}\text{We have}\mathrm{A}+\mathrm{X}=\mathrm{B}\\ ⇒\left[\begin{array}{cc}\begin{array}{c}8\\ 4\\ 3\end{array}& \begin{array}{c}0\\ -2\\ 6\end{array}\end{array}\right]+\mathrm{X}=\left[\begin{array}{cc}\begin{array}{c}2\\ 4\\ -5\end{array}& \begin{array}{c}-2\\ 2\\ 1\end{array}\end{array}\right]\\ ⇒\mathrm{X}=\left[\begin{array}{cc}\begin{array}{c}\text{2}\\ \text{4}\\ -\text{5}\end{array}& \begin{array}{c}-\text{2}\\ \text{2}\\ \text{1}\end{array}\end{array}\right]-\left[\begin{array}{cc}\begin{array}{c}\text{8}\\ \text{4}\\ \text{3}\end{array}& \begin{array}{c}\text{0}\\ -2\\ 6\end{array}\end{array}\right]\\ \text{}=\left[\begin{array}{cc}\begin{array}{c}\text{2}-8\\ 4-4\\ -5-3\end{array}& \begin{array}{c}-2-0\\ 2+2\\ 1-6\end{array}\end{array}\right]\\ \text{\hspace{0.17em}}=\left[\begin{array}{cc}\begin{array}{c}-6\\ 0\\ -8\end{array}& \begin{array}{c}-2\\ 4\\ -5\end{array}\end{array}\right]\end{array}$

Q.11

$\mathrm{Given}\mathrm{A}=\left[\begin{array}{l}5-10\\ 734\end{array}\right]\mathrm{and}\mathrm{B}=\left[\begin{array}{l}-320\\ -562\end{array}\right].\mathrm{Find}\mathrm{A}+\mathrm{B},\mathrm{A}–\mathrm{B}.$

Ans

$\begin{array}{l}\mathrm{Here},\mathrm{A}+\mathrm{B}=\left[\begin{array}{l}5-10\\ 734\end{array}\right]+\mathrm{}\left[\begin{array}{l}-320\\ -562\end{array}\right]\\ =\mathrm{}\left[\begin{array}{l}210\\ 296\end{array}\right]\\ \mathrm{A}-\mathrm{B}\mathrm{}=\mathrm{}\left[\begin{array}{l}5-10\\ 734\end{array}\right]-\mathrm{}\left[\begin{array}{l}-320\\ -562\end{array}\right]\\ =\mathrm{}\left[\begin{array}{l}5+3-1-20\mathrm{}-0\\ 7\mathrm{}+33-64-2\end{array}\right]\\ =\mathrm{}\left[\begin{array}{l}8-30\\ 12-32\end{array}\right]\end{array}$

Q.12

$\mathrm{If}\left[\begin{array}{l}\mathrm{x}+\mathrm{y}4\\ 7\mathrm{xy}\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{l}64\\ 78\end{array}\right],\mathrm{find}\mathrm{the}\mathrm{values}\mathrm{of}\mathrm{x}\mathrm{and}\mathrm{y}.$

Ans

$\begin{array}{l}\mathrm{x}+\mathrm{y}\mathrm{}=6\\ \mathrm{xy}=8\mathrm{}⇒\mathrm{}\mathrm{y}\mathrm{}=\mathrm{}\frac{8}{\mathrm{x}}\\ \mathrm{so},\mathrm{x}+\frac{8}{\mathrm{x}}\mathrm{}=\mathrm{}6\\ ⇒\mathrm{}{\mathrm{x}}^{2}-6\mathrm{x}+6=0\\ ⇒\mathrm{}\left(\mathrm{x}-2\right)\left(\mathrm{x}-4\right)\mathrm{}=\mathrm{}4\\ \mathrm{when}\mathrm{x}= 2,\mathrm{y}= 4\\ \mathrm{or}\mathrm{x}= 4,\mathrm{y}= 2\end{array}$

Q.13

$\mathrm{Construct}\mathrm{a}22\mathrm{matrix}\mathrm{A}=\left[{\mathrm{a}}_{\mathrm{ij}}\right]\mathrm{whose}\mathrm{elements}\mathrm{are}\mathrm{given}\mathrm{by}{\mathrm{a}}_{\mathrm{ij}}\mathrm{}=\mathrm{}\frac{{\left(\mathrm{i}+\mathrm{j}\right)}^{2}}{2}$

Ans

$\begin{array}{l}\mathrm{Here}\mathrm{A}=\left[\begin{array}{l}{\mathrm{a}}_{11}{\mathrm{a}}_{12}\\ {\mathrm{a}}_{21}{\mathrm{a}}_{22}\end{array}\right]\\ {\mathrm{a}}_{11}\mathrm{}=\mathrm{}\frac{{\left(1+1\right)}^{2}}{2}\mathrm{}=\mathrm{}\frac{{2}^{2}}{2}\mathrm{}=\mathrm{}\frac{4}{2}\mathrm{}=\mathrm{}2{\mathrm{a}}_{12}\mathrm{}=\mathrm{}\frac{{\left(1+2\right)}^{2}}{2}\mathrm{}=\mathrm{}\frac{{3}^{2}}{2}\mathrm{}=\mathrm{}\frac{9}{2}\mathrm{}\\ {\mathrm{a}}_{21}\mathrm{}=\mathrm{}\frac{{\left(2+1\right)}^{2}}{2}\mathrm{}=\mathrm{}\frac{{3}^{2}}{2}\mathrm{}=\mathrm{}\frac{9}{2}{\mathrm{a}}_{22}=\mathrm{}\frac{{\left(2+2\right)}^{2}}{2}\mathrm{}=\mathrm{}\frac{16}{2}\mathrm{}=\mathrm{}8\\ \mathrm{Hence}\mathrm{A}=\left[\begin{array}{l}2\frac{9}{2}\\ \frac{9}{2}8\end{array}\right]\end{array}$

Q.14

$\begin{array}{l}\mathrm{Consider}\mathrm{the}\mathrm{matrix}\\ \mathrm{A}=\left[\begin{array}{l}25\sqrt{7}3\\ 356\frac{5}{3}-1\\ \sqrt{3}160\end{array}\right]\end{array}$

Ans

(i) Order of the matrix = 3 4

(ii) a13 = 7, a21 = 35

a23 = 5/3, a34 = 0

Q.15

$\begin{array}{l}\mathrm{If}\mathrm{A}=\left[\begin{array}{l}31\\ 75\end{array}\right]\mathrm{},\mathrm{find}\mathrm{x}\mathrm{and}\mathrm{y}\mathrm{such}\mathrm{that}{\mathrm{A}}^{\mathrm{2}}+\mathrm{XI}=\mathrm{YA}.\\ \mathrm{Hence}\mathrm{find}{\mathrm{A}}^{-1}.\end{array}$

Ans

$\begin{array}{l}\mathrm{Given}\mathrm{A}=\left[\begin{array}{l}31\\ 75\end{array}\right]\mathrm{}\\ \therefore \mathrm{}{\mathrm{A}}^{2}=\mathrm{}\mathrm{A}.\mathrm{A}\mathrm{}=+\left[\begin{array}{l}31\\ 75\end{array}\right]\mathrm{}\left[\begin{array}{l}31\\ 75\end{array}\right]\mathrm{}\\ ⇒\mathrm{}{\mathrm{A}}^{2}\mathrm{}=\mathrm{}\left[\begin{array}{l}9+7 3+5\\ 21+357+25\end{array}\right]\mathrm{}\\ ⇒\mathrm{}{\mathrm{A}}^{2}\mathrm{}=\mathrm{}\left[\begin{array}{l}168\\ 5632\end{array}\right].\dots ..\left(1\right)\\ \mathrm{and}\mathrm{xI}=\mathrm{x}\left[\begin{array}{l}10\\ 01\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{l}\mathrm{x}0\\ 0\mathrm{x}\end{array}\right]\\ \mathrm{RHS}=\mathrm{yA}=\mathrm{y}\left[\begin{array}{l}31\\ 75\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{l}3\mathrm{y}\mathrm{y}\\ 7\mathrm{y}5\mathrm{y}\end{array}\right].\dots ..\left(2\right)\\ \mathrm{LHS}={\mathrm{A}}^{2}+\mathrm{xI}=\left[\begin{array}{l}168\\ 5632\end{array}\right]\mathrm{}+\mathrm{}\left[\begin{array}{l}\mathrm{x}0\\ 0\mathrm{x}\end{array}\right]\\ \mathrm{By}\mathrm{using}\mathrm{equality}\mathrm{of}\mathrm{matrices}\\ 16+\mathrm{x}= 3\mathrm{y}\\ ⇒56= 7\mathrm{y}\\ ⇒\mathrm{y}= 8\\ \mathrm{And},\mathrm{x}= 8\\ \mathrm{Substituting}\mathrm{the}\mathrm{values}\mathrm{of}\mathrm{x}\mathrm{and}\mathrm{y},\mathrm{we}\mathrm{get}\\ {\mathrm{A}}^{2}+8\mathrm{I}\mathrm{}=\mathrm{}8\mathrm{A}\\ ⇒8\mathrm{I}\mathrm{}=\mathrm{}8\mathrm{A}-{\mathrm{A}}^{2}\\ ⇒\mathrm{I}\mathrm{}=\mathrm{}\frac{1}{8}\left[8\mathrm{A}-{\mathrm{A}}^{2}\right]\\ ⇒{\mathrm{IA}}^{-1}=\frac{1}{8}\left[8{\mathrm{AA}}^{-1}-\mathrm{A}\left({\mathrm{AA}}^{-1}\right)\right]\\ ⇒{\mathrm{IA}}^{-1}=\frac{1}{8}\left[8\mathrm{I}-\mathrm{AI}\right]\\ ⇒{\mathrm{IA}}^{-1}=\frac{1}{8}\left[8{\mathrm{AA}}^{-1}-\mathrm{A}\left({\mathrm{AA}}^{-1}\right)\right]\\ ⇒{\mathrm{IA}}^{-1}=\frac{1}{8}\left[8\mathrm{I}\mathrm{}-\mathrm{AI}\right]\\ ⇒=\frac{1}{8}\left[8\mathrm{I}-\mathrm{A}\right]\\ ⇒=\frac{1}{8}\left[\left[\begin{array}{l}80\\ 08\end{array}\right]-\left[\begin{array}{l}31\\ 75\end{array}\right]\right]\\ ⇒=\frac{1}{8}\left[\begin{array}{l}5-1\\ –73\end{array}\right]\\ ⇒\mathrm{x}= 8,\mathrm{y}= 8\mathrm{and}{\mathrm{A}}^{-1}\mathrm{}=\frac{1}{8}\mathrm{}\left[\begin{array}{l}5-1\\ –73\end{array}\right]\end{array}$

Q.16 Find the order of matrix A = [a ub>ij]m x n.

AnsOrder of the matrix is m x n.

Q.17 A matrix has 24 elements. What re the possible orders it can have?

AnsThe possible orders of the matrix are:

1 24, 2 12, 3 8, 4 6, 6 4, 8 3, 12 2 and 24 1.

Q.18

$\begin{array}{l}\mathrm{If}\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{are}\mathrm{two}\mathrm{equal}\mathrm{square}\mathrm{matrices}\mathrm{such}\mathrm{that}\\ \mathrm{A}=\left[\begin{array}{l}\mathrm{x}\mathrm{y}\\ 23\end{array}\right]\mathrm{and}\mathrm{B}=\left[\begin{array}{l}14\\ 23\end{array}\right]\mathrm{},\mathrm{then}\mathrm{find}\mathrm{the}\mathrm{values}\mathrm{of}\mathrm{x}\mathrm{and}\mathrm{y}.\end{array}$

Ans

x = 1 and y = 4 (Equality matrices property)

Q.19 If A and B are two square atrices and K is a scalar quantity then K(A+B) = ______.

Ans K(A+B) = KA+KB

Q.20

$\text{Write the general formula for elements of matrix A=}\left[\begin{array}{l}\frac{1}{2}\frac{1}{2}\mathrm{}\\ \frac{2}{1}\frac{2}{2}\end{array}\right]=\mathrm{}\left[\begin{array}{l}1\frac{1}{2}\mathrm{}\\ 21\end{array}\right].$

Ans

$\mathrm{A}=\left[\mathrm{aij}\right]\mathrm{is}\mathrm{a}2×2\mathrm{matrix}\mathrm{whose}\mathrm{elements}\mathrm{are}\mathrm{given}\mathrm{by}=\mathrm{aij}=\frac{\mathrm{i}}{\mathrm{j}}$

Q.21 A square matrix in which elements in the diagonal are all 1 and rest are all zero is called an ________.

Ans

Identity Matrix

Q.22 Define identity matrix.

Ans

A square matrix in which elements in the diagonal are all 1 and rest all are zero is called an identity matrix.

Q.23

$\mathrm{If}\mathrm{A}=\left[\begin{array}{l}10\\ 24\end{array}\right]\mathrm{and}\mathrm{B}=\left[\begin{array}{l}2-1\\ 37\end{array}\right],\mathrm{find}\mathrm{A}+\mathrm{B}.$

Ans

$\begin{array}{l}\mathrm{A}+\mathrm{B}=\left[\begin{array}{l}10\\ 24\end{array}\right]+\mathrm{}\left[\begin{array}{l}2-1\\ 37\end{array}\right]\\ =\mathrm{}\left[\begin{array}{l}1+20 – 7\\ 2+34 +7\end{array}\right]\\ =\mathrm{}\left[\begin{array}{l}3-1\mathrm{}\\ 511\end{array}\right]\end{array}$

Q.24

$\mathrm{If}\mathrm{A}=\left[\begin{array}{l}13 5\\ 24 6\\ 35 7\end{array}\right]\mathrm{},\mathrm{find}\mathrm{the}\mathrm{diagonal}\mathrm{elements}\mathrm{of}\mathrm{A}.$

Ans

Diagonal elements of A are (1, 4, 7).

Q.25

$\mathrm{If}\mathrm{A}=\left[\begin{array}{l}\mathrm{a}0\\ 0\mathrm{a}\end{array}\right]\mathrm{and}\mathrm{B}=\left[\begin{array}{l}0\mathrm{a}\\ \mathrm{a}0\end{array}\right]\mathrm{},\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{AB}.$

Ans

$\begin{array}{l}\mathrm{AB}=\left[\begin{array}{l}\mathrm{a}0\\ 0–\mathrm{a}\end{array}\right]\mathrm{}+\mathrm{}\left[\begin{array}{l}\mathrm{a}0\\ 0–\mathrm{a}\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{l}\mathrm{a}×0+ 0×\mathrm{a}\mathrm{a}×\mathrm{a}+ 0×0\\ 0×0+\left(-\mathrm{a}\right)×\mathrm{a}0×\mathrm{a}+\left(–\mathrm{a}\right)×0\end{array}\right]\\ =\mathrm{}\left[\begin{array}{l}0{\mathrm{a}}^{2}\\ -{\mathrm{a}}^{2}0\end{array}\right]\end{array}$

Q.26

$\begin{array}{l}\mathrm{If}\mathrm{A}=\left[\begin{array}{l}10\\ 0-1\end{array}\right]\mathrm{and}\mathrm{B}=\left[\begin{array}{l}\mathrm{a}0\\ 0–\mathrm{a}\end{array}\right]\mathrm{satisfying}2\mathrm{A}+2\mathrm{B}=0,\\ \mathrm{then}\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{a}.\end{array}$

Ans

$\begin{array}{l}2\mathrm{A}+\mathrm{B}=2\left[\begin{array}{l}10\\ 0-1\end{array}\right]\mathrm{}+\mathrm{}\left[\begin{array}{l}\mathrm{a}0\\ 0–\mathrm{a}\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{l}2+\mathrm{a}0\\ 0-2 –\mathrm{a}\end{array}\right]\\ \mathrm{Given}, 2\mathrm{A}+\mathrm{B}= 0\\ =\left[\begin{array}{l}2+\mathrm{a}0\\ 0-2\mathrm{a}\end{array}\right]\\ ⇒\mathrm{}\mathrm{a}\mathrm{}=-2\end{array}$

Q.27

$\mathrm{If}\mathrm{A}=\left[\begin{array}{l}\mathrm{a}–\mathrm{a}\\ \mathrm{b}–\mathrm{b}\end{array}\right]\mathrm{and}\mathrm{B}=\left[\begin{array}{l}\mathrm{b}\mathrm{a}\\ –\mathrm{b}–\mathrm{a}\end{array}\right],\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}2\mathrm{A}-2\mathrm{B}.$

Ans

$\begin{array}{l}2\mathrm{A}– 2\mathrm{B}=\left[\begin{array}{l}2\mathrm{a}-2\mathrm{a}\\ 2\mathrm{b}-2\mathrm{b}\end{array}\right]–\left[\begin{array}{l}2\mathrm{b}2\mathrm{a}\\ –2\mathrm{b}-2\mathrm{a}\end{array}\right]\\ =\left[\begin{array}{l}2\mathrm{a}-2\mathrm{b}-2\mathrm{a}-2\mathrm{a}\\ 2\mathrm{b}–\left(–2\mathrm{b}\right)-2\mathrm{b}–\left(–2\mathrm{a}\right)\end{array}\right]\mathrm{}\\ =\left[\begin{array}{l}2\mathrm{a}-2\mathrm{b}-4\mathrm{a}\\ 4\mathrm{b}2\mathrm{a}-2\mathrm{b}\end{array}\right]\end{array}$

Q.28

$\begin{array}{l}\mathrm{Let}\mathrm{A}=\left[\begin{array}{l}0-\mathrm{tan}\frac{\mathrm{\alpha }}{2}\\ \mathrm{tan}\frac{\mathrm{\alpha }}{2}0\end{array}\right]\mathrm{and}\mathrm{I}\mathrm{be}\mathrm{the}\mathrm{identity}\mathrm{matrix}\mathrm{of}\mathrm{order}2.\\ \mathrm{show}\mathrm{that}:\mathrm{I}+\mathrm{A}\left(\mathrm{I}–\mathrm{A}\right)\mathrm{}\left[\begin{array}{l}\mathrm{cos\alpha }\mathrm{}-\mathrm{}\mathrm{sin\alpha }\\ \mathrm{sin\alpha }\mathrm{cos\alpha }\end{array}\right]\end{array}$

Ans

$\begin{array}{l}\mathrm{I}+\mathrm{A}=\left[\begin{array}{l}10\\ 01\end{array}\right]+\left[\begin{array}{l}0-\mathrm{tan}\frac{\mathrm{\alpha }}{2}\\ \mathrm{tan}\frac{\mathrm{\alpha }}{2}0\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{l}1-\mathrm{tan}\frac{\mathrm{\alpha }}{2}\\ \mathrm{tan}\frac{\mathrm{\alpha }}{2}1\end{array}\right]\\ \mathrm{I}-\mathrm{A}=\left[\begin{array}{l}10\\ 01\end{array}\right]-\mathrm{}\left[\begin{array}{l}0-\mathrm{tan}\frac{\mathrm{\alpha }}{2}\\ \mathrm{tan}\frac{\mathrm{\alpha }}{2}0\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{l}1\mathrm{tan}\frac{\mathrm{\alpha }}{2}\\ -\mathrm{tan}\frac{\mathrm{\alpha }}{2}1\end{array}\right]\\ \therefore \mathrm{}\left(\mathrm{I}–\mathrm{A}\right)\mathrm{}\left[\begin{array}{l}\mathrm{cos\alpha }\mathrm{}-\mathrm{}\mathrm{sin\alpha }\\ \mathrm{sin\alpha }\mathrm{cos\alpha }\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{l}1\mathrm{tan}\frac{\mathrm{\alpha }}{2}\\ -\mathrm{tan}\frac{\mathrm{\alpha }}{2}1\end{array}\right]\left[\begin{array}{l}\mathrm{cos\alpha }\mathrm{}-\mathrm{}\mathrm{sin\alpha }\\ \mathrm{sin\alpha }\mathrm{cos\alpha }\end{array}\right]\\ ⇒\mathrm{}\left(\mathrm{I}–\mathrm{A}\right)\mathrm{}\left[\begin{array}{l}\mathrm{cos\alpha }\mathrm{}-\mathrm{}\mathrm{sin\alpha }\\ \mathrm{sin\alpha }\mathrm{cos\alpha }\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{l}1\mathrm{tan}\frac{\mathrm{\alpha }}{2}\\ -\mathrm{tan}\frac{\mathrm{\alpha }}{2}1\end{array}\right]\mathrm{}\left[\frac{\frac{1-{\mathrm{tan}}^{2}\frac{\mathrm{\alpha }}{2}}{1+{\mathrm{tan}}^{2}\frac{\mathrm{\alpha }}{2}}-\mathrm{}\frac{2\mathrm{tan}\frac{\mathrm{\alpha }}{2}}{1+{\mathrm{tan}}^{2}\frac{\mathrm{\alpha }}{2}}}{\frac{2\mathrm{tan}\frac{\mathrm{\alpha }}{2}}{1+{\mathrm{tan}}^{2}\frac{\mathrm{\alpha }}{2}}\frac{1-{\mathrm{tan}}^{2}\frac{\mathrm{\alpha }}{2}}{1+{\mathrm{tan}}^{2}\frac{\mathrm{\alpha }}{2}}}\right]\\ ⇒\mathrm{}\left(\mathrm{I}–\mathrm{A}\right)\mathrm{}\left[\begin{array}{l}\mathrm{cos\alpha }\mathrm{}-\mathrm{}\mathrm{sin\alpha }\\ \mathrm{sin\alpha }\mathrm{cos\alpha }\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{l}1\mathrm{t}\\ -\mathrm{t}1\end{array}\right]\mathrm{}\left[\frac{\frac{1-{\mathrm{t}}^{2}}{1+{\mathrm{t}}^{2}}-\mathrm{}\frac{2\mathrm{t}}{1+{\mathrm{t}}^{2}}}{\frac{2\mathrm{t}}{1+{\mathrm{t}}^{2}}\frac{1-{\mathrm{t}}^{2}}{1+{\mathrm{t}}^{2}}}\right],\mathrm{where}\mathrm{t}=\frac{\mathrm{tan\alpha }}{2}\\ ⇒\mathrm{}\left(\mathrm{I}–\mathrm{A}\right)\mathrm{}\left[\begin{array}{l}\mathrm{cos\alpha }\mathrm{}-\mathrm{}\mathrm{sin\alpha }\\ \mathrm{sin\alpha }\mathrm{cos\alpha }\end{array}\right]\mathrm{}=\left[\frac{\frac{1-{\mathrm{t}}^{2}+2{\mathrm{t}}^{2}}{1+{\mathrm{t}}^{2}}\frac{-2\mathrm{t}+\mathrm{t}-{\mathrm{t}}^{3}}{1+{\mathrm{t}}^{2}}}{\frac{-\mathrm{t}+{\mathrm{t}}^{3}+2\mathrm{t}}{1+{\mathrm{t}}^{2}}\frac{2{\mathrm{t}}^{2}+1-{\mathrm{t}}^{2}}{1+{\mathrm{t}}^{2}}}\right]\\ ⇒\mathrm{}\left(\mathrm{I}–\mathrm{A}\right)\mathrm{}\left[\begin{array}{l}\mathrm{cos\alpha }\mathrm{}-\mathrm{}\mathrm{sin\alpha }\\ \mathrm{sin\alpha }\mathrm{cos\alpha }\end{array}\right]\mathrm{}=\left[\frac{\frac{1-{\mathrm{t}}^{2}}{1+{\mathrm{t}}^{2}}\frac{-\mathrm{t}+\left(1-{\mathrm{t}}^{2}\right)}{1+{\mathrm{t}}^{2}}}{\frac{\mathrm{t}\left(1+{\mathrm{t}}^{2}\right)}{1+{\mathrm{t}}^{2}}\frac{1+{\mathrm{t}}^{2}}{1+{\mathrm{t}}^{2}}}\right]\mathrm{}=\mathrm{}\left[\begin{array}{l}1-\mathrm{t}\\ \mathrm{t}1\end{array}\right]\\ ⇒\mathrm{}\left(\mathrm{I}–\mathrm{A}\right)\mathrm{}\left[\begin{array}{l}\mathrm{cos\alpha }\mathrm{}-\mathrm{}\mathrm{sin\alpha }\\ \mathrm{sin\alpha }\mathrm{cos\alpha }\end{array}\right]\mathrm{}=\left[\begin{array}{l}1-\mathrm{tan}\frac{\mathrm{\alpha }}{2}\\ \mathrm{tan}\frac{\mathrm{\alpha }}{2}1\mathrm{}\end{array}\right]\mathrm{}=\mathrm{I}+\mathrm{A}\end{array}$

Q.29

$\text{Find the matrix x so that x}\left[\begin{array}{l}12 3\\ 45 6\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{l}–7-8 -9\\ 45 6\end{array}\right].$

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{x}=\left[\begin{array}{l}\mathrm{a}\mathrm{b}\\ \mathrm{c}\mathrm{d}\end{array}\right]\mathrm{}\\ \mathrm{}\therefore \mathrm{}\left[\begin{array}{l}\mathrm{a}\mathrm{b}\\ \mathrm{c}\mathrm{d}\end{array}\right]\mathrm{}\left[\begin{array}{l}12 3\\ 45 6\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{l}–7-8 -9\\ 45 6\end{array}\right]\\ ⇒\mathrm{}\left[\begin{array}{l}\mathrm{a}×1+\mathrm{b}×4\mathrm{a}×2+\mathrm{b}×5\mathrm{a}×3+\mathrm{b}×6\\ \mathrm{c}×1+\mathrm{d}×4\mathrm{c}×2+\mathrm{d}×5\mathrm{c}×3+\mathrm{d}×6\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{l}–7-8 -9\\ 45 6\end{array}\right]\\ ⇒\mathrm{}\left[\begin{array}{l}\mathrm{a}+4\mathrm{b}2\mathrm{a}+5\mathrm{b}3\mathrm{a}+6\mathrm{b}\\ \mathrm{c}+4\mathrm{d}2\mathrm{c}+5\mathrm{d}3\mathrm{c}+6\mathrm{d}\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{l}–7-8 -9\\ 24 6\end{array}\right]\mathrm{}\\ \mathrm{By}\mathrm{using}\mathrm{equality}\mathrm{and}\mathrm{solving}\\ \begin{array}{l}\mathrm{a}+4\mathrm{b}= -7\\ 2\mathrm{a}+5\mathrm{b}= – 8\end{array}}\mathrm{}.\dots \dots ..\left(1\right)\begin{array}{l}\mathrm{c}+4\mathrm{d}= 2\\ 2\mathrm{c}+5\mathrm{d}= 4\end{array}}\mathrm{}.\dots \dots ..\left(2\right)\\ \mathrm{Now},\mathrm{by}\left(1\right)\mathrm{a}=1\mathrm{b}=-2\\ \mathrm{by}\mathrm{}\left(2\right)\mathrm{c}\mathrm{}=2\mathrm{d}\mathrm{}=0\\ \mathrm{Hence}\mathrm{}\mathrm{x}\mathrm{}=\mathrm{}\left[\begin{array}{l}1-2\\ 20\end{array}\right]\end{array}$

Q.30 Show that for any square matrix A, A + A’ is a symmetric matrix and A – A’ is a skew symmetric matrix.

Ans

Let B = A + A’
B’ = (A + A’)’
= A’ + (A’)’ [as (A + B)’ = A’ + B’]
= A’ + A [ (A’)’ = A]
= B
∴ B = A + A’ is a symmetric matrix.

Now let C = A – A’
C’ = (A – A’)’
= A’ – (A’)’
= A’ – A
= -(A – A’)
∴ C = A – A’ is a skew symmetric matrix.

Q.31 Find the order of matrix A = [ij]m x n.

AnsOrder of the matrix is m x n.

Q.32

$\begin{array}{l}\mathrm{If}\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{are}\mathrm{two}\mathrm{equal}\mathrm{square}\mathrm{matrices}\mathrm{such}\mathrm{that}\\ \mathrm{A}=\left[\begin{array}{l}\mathrm{x}\mathrm{y}\\ 23\end{array}\right]\mathrm{and}\mathrm{B}=\left[\begin{array}{l}14\\ 23\end{array}\right],\mathrm{then}\mathrm{find}\mathrm{the}\mathrm{values}\mathrm{of}\mathrm{x}\mathrm{and}\mathrm{y}.\end{array}$

Ans

$\mathrm{x}= 1\mathrm{and}\mathrm{y}= 4\left(\mathrm{Equality}\mathrm{of}\mathrm{matrices}\mathrm{property}\right)\mathrm{.}$

Q.33

$\text{Construct a 2 ×2 matrix A =}\left[\mathrm{aij}\right]\mathrm{whose}\mathrm{elements}\mathrm{are}\mathrm{given}\mathrm{by}\mathrm{aij}\frac{\mathrm{i}}{\mathrm{j}}.$

Ans

$\mathrm{A}=\left[\begin{array}{l}\frac{1}{1}\frac{1}{2}\\ \frac{2}{1}\frac{2}{2}\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{l}1\frac{1}{2}\\ 21\end{array}\right]$

Q.34

$\begin{array}{l}\mathrm{If}\mathrm{A}=\left[\begin{array}{l}12 3\\ 3-2 1\\ 42 1\end{array}\right],\mathrm{that}\mathrm{show}\mathrm{that}{\mathrm{A}}^{3}-23\mathrm{A}-40\mathrm{I}\mathrm{}=\mathrm{}0.\\ \end{array}$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have}{\mathrm{A}}^{2}=\mathrm{A}.\mathrm{A}=\left[\begin{array}{l}12 3\\ 3-2 1\\ 42 1\end{array}\right]\mathrm{}\left[\begin{array}{l}12 3\\ 3-2 1\\ 42 1\end{array}\right]\\ =\mathrm{}\left[\begin{array}{l}194 8\\ 112 8\\ 146 15\end{array}\right]\mathrm{}\\ \mathrm{So},{\mathrm{A}}^{3}={\mathrm{A}}^{2}.\mathrm{A}=\left[\begin{array}{l}194 8\\ 112 8\\ 146 15\end{array}\right]\mathrm{}\left[\begin{array}{l}12 3\\ 3-2 1\\ 42 1\end{array}\right]\\ =\mathrm{}\left[\begin{array}{l}6346 69\\ 69-6 23\\ 9246 63\end{array}\right]\\ \mathrm{Now},\\ {\mathrm{A}}^{3}-23\mathrm{A}-40\mathrm{I}=\left[\begin{array}{l}6346 69\\ 69-6 23\\ 9246 63\end{array}\right]\mathrm{}-23\mathrm{}\left[\begin{array}{l}12 3\\ 3-2 1\\ 42 1\end{array}\right]-40\mathrm{}\left[\begin{array}{l}10 0\\ 01 0\\ 00 1\end{array}\right]\\ =\left[\begin{array}{l}6346 69\\ 69-6 23\\ 9246 63\end{array}\right]\mathrm{}+\mathrm{}\left[\begin{array}{l}–23-46 -69\\ –6946 -23\\ –92-46 -23\end{array}\right]+\mathrm{}\left[\begin{array}{l}–400 0\\ 0-40 0\\ 00 -40\end{array}\right]\\ =\left[\begin{array}{l}63-23 -40 46-46+0 69-69+0\\ 69-69 +0 -6+46-40 23-23+0\\ 92-92 +0 46-46+0 63-23-40\end{array}\right]\\ \mathrm{}=\mathrm{}\left[\begin{array}{l}00 0\\ 00 0\\ 00 0\end{array}\right]=\mathrm{}0=\mathrm{RHS}\end{array}$

Q.35

$\text{If A =}\left[\begin{array}{l}10 2\\ 02 1\\ 20 3\end{array}\right],\mathrm{prove}\mathrm{that}{\mathrm{A}}^{3}-6{\mathrm{A}}^{2}+7\mathrm{A}+2\mathrm{I}\mathrm{}=\mathrm{}0.$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have},\mathrm{A}=\left[\begin{array}{l}10 2\\ 02 1\\ 20 3\end{array}\right]\\ \mathrm{}{\mathrm{A}}^{2}\mathrm{}=\mathrm{}\mathrm{A}.\mathrm{A}\mathrm{}=\mathrm{}\left[\begin{array}{l}10 2\\ 02 1\\ 20 3\end{array}\right].\mathrm{}\left[\begin{array}{l}10 2\\ 02 1\\ 20 3\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{l}50 8\\ 24 5\\ 80 13\end{array}\right]\\ {\mathrm{A}}^{3}\mathrm{}=\mathrm{}{\mathrm{A}}^{2}.\mathrm{A}\mathrm{}=\mathrm{}\left[\begin{array}{l}50 8\\ 24 5\\ 80 13\end{array}\right].\mathrm{}\left[\begin{array}{l}10 2\\ 02 1\\ 20 3\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{l}210 34\\ 128 23\\ 340 55\end{array}\right]\\ \mathrm{LHS}:\\ {\mathrm{A}}^{3}-6{\mathrm{A}}^{2}+7\mathrm{A}+2\mathrm{I}\\ \mathrm{}=\mathrm{}\left[\begin{array}{l}210 34\\ 128 23\\ 340 55\end{array}\right]-6\mathrm{}\left[\begin{array}{l}50 8\\ 24 5\\ 80 13\end{array}\right]+7\mathrm{}\left[\begin{array}{l}10 2\\ 02 1\\ 20 3\end{array}\right]+2\left[\begin{array}{l}10 0\\ 01 0\\ 00 1\end{array}\right]\\ \mathrm{}=\mathrm{}\left[\begin{array}{l}210 34\\ 128 23\\ 340 55\end{array}\right]+\mathrm{}\left[\begin{array}{l}300 -48\\ –12-24 -30\\ –480 -78\end{array}\right]+\mathrm{}\left[\begin{array}{l}70 14\\ 014 7\\ 140 21\end{array}\right]+\left[\begin{array}{l}20 0\\ 02 0\\ 00 2\end{array}\right]\\ \mathrm{}=\mathrm{}\left[\begin{array}{l}21-30+7+2 0+0+0+0 34-48+14+0\\ 12-12+0+0 8-24+14+2 23-30+7+0\\ 34-48+14+0 0+0+0+0 55-78+21+2\end{array}\right]\\ =\mathrm{}\left[\begin{array}{l}00 0\\ 00 0\\ 00 0\end{array}\right]\\ \mathrm{}=\mathrm{}\mathrm{RHS}\end{array}$

Q.36

$\begin{array}{l}\mathrm{Express}\mathrm{the}\mathrm{following}\mathrm{matrix}\mathrm{as}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{a}\mathrm{symmetric}\mathrm{and}\mathrm{a}\mathrm{skew}\mathrm{symmetric}\mathrm{matrix}:\\ \left[\begin{array}{l}6-2 2\\ –23 -1\\ 2-1 3\end{array}\right]\end{array}$

Ans

$\begin{array}{l}\mathrm{P}=\frac{1}{2}\mathrm{}\left(\mathrm{A}+\mathrm{A}‘\right)\\ =\mathrm{}\frac{1}{2}\left\{\left[\begin{array}{l}6-2 2\\ –23 -1\\ 2-1 3\end{array}\right]+\left[\begin{array}{l}6-2 2\\ –23 -1\\ 2-1 3\end{array}\right]\right\}\\ =\mathrm{}\frac{1}{2}\left[\begin{array}{l}12-4 4\\ –46 -2\\ 4-2 6\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{l}6-2 2\\ –23 -1\\ 2-1 3\end{array}\right]\\ \mathrm{Thus},\mathrm{P}=\frac{1}{2}\left(\mathrm{A}–\mathrm{A}‘\right)\mathrm{is}\mathrm{symmetric}\mathrm{matrix}.\\ \mathrm{}\mathrm{Q}=\frac{1}{2}\left(\mathrm{A}–\mathrm{A}‘\right)\\ =\mathrm{}\frac{1}{2}\mathrm{}\left\{\left[\begin{array}{l}6-2 2\\ –23 -1\\ 2-1 3\end{array}\right]+\left[\begin{array}{l}6-2 2\\ –23 -1\\ 2-1 3\end{array}\right]\right\}\mathrm{}=\mathrm{}\left[\begin{array}{l}00 0\\ 00 0\\ 00 0\end{array}\right]\\ \mathrm{}\mathrm{Q}‘\mathrm{}=\mathrm{}\left[\begin{array}{l}00 0\\ 00 0\\ 00 0\end{array}\right]\\ \mathrm{Thus},\mathrm{Q}=\frac{1}{2}\left(\mathrm{A}-\mathrm{A}‘\right)\mathrm{is}\mathrm{skew}\mathrm{symmetric}\mathrm{matrix}.\\ \mathrm{P}+\mathrm{Q}=\frac{\mathrm{A}+\mathrm{A}‘}{2}+\frac{\mathrm{A}+\mathrm{A}‘}{2}\mathrm{}=\mathrm{}\frac{2\mathrm{A}}{2}\mathrm{}=\mathrm{}\mathrm{A}\\ \mathrm{Thus},\mathrm{A}\mathrm{is}\mathrm{expressed}\mathrm{as}\mathrm{a}\mathrm{sum}\mathrm{of}\mathrm{a}\mathrm{symmetric}\mathrm{and}\mathrm{a}\mathrm{skew}\\ \mathrm{symmetric}\mathrm{matrix}.\end{array}$

Q.37

$\begin{array}{l}\mathrm{Let}\mathrm{A}=\left[\begin{array}{l}01\\ 00\end{array}\right],\mathrm{show}\mathrm{that}{\left(\mathrm{aI}+\mathrm{bA}\right)}^{\mathrm{n}}\mathrm{}={\mathrm{a}}^{\mathrm{n}}\mathrm{I}+{\mathrm{na}}^{\mathrm{n}-1}\mathrm{bA},\mathrm{where}\mathrm{I}\mathrm{is}\mathrm{the}\mathrm{identity}\\ \mathrm{matrix}\mathrm{of}\mathrm{order}2\mathrm{and}\mathrm{n}\in \mathrm{N}.\end{array}$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{shall}\mathrm{prove}\mathrm{the}\mathrm{result}\mathrm{by}\mathrm{using}\mathrm{the}\mathrm{principle}\mathrm{of}\mathrm{mathematical}\mathrm{induction}.\\ \mathrm{For}\mathrm{n}= 1,\mathrm{we}\mathrm{have}\\ \mathrm{P}\left(1\right):\mathrm{}\left(\mathrm{al}+\mathrm{bA}\right)\mathrm{}=\mathrm{al}+{\mathrm{ba}}^{0}\mathrm{A}\\ =\mathrm{al}+\mathrm{bA}\\ \mathrm{Therefore},\mathrm{the}\mathrm{result}\mathrm{is}\mathrm{true}\mathrm{for}\mathrm{n}= 1.\\ \mathrm{Let}\mathrm{the}\mathrm{result}\mathrm{be}\mathrm{true}\mathrm{for}\mathrm{n}=\mathrm{k}\\ \mathrm{so},\\ \mathrm{P}{\left(\mathrm{al}+\mathrm{bA}\right)}^{\mathrm{k}}={\mathrm{a}}^{\mathrm{k}}\mathrm{l}+{\mathrm{ka}}^{\mathrm{k}-1}\mathrm{bA}\mathrm{Now},\mathrm{we}\mathrm{will}\mathrm{prove}\mathrm{that}\mathrm{it}\mathrm{is}\mathrm{true}\mathrm{for}\mathrm{n}=\mathrm{k}+1\\ {\left(\mathrm{al}+\mathrm{bA}\right)}^{\mathrm{k}+1}={\left(\mathrm{al}+\mathrm{bA}\right)}^{\mathrm{k}}\left(\mathrm{al}+\mathrm{bA}\right)\\ =\mathrm{}\left({\mathrm{a}}^{\mathrm{k}}+{\mathrm{ka}}^{\mathrm{k}-1}\mathrm{bA}\right)\mathrm{}\left(\mathrm{al}+\mathrm{bA}\right)\\ =\mathrm{}{\mathrm{a}}^{\mathrm{k}+1}\mathrm{l}+{\mathrm{ka}}^{\mathrm{k}}\mathrm{bA}+{\mathrm{a}}^{\mathrm{k}}\mathrm{bAl}+{\mathrm{ka}}^{\mathrm{k}-1}{\mathrm{b}}^{2}{\mathrm{A}}^{2}\\ {\left(\mathrm{al}+\mathrm{bA}\right)}^{\mathrm{k}+1}={\mathrm{a}}^{\mathrm{k}+1}+\left(\mathrm{k}+1\right){\mathrm{a}}^{\mathrm{k}}\mathrm{bA}+{\mathrm{ka}}^{\mathrm{k}-1}{\mathrm{b}}^{2}{\mathrm{A}}^{2}.\dots ..\left(1\right)\\ \mathrm{Now},{\mathrm{A}}^{2}=\mathrm{}\left[\begin{array}{l}01\\ 00\end{array}\right]\mathrm{}\left[\begin{array}{l}01\\ 00\end{array}\right]\\ =\mathrm{}\left[\begin{array}{l}01\\ 00\end{array}\right]\mathrm{}=\mathrm{}0\mathrm{}\\ \mathrm{From}\mathrm{equation}\left(1\right),\mathrm{we}\mathrm{have}\\ {\left(\mathrm{al}+\mathrm{bA}\right)}^{\mathrm{k}+1}={\mathrm{a}}^{\mathrm{k}+1}+\left(\mathrm{k}+1\right){\mathrm{a}}^{\mathrm{k}}\mathrm{bA}+{\mathrm{ka}}^{\mathrm{k}-1}{\mathrm{b}}^{2}\left(0\right)\\ =\mathrm{}{\mathrm{a}}^{\mathrm{k}+1}+\left(\mathrm{k}+1\right)\mathrm{}{\mathrm{a}}^{\mathrm{k}}\mathrm{bA}\\ \mathrm{Therefore},\mathrm{the}\mathrm{result}\mathrm{is}\mathrm{true}\mathrm{for}\mathrm{n}=\mathrm{k}+1.\\ \mathrm{Thus},\mathrm{by}\mathrm{the}\mathrm{Principle}\mathrm{of}\mathrm{Mathematical}\mathrm{Induction},\mathrm{we}\mathrm{have}\\ {\left(\mathrm{aI}+\mathrm{bA}\right)}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{n}}\mathrm{I}+{\mathrm{na}}^{\mathrm{n}-1}\mathrm{bA},\mathrm{for}\mathrm{all}\mathrm{n}\in \mathrm{N}.\end{array}$

Q.38

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{inverse}\mathrm{of}\mathrm{the}\mathrm{following}\mathrm{matrix},\mathrm{if}\mathrm{it}\mathrm{exists}.\\ \left[\begin{array}{c}\begin{array}{l}2-33\\ 223\end{array}\\ 3-22\end{array}\right]\mathrm{}\end{array}$

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{A}=\left[\begin{array}{c}\begin{array}{l}2-33\\ 223\end{array}\\ 3-22\end{array}\right]\mathrm{}\\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{A}=\mathrm{IA}\\ \mathrm{}\therefore \mathrm{}\left[\begin{array}{c}\begin{array}{l}2-33\\ 223\end{array}\\ 3-22\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{c}\begin{array}{l}100\\ 010\end{array}\\ 001\end{array}\right]\mathrm{A}\\ ⇒\left[\begin{array}{c}\begin{array}{l}2-33\\ 050\end{array}\\ 3-22\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{c}\begin{array}{l}\mathrm{}100\\ -110\end{array}\\ \mathrm{}001\end{array}\right]\mathrm{A}\left({\mathrm{R}}_{2}\to {\mathrm{R}}_{2}-{\mathrm{R}}_{1}\right)\\ ⇒\left[\begin{array}{c}\begin{array}{l}2-33\\ 010\end{array}\\ 3-22\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{c}\begin{array}{l}\mathrm{}100\\ -\frac{1}{5}\frac{1}{5}0\end{array}\\ \mathrm{}001\end{array}\right]\mathrm{A}\left({\mathrm{R}}_{2}\to \frac{1}{5}{\mathrm{R}}_{2}\right)\\ ⇒\left[\begin{array}{c}\begin{array}{l}-1-11\\ \mathrm{}010\end{array}\\ 3-22\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{c}\begin{array}{l}\mathrm{}10-1\\ -\frac{1}{5}\frac{1}{5}0\end{array}\\ \mathrm{}001\end{array}\right]\mathrm{A}\left({\mathrm{R}}_{1}\to {\mathrm{R}}_{1}-{\mathrm{R}}_{3}\right)\\ ⇒\left[\begin{array}{c}\begin{array}{l}-101\\ \mathrm{}010\end{array}\\ 302\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{c}\begin{array}{l}\mathrm{}\frac{4}{5}\frac{1}{5}-1\\ -\frac{1}{5}\frac{1}{5}0\end{array}\\ -\mathrm{}\frac{2}{5}\frac{2}{5}1\end{array}\right]\mathrm{A}\left(\begin{array}{l}{\mathrm{R}}_{1}\to {\mathrm{R}}_{1}+{\mathrm{R}}_{2}\mathrm{and}\\ {\mathrm{R}}_{3}\mathrm{}\to {\mathrm{R}}_{3}+2{\mathrm{R}}_{2}\end{array}\right)\\ ⇒\left[\begin{array}{c}\begin{array}{l}-101\\ \mathrm{}010\end{array}\\ 005\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{c}\begin{array}{l}\mathrm{}\frac{4}{5}\frac{1}{5}-1\\ -\frac{1}{5}\frac{1}{5}0\end{array}\\ 21-2\end{array}\right]\mathrm{A}\left({\mathrm{R}}_{3}\mathrm{}\to {\mathrm{R}}_{3}+3{\mathrm{R}}_{1}\right)\\ ⇒\left[\begin{array}{c}\begin{array}{l}-101\\ \mathrm{}010\end{array}\\ 001\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{c}\begin{array}{l}\mathrm{}\frac{4}{5}\frac{1}{5}-1\\ -\frac{1}{5}\frac{1}{5}0\end{array}\\ \frac{2}{5}\frac{1}{5}\frac{-2}{5}\end{array}\right]\mathrm{A}\left({\mathrm{R}}_{3}\to \frac{1}{5}{\mathrm{R}}_{3}\right)\\ ⇒\left[\begin{array}{c}\begin{array}{l}-101\\ \mathrm{}010\end{array}\\ 001\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{c}\begin{array}{l}\mathrm{}\frac{2}{5}0\frac{-3}{5}\\ -\frac{1}{5}\frac{1}{5}0\end{array}\\ \frac{2}{5}\frac{1}{5}\frac{-2}{5}\end{array}\right]\mathrm{A}\left({\mathrm{R}}_{3}\mathrm{}\to {\mathrm{R}}_{1}-{\mathrm{R}}_{3}\right)\\ ⇒\left[\begin{array}{c}\begin{array}{l}100\\ \mathrm{}010\end{array}\\ 001\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{c}\begin{array}{l}\mathrm{}-\frac{2}{5}0\frac{3}{5}\\ -\frac{1}{5}\frac{1}{5}0\end{array}\\ \frac{2}{5}\frac{1}{5}\frac{-2}{5}\end{array}\right]\mathrm{A}\left({\mathrm{R}}_{1}\mathrm{}\to \mathrm{}\left(-1\right)\mathrm{}{\mathrm{R}}_{1}\right)\\ {\mathrm{A}}^{-1}=\mathrm{}\left[\begin{array}{c}\begin{array}{l}\mathrm{}-\frac{2}{5}0\frac{3}{5}\\ -\frac{1}{5}\frac{1}{5}0\end{array}\\ \frac{2}{5}\frac{1}{5}\frac{-2}{5}\end{array}\right]\end{array}$

Q.39

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{values}\mathrm{of}\mathrm{x},\mathrm{y},\mathrm{z}\mathrm{if}\mathrm{the}\mathrm{matrix}\\ \mathrm{A}=\left[\begin{array}{c}02\mathrm{y}\mathrm{z}\\ \mathrm{x}\mathrm{y}-\mathrm{z}\\ \mathrm{x}-\mathrm{y}\mathrm{z}\end{array}\right]\mathrm{satisfy}\mathrm{the}\mathrm{equation}\mathrm{A}‘\mathrm{A}={\mathrm{I}}_{3}.\end{array}$

Ans

$\begin{array}{l}\mathrm{A}=\left[\begin{array}{c}02\mathrm{y}\mathrm{z}\\ \mathrm{x}\mathrm{y}-\mathrm{z}\\ \mathrm{x}-\mathrm{y}\mathrm{z}\end{array}\right]\mathrm{and}\mathrm{A}‘ =\left[\begin{array}{c}0\mathrm{x}\mathrm{x}\\ 2\mathrm{y}\mathrm{y}-\mathrm{y}\\ \mathrm{z}-\mathrm{z}\mathrm{z}\end{array}\right]\\ \mathrm{Now},\\ \mathrm{A}‘\mathrm{A}={\mathrm{I}}_{3}\mathrm{}⇒\mathrm{}\left[\begin{array}{c}0\mathrm{x}\mathrm{x}\\ 2\mathrm{y}\mathrm{y}-\mathrm{y}\\ \mathrm{z}-\mathrm{z}\mathrm{z}\end{array}\right].\mathrm{}\left[\begin{array}{c}02\mathrm{y}\mathrm{z}\\ \mathrm{x}\mathrm{y}-\mathrm{z}\\ \mathrm{x}-\mathrm{y}\mathrm{z}\end{array}\right]=\mathrm{}\left[\begin{array}{c}100\\ 010\\ 001\end{array}\right]\\ \left[\begin{array}{c}0+{\mathrm{x}}^{2}+{\mathrm{x}}^{2}0+\mathrm{xy}-\mathrm{}\mathrm{xy}0-\mathrm{xz}+\mathrm{xz}\mathrm{}\\ 0+\mathrm{yx}-\mathrm{yx}4{\mathrm{y}}^{2}+{\mathrm{y}}^{2}+{\mathrm{y}}^{2}2\mathrm{yz}-\mathrm{yz}-\mathrm{yz}\\ 0-\mathrm{zx}+\mathrm{zx}2\mathrm{yz}-\mathrm{yz}-\mathrm{yz}{\mathrm{z}}^{2}+{\mathrm{z}}^{2}+{\mathrm{z}}^{2}\end{array}\right]=\mathrm{ }\left[\begin{array}{c}100\\ 010\\ 001\end{array}\right]\\ \left[\begin{array}{c}2{\mathrm{x}}^{2}00\\ 06{\mathrm{y}}^{2}0\\ 003{\mathrm{z}}^{2}\end{array}\right]\mathrm{}=\mathrm{ }\left[\begin{array}{c}100\\ 010\\ 001\end{array}\right]\mathrm{}\\ ⇒\mathrm{}2{\mathrm{x}}^{2}=\mathrm{}1,\mathrm{}6{\mathrm{y}}^{2}\mathrm{}=\mathrm{}1\mathrm{and}3{\mathrm{z}}^{2}=1\\ ⇒\mathrm{x}\mathrm{}=\mathrm{}±\mathrm{}\frac{1}{\sqrt{6}},\mathrm{}\mathrm{y}\mathrm{}=\mathrm{}±\mathrm{}\frac{1}{\sqrt{6}}\mathrm{and}\mathrm{z}=±\mathrm{}\frac{1}{\sqrt{3}}\end{array}$

Q.40

$\mathrm{If}\mathrm{A}=\left[\begin{array}{c}\begin{array}{l}122\\ 212\end{array}\\ 221\end{array}\right],\mathrm{prove}\mathrm{that}{\mathrm{A}}^{2}-4\mathrm{A}-5\mathrm{I}= 0.\mathrm{Hence},\mathrm{find}{\mathrm{A}}^{-1}\mathrm{.}$

Ans

$\begin{array}{l}\mathrm{Given}\mathrm{A}=\mathrm{If}\mathrm{A}=\left[\begin{array}{c}122\\ 212\\ 221\end{array}\right]\\ ⇒{\mathrm{A}}^{2}\mathrm{}=\left[\begin{array}{c}\begin{array}{l}122\\ 212\end{array}\\ 221\end{array}\right].\left[\begin{array}{c}\begin{array}{l}122\\ 212\end{array}\\ 221\end{array}\right]\\ \\ {\mathrm{A}}^{2}\mathrm{}=\mathrm{}\left[\begin{array}{l}1+4+42+2+42+4+2\\ 2+2+44+1+44+2+2\\ 2+4+24+2+24+4+1\end{array}\right]\\ =\mathrm{}\left[\begin{array}{c}\begin{array}{l}988\\ 898\end{array}\\ 889\end{array}\right]\\ \mathrm{Hence},{\mathrm{A}}^{2}\mathrm{}-\mathrm{}4\mathrm{A}-5\mathrm{I}=\left[\begin{array}{c}\begin{array}{l}988\\ 898\end{array}\\ 889\end{array}\right]-4\left[\begin{array}{c}\begin{array}{l}122\\ 212\end{array}\\ 221\end{array}\right]-5\mathrm{}\left[\begin{array}{c}\begin{array}{l}100\\ 010\end{array}\\ 001\end{array}\right]\\ =\left[\begin{array}{c}\begin{array}{l}988\\ 898\end{array}\\ 889\end{array}\right]+\left[\begin{array}{ccc}-4& -8& -8\\ -8& -4& -8\\ -8& -8& -4\end{array}\right]+\mathrm{}\left[\begin{array}{c}\begin{array}{l}-500\\ 0-50\end{array}\\ 005\end{array}\right]\\ ⇒{\mathrm{A}}^{2}-4\mathrm{A}-5\mathrm{I}= 0\\ {\mathrm{A}}^{2}-4\mathrm{A}=5\mathrm{I}\\ \mathrm{A}\left(\mathrm{A}-4\mathrm{I}\right)\mathrm{}=\mathrm{}5\mathrm{I}\\ {\mathrm{A}}^{-1\mathrm{}}=\frac{1}{5}\left\{\left[\begin{array}{l}1-42-02-\mathrm{0}\\ \mathrm{2}-01-42-\mathrm{0}\\ \mathrm{2}-02-01-\mathrm{4}\end{array}\right]\right\}\\ {\mathrm{A}}^{-1\mathrm{}}=\frac{1}{5}\left[\begin{array}{c}\begin{array}{l}-322\\ 2-32\end{array}\\ 22-3\end{array}\right]\end{array}$

Q.41

$\mathrm{A}=\left[\begin{array}{ccc}1& 0& 2\\ 0& 2& 1\\ 2& 0& 3\end{array}\right]\mathrm{and}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^{3}–6{\mathrm{x}}^{2}+7\mathrm{x}+ 2,\mathrm{verify}\mathrm{that}\mathrm{f}\left(\mathrm{A}\right)=0.$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have}\mathrm{A}=\left[\begin{array}{ccc}1& 0& 2\\ 0& 2& 1\\ 2& 0& 3\end{array}\right]\\ \mathrm{So},{\mathrm{A}}^{2}\mathrm{}=\mathrm{A},\mathrm{A}\left[\begin{array}{ccc}1& 0& 2\\ 0& 2& 1\\ 2& 0& 3\end{array}\right].\left[\begin{array}{ccc}1& 0& 2\\ 0& 2& 1\\ 2& 0& 3\end{array}\right]\\ =\mathrm{}\left[\begin{array}{l}1+0+40+0+02+0+6\\ 0+0+20+4+00+2+3\\ 2+0+60+0+04+0+9\end{array}\right]\\ {\mathrm{A}}^{2}\mathrm{}=\mathrm{}\left[\begin{array}{ccc}5& 0& \mathrm{}8\\ 2& 4& \mathrm{}5\\ 8& 0& 13\end{array}\right]\\ {\mathrm{A}}^{3}\mathrm{}={\mathrm{A}}^{2},\mathrm{A}\left[\begin{array}{ccc}5& 0& \mathrm{}8\\ 2& 4& \mathrm{}5\\ 8& 0& 13\end{array}\right].\left[\begin{array}{ccc}1& 0& 2\\ 0& 2& 1\\ 2& 0& 3\end{array}\right]=\mathrm{}\left[\begin{array}{ccc}21& 0& 34\\ 12& 8& 23\\ 34& 0& 55\end{array}\right]\\ \mathrm{Since},\mathrm{f}\left(\mathrm{x}\right)\mathrm{}=\mathrm{}{\mathrm{x}}^{3}-6{\mathrm{x}}^{2}+7\mathrm{x}+2\\ \mathrm{Putting}\mathrm{x}=\mathrm{A},\mathrm{we}\mathrm{get}\\ \mathrm{f}\left(\mathrm{A}\right)\mathrm{}={\mathrm{A}}^{3}\mathrm{}-6{\mathrm{A}}^{2}+7\mathrm{A}+2\mathrm{I}\\ =\left[\begin{array}{ccc}21& 0& \mathrm{}34\\ 12& 8& \mathrm{}23\\ 34& 0& 55\end{array}\right]-6\left[\begin{array}{ccc}5& 0& 8\\ 2& 4& 5\\ 8& 0& 13\end{array}\right]+7\mathrm{}\left[\begin{array}{ccc}1& 0& 2\\ 0& 2& 1\\ 2& 0& 3\end{array}\right]\mathrm{}+2\mathrm{}\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\\ =\left[\begin{array}{ccc}21& 0& \mathrm{}34\\ 12& 8& \mathrm{}23\\ 34& 0& 55\end{array}\right]+\left[\begin{array}{ccc}30& 0& -48\\ 12& -24& -30\\ -48& 0& -78\end{array}\right]+\left[\begin{array}{ccc}7& 0& 14\\ 0& 14& 7\\ 14& 0& 21\end{array}\right]\mathrm{}+\mathrm{}\left[\begin{array}{ccc}2& 0& 0\\ 0& 2& 0\\ 0& 0& 2\end{array}\right]\\ =\mathrm{}\left[\begin{array}{c}21-30+7+20 + 0 + 0 + 0 34-48+14+0\\ 12-12+0+08-24+14+223-30+7+0\\ 34-48+14+00\mathrm{}+\mathrm{}0\mathrm{}+\mathrm{}0\mathrm{}+\mathrm{}055-78+21+2\end{array}\right]\\ =\mathrm{}\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]\\ \mathrm{}\therefore \mathrm{f}\left(\mathrm{A}\right)\mathrm{}=\mathrm{}0\end{array}$

Q.42

$\begin{array}{l}\mathrm{Find}{\mathrm{A}}^{-1},\mathrm{when}\mathrm{A}=\left[\begin{array}{ccc}1& 2& -3\\ 2& 3& 2\\ 3& -3& -4\end{array}\right]\mathrm{Hence},\mathrm{solve}\mathrm{the}\mathrm{following}\mathrm{system}\mathrm{of}\mathrm{equations}:\\ \mathrm{x}+2\mathrm{y}– 3\mathrm{z}=-4\\ 2\mathrm{x}+3\mathrm{y}+2\mathrm{z}= 2\\ 3\mathrm{x}-3\mathrm{y}-4\mathrm{z}=11\end{array}$

Ans

$\begin{array}{l}\mathrm{Since}\mathrm{A}=\mathrm{IA},\mathrm{So}\\ \left[\begin{array}{ccc}1& 2& -3\\ 2& 3& 2\\ 3& -3& -4\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\mathrm{A}\\ \left[\begin{array}{ccc}1& 2& -3\\ 0& -1& 8\\ 0& -9& 5\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{ccc}1& 0& 0\\ 2& 1& 0\\ 3& 0& 1\end{array}\right]\mathrm{A}\left[\begin{array}{l}{\mathrm{R}}_{2}\to {\mathrm{R}}_{2}\mathrm{}+\mathrm{}\left(-2\right){\mathrm{R}}_{1}\\ {\mathrm{R}}_{3}\to {\mathrm{R}}_{3}\mathrm{}+\mathrm{}\left(-3\right){\mathrm{R}}_{1}\end{array}\right]\\ \left[\begin{array}{ccc}1& 2& -3\\ 0& -1& 8\\ 0& -9& -5\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{ccc}1& 0& 0\\ 2& -1& 0\\ 3& 0& -1\end{array}\right]\mathrm{A}\left[\begin{array}{l}{\mathrm{R}}_{2}\to \mathrm{}\left(-2\right){\mathrm{R}}_{2}\\ {\mathrm{R}}_{3}\to \mathrm{}\left(-3\right){\mathrm{R}}_{3}\end{array}\right]\\ \left[\begin{array}{ccc}1& 0& 13\\ 0& 1& -8\\ 0& 0& 67\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{ccc}-3& 2& \mathrm{}0\\ 2& -1& \mathrm{}0\\ -15& 9& -1\end{array}\right]\mathrm{A}\left[\begin{array}{l}{\mathrm{R}}_{1}\to {\mathrm{R}}_{1}\mathrm{}+\mathrm{}\left(-2\right){\mathrm{R}}_{2}\\ {\mathrm{R}}_{3}\to {\mathrm{R}}_{3}\mathrm{}+\mathrm{}\left(-9\right){\mathrm{R}}_{2}\end{array}\right]\\ \left[\begin{array}{ccc}1& 0& 13\\ 0& 1& -8\\ 0& 0& \mathrm{}1\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{ccc}-3& 2& \mathrm{}0\\ 2& -1& \mathrm{}0\\ \frac{-15}{67}& \frac{9}{67}& \frac{-1}{67}\end{array}\right]\mathrm{A}\left[{\mathrm{R}}_{3}\to \frac{1}{67}{\mathrm{R}}_{3}\right]\\ \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& \mathrm{}1\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{ccc}\frac{-3}{67}& \frac{17}{67}& \mathrm{}\frac{13}{67}\\ \frac{14}{67}& \frac{5}{67}& \mathrm{}\frac{-8}{67}\\ \frac{-15}{67}& \frac{9}{67}& \frac{-1}{67}\end{array}\right]\mathrm{A}\left[\begin{array}{l}{\mathrm{R}}_{1}\to {\mathrm{R}}_{1}\mathrm{}+\mathrm{}\left(-13\right){\mathrm{R}}_{3}\\ {\mathrm{R}}_{2}\to {\mathrm{R}}_{2}\mathrm{}+\mathrm{}8{\mathrm{R}}_{3}\end{array}\right]\\ \therefore {\mathrm{A}}^{-1}\mathrm{}=\mathrm{}\left[\begin{array}{ccc}\frac{-6}{67}& \frac{17}{67}& \mathrm{}\frac{13}{67}\\ \frac{14}{67}& \frac{5}{67}& \mathrm{}\frac{-8}{67}\\ \frac{-15}{67}& \frac{9}{67}& \frac{-1}{67}\end{array}\right]\end{array}$ $\begin{array}{l}\mathrm{x}+2\mathrm{y}– 3\mathrm{z}=-4\\ 2\mathrm{x}+3\mathrm{y}+2\mathrm{z}= 2\\ 3\mathrm{x}-3\mathrm{y}-4\mathrm{z}=11\\ \\ \mathrm{A}=\left[\begin{array}{ccc}1& 2& -3\\ 2& 3& 2\\ 3& -3& -4\end{array}\right],\mathrm{x}=\left[\begin{array}{c}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]\mathrm{and}\mathrm{B}\left[\begin{array}{c}-4\\ 2\\ 11\end{array}\right]\\ \mathrm{So},\\ \mathrm{}\mathrm{x}\mathrm{}={\mathrm{A}}^{-1}\mathrm{B}=\left[\begin{array}{ccc}\frac{-6}{67}& \frac{17}{67}& \mathrm{}\frac{13}{67}\\ \frac{14}{67}& \frac{5}{67}& \mathrm{}\frac{-8}{67}\\ \frac{-15}{67}& \frac{9}{67}& \frac{-1}{67}\end{array}\right]\mathrm{}\left[\begin{array}{c}-4\\ 2\\ 11\end{array}\right]\\ =\mathrm{}\frac{1}{67}\mathrm{}\left[\begin{array}{ccc}-6& 17& \mathrm{}13\\ 14& 5& \mathrm{}-8\\ -15& 9& -1\end{array}\right]\mathrm{}\left[\begin{array}{c}-4\\ 2\\ 11\end{array}\right]\\ \mathrm{}=\mathrm{}\frac{1}{67}\mathrm{}\left[\begin{array}{l}201\\ -134\\ 67\end{array}\right]\mathrm{}\left[\begin{array}{c}\frac{201}{67}\\ \frac{-134}{67}\\ \frac{67}{67}\end{array}\right]\\ \left[\begin{array}{c}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{c}3\\ -2\\ 1\end{array}\right]\mathrm{}⇒\mathrm{}\mathrm{x}\mathrm{}=\mathrm{}3,\mathrm{}\mathrm{y}\mathrm{}=\mathrm{}-2\mathrm{and}\mathrm{z}= 1\end{array}$

Q.43

$\begin{array}{l}\mathrm{Find}{\mathrm{A}}^{-1}\mathrm{by}\mathrm{using}\mathrm{elementary}\mathrm{transformations},\mathrm{where}\\ \mathrm{A}=\left[\begin{array}{ccc}2& -1& 4\\ 4& \mathrm{}0& 2\\ 3& -2& 7\end{array}\right]\end{array}$

Ans

$\begin{array}{l}\mathrm{Since}\mathrm{A}=\mathrm{IA},\mathrm{So}\\ \left[\begin{array}{ccc}2& -1& 4\\ 4& \mathrm{}0& 2\\ 3& -2& 7\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\mathrm{A}\\ \left[\begin{array}{ccc}3& -2& 7\\ 4& \mathrm{}0& 2\\ 2& -1& 4\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{ccc}0& 0& 1\\ 0& 1& 0\\ 1& 0& 0\end{array}\right]\mathrm{A}\left[{\mathrm{R}}_{1}↔{\mathrm{R}}_{3}\right]\\ \left[\begin{array}{ccc}1& -1& 3\\ 4& \mathrm{}0& 2\\ 2& -1& 4\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{ccc}-1& 0& 1\\ 0& 1& 0\\ 1& 0& 0\end{array}\right]\mathrm{A}\left[{\mathrm{R}}_{1}\to {\mathrm{R}}_{3}-{\mathrm{R}}_{3}\right]\\ \left[\begin{array}{ccc}1& -1& 3\\ 0& \mathrm{}4& -10\\ 0& 1& -2\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{ccc}-1& 0& 1\\ 4& 1& -4\\ 3& 0& -2\end{array}\right]\mathrm{A}\left[\begin{array}{l}{\mathrm{R}}_{2}\to {\mathrm{R}}_{2}\mathrm{+}\left(-4\right){\mathrm{R}}_{1}\\ {\mathrm{R}}_{3}\to {\mathrm{R}}_{3}+\left(-2\right){\mathrm{R}}_{1}\end{array}\right]\\ \left[\begin{array}{ccc}1& -1& 3\\ 0& \mathrm{}1& \mathrm{}-2\\ 0& 4& -10\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{ccc}-1& 0& 1\\ 3& 0& -2\\ 4& 1& -4\end{array}\right]\mathrm{A}\left[{\mathrm{R}}_{2}↔{\mathrm{R}}_{3}\right]\\ \left[\begin{array}{ccc}1& 0& 1\\ 0& \mathrm{}1& \mathrm{}-2\\ 0& 0& \mathrm{}-2\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{ccc}\mathrm{}2& 0& \mathrm{}-1\\ \mathrm{}3& 0& -2\\ -8& 1& \mathrm{}4\end{array}\right]\mathrm{A}\left[\begin{array}{l}{\mathrm{R}}_{1}\to {\mathrm{R}}_{1}+{\mathrm{R}}_{2}\\ {\mathrm{R}}_{3}\to {\mathrm{R}}_{3}+\left(-4\right){\mathrm{R}}_{2}\end{array}\right]\\ \left[\begin{array}{ccc}1& 0& 1\\ 0& \mathrm{}1& \mathrm{}-2\\ 0& 0& 1\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{ccc}\mathrm{}2& 0& -1\\ \mathrm{}3& 0& -2\\ 4& \frac{-1}{2}& -2\end{array}\right]\mathrm{A}\left[{\mathrm{R}}_{3}\to \mathrm{}\left(-\frac{1}{2}\right){\mathrm{R}}_{3}\right]\\ \left[\begin{array}{ccc}1& 0& \mathrm{}0\\ 0& \mathrm{}1& 0\\ 0& 0& 1\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{ccc}\mathrm{}-2& \frac{1}{2}& \mathrm{}-1\\ 11& -1& -6\\ 4& \frac{1}{2}& \mathrm{}-2\end{array}\right]\mathrm{A}\left[\begin{array}{l}{\mathrm{R}}_{1}\to {\mathrm{R}}_{1}\mathrm{}-{\mathrm{R}}_{3}\\ {\mathrm{R}}_{2}\to {\mathrm{R}}_{2}\mathrm{}+2{\mathrm{R}}_{3}\end{array}\right]\\ \mathrm{Hence},{\mathrm{A}}^{-1}\mathrm{}=\mathrm{}\left[\begin{array}{ccc}-2& \mathrm{}\frac{1}{2}& \mathrm{}-1\\ 11& -1& -6\\ 4& \frac{-1}{2}& -2\end{array}\right].\end{array}$

Q.44

$\text{If F}\left(\mathrm{x}\right)\mathrm{}=\mathrm{}\left[\begin{array}{ccc}\mathrm{cosx}& -\mathrm{sinx}& 0\\ \mathrm{sinx}& \mathrm{cosx}& 0\\ 0& 0& 1\end{array}\right],\mathrm{show}\mathrm{that}\mathrm{F}\left(\mathrm{x}\right)\mathrm{F}\left(\mathrm{y}\right)\mathrm{}=\mathrm{F}\left(\mathrm{x}+\mathrm{y}\right).$

Ans

$\begin{array}{l}\mathrm{Given}:\mathrm{F}\left(\mathrm{x}\right)=\mathrm{F}\left(\mathrm{x}\right)\mathrm{}=\mathrm{}\left[\begin{array}{ccc}\mathrm{cosx}& -\mathrm{sinx}& 0\\ \mathrm{sinx}& \mathrm{cosx}& 0\\ 0& 0& 1\end{array}\right]\\ \mathrm{F}\left(\mathrm{y}\right)\mathrm{}=\mathrm{}\left[\begin{array}{ccc}\mathrm{cosy}& -\mathrm{siny}& 0\\ \mathrm{siny}& \mathrm{cosy}& 0\\ 0& 0& 1\end{array}\right]\\ \mathrm{L}.\mathrm{H}.\mathrm{S}. :\\ \mathrm{F}\left(\mathrm{x}\right),\mathrm{F}\left(\mathrm{y}\right)\mathrm{}=\mathrm{}\left[\begin{array}{ccc}\mathrm{cosx}& -\mathrm{sinx}& 0\\ \mathrm{sinx}& \mathrm{cosx}& 0\\ 0& 0& 1\end{array}\right]\mathrm{}\left[\begin{array}{ccc}\mathrm{cosy}& -\mathrm{siny}& 0\\ \mathrm{siny}& \mathrm{cosy}& 0\\ 0& 0& 1\end{array}\right]\\ \mathrm{}=\mathrm{}\left[\begin{array}{c}\mathrm{cosxcosy}-\mathrm{sinx}\mathrm{}\mathrm{csiny}-\mathrm{sinycosx}\mathrm{}-\mathrm{sinxcosy}0\\ \mathrm{sinxcosy}+\mathrm{cosx}\mathrm{}\mathrm{siny}-\mathrm{sinxsiny}+\mathrm{cosxcosy}0\\ 001\end{array}\right]\\ \mathrm{}=\mathrm{}\left[\begin{array}{c}\mathrm{cos}\left(\mathrm{x}+\mathrm{y}\right)-\mathrm{sin}\left(\mathrm{x}+\mathrm{y}\right)0\\ \mathrm{sin}\left(\mathrm{x}+\mathrm{y}\right)-\mathrm{cos}\left(\mathrm{x}+\mathrm{y}\right)0\\ 001\end{array}\right]\\ \mathrm{}=\mathrm{F}\left(\mathrm{x}+\mathrm{y}\right)\\ \mathrm{}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.45

$\mathrm{If}\mathrm{x}\left[\begin{array}{c}5\\ 6\end{array}\right]\mathrm{}+\mathrm{}\mathrm{y}\mathrm{}\left[\begin{array}{c}-1\\ \mathrm{}1\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{c}15\\ 7\end{array}\right],\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{x}\mathrm{and}\mathrm{y}.$

Ans

$\begin{array}{l}\mathrm{Given}:\\ \mathrm{x}\left[\begin{array}{c}5\\ 6\end{array}\right]\mathrm{}+\mathrm{}\mathrm{y}\mathrm{}\left[\begin{array}{c}-1\\ 1\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{c}15\\ 7\end{array}\right]\\ ⇒\mathrm{}\left[\begin{array}{c}5\mathrm{x}\\ 6\mathrm{x}\end{array}\right]\mathrm{}+\mathrm{}\mathrm{y}\mathrm{}\left[\begin{array}{c}-\mathrm{y}\\ \mathrm{y}\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{c}15\\ 7\end{array}\right]\\ ⇒\mathrm{}\left[\begin{array}{c}5\mathrm{x}-\mathrm{y}\\ 6\mathrm{x}+\mathrm{y}\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{c}15\\ 7\end{array}\right]\\ ⇒\mathrm{}5\mathrm{x}-\mathrm{y}=\mathrm{}15,\mathrm{}6\mathrm{x}\mathrm{}+\mathrm{}\mathrm{y}\mathrm{}=7\\ ⇒\mathrm{}\mathrm{x}=2\mathrm{and}\mathrm{y}\mathrm{}=-5\end{array}$

Q.46

$\mathrm{Find}\mathrm{X},\mathrm{if}\mathrm{Y}=\left[\begin{array}{cc}\begin{array}{l}5\\ 1\end{array}& \begin{array}{l}\mathrm{}7\\ \mathrm{}3\end{array}\end{array}\right]\mathrm{and}2\mathrm{X}+\mathrm{Y}=\left[\begin{array}{cc}\begin{array}{l}1\\ -3\end{array}& \begin{array}{l}\mathrm{}0\\ \mathrm{}2\end{array}\end{array}\right]\mathrm{.}$

Ans

$\begin{array}{l}\mathrm{Given},\mathrm{Y}=\left[\begin{array}{cc}\begin{array}{l}5\\ 1\end{array}& \begin{array}{l}\mathrm{}7\\ \mathrm{}4\end{array}\end{array}\right]\\ \mathrm{and}2\mathrm{X}+\mathrm{Y}=\left[\begin{array}{cc}\begin{array}{l}1\\ -3\end{array}& \begin{array}{l}\mathrm{}0\\ \mathrm{}2\end{array}\end{array}\right].\dots \dots \left(\mathrm{i}\right)\\ \mathrm{Putting}\mathrm{value}\mathrm{of}\mathrm{Y}\mathrm{in}\mathrm{equation}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get}\\ 2\mathrm{X}+\left[\begin{array}{cc}\begin{array}{l}5\\ 1\end{array}& \begin{array}{l}\mathrm{}7\\ \mathrm{}4\end{array}\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{cc}\begin{array}{l}1\\ -3\end{array}& \begin{array}{l}\mathrm{}0\\ \mathrm{}2\end{array}\end{array}\right]\mathrm{}\\ 2\mathrm{X}=\left[\begin{array}{cc}\begin{array}{l}1\\ -3\end{array}& \begin{array}{l}\mathrm{}0\\ \mathrm{}2\end{array}\end{array}\right]-\left[\begin{array}{cc}\begin{array}{l}5\\ 1\end{array}& \begin{array}{l}\mathrm{}7\\ \mathrm{}3\end{array}\end{array}\right]\\ =\mathrm{}\left[\begin{array}{cc}1& -5\\ -3& -1\end{array}\begin{array}{cc}0& -7\\ 2& -4\end{array}\right]\\ \mathrm{X}=\frac{1}{2}\mathrm{}\left[\begin{array}{cc}\begin{array}{l}-4\\ -4\end{array}& \begin{array}{l}\mathrm{}-7\\ \mathrm{}-2\end{array}\end{array}\right]\\ =\mathrm{}\left[\begin{array}{cc}-2& -\frac{7}{2}\end{array}\right]\mathrm{}\end{array}$

Q.47

$\begin{array}{l}\mathrm{Using}\mathrm{elementary}\mathrm{transformations},\mathrm{find}\mathrm{the}\mathrm{inverse}\\ \mathrm{of}\mathrm{the}\mathrm{matrix}\left[\begin{array}{cc}\begin{array}{l}1\\ 2\end{array}& \begin{array}{l}-2\\ 3\end{array}\end{array}\right]\mathrm{if}\mathrm{it}\mathrm{exists}.\end{array}$

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{A}=\left[\begin{array}{l}\begin{array}{cc}1& \end{array}-5\\ 33\end{array}\right]\\ \mathrm{Since},\mathrm{A}=\mathrm{IA}\\ \left[\begin{array}{cc}\begin{array}{l}1\\ 3\end{array}& \begin{array}{l}-5\\ 3\end{array}\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{cc}\begin{array}{l}1\\ 0\end{array}& \begin{array}{l}\mathrm{}0\\ \mathrm{}1\end{array}\end{array}\right]\mathrm{A}\\ \mathrm{Apply}{\mathrm{R}}_{2}\mathrm{}\to {\mathrm{R}}_{2}-3{\mathrm{R}}_{1}\\ \mathrm{}\left[\begin{array}{cc}\begin{array}{l}1\\ 3\end{array}& \begin{array}{l}-5\\ 18\end{array}\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{cc}\begin{array}{l}1\\ -3\end{array}& \begin{array}{l}\mathrm{}0\\ \mathrm{}1\end{array}\end{array}\right]\mathrm{A}\\ {\mathrm{R}}_{2}\to \mathrm{}\frac{1}{18}{\mathrm{R}}_{2}\\ \therefore \left[\begin{array}{cc}\begin{array}{l}1\\ 0\end{array}& \begin{array}{l}-5\\ 1\end{array}\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{cc}\begin{array}{l}\mathrm{}1\\ -\frac{1}{6}\end{array}& \begin{array}{l}\mathrm{}0\\ \mathrm{}\frac{1}{18}\end{array}\end{array}\right]\mathrm{A}\\ \mathrm{Apply}{\mathrm{R}}_{1}\mathrm{}\to {\mathrm{R}}_{1}\mathrm{}+\mathrm{}5{\mathrm{R}}_{2}\\ \mathrm{}\left[\begin{array}{cc}\begin{array}{l}1\\ 0\end{array}& \begin{array}{l}0\\ 1\end{array}\end{array}\right]\mathrm{}=\mathrm{}\left[\begin{array}{cc}\begin{array}{l}\mathrm{}\frac{1}{6}\\ -\frac{1}{6}\end{array}& \begin{array}{l}\mathrm{}\frac{5}{18}\\ \mathrm{}\frac{1}{18}\end{array}\end{array}\right]\mathrm{A}\\ {\mathrm{A}}^{-1}=\left[\begin{array}{cc}\begin{array}{l}\mathrm{}\frac{1}{6}\\ -\frac{1}{6}\end{array}& \begin{array}{l}\mathrm{}\frac{5}{18}\\ \mathrm{}\frac{1}{18}\end{array}\end{array}\right]\end{array}$