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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}.

	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,

Addition:

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 = xij– yijm 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.

(c) Existence of additive identity:

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.

(d) Existence of additive inverse:

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:

Non-Commutative Law:

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

Associative Law:

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

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:

If k is a scalar quantity, then k(A + B) = kA + kB.
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:

(PT )T = P
(kP)T =kPT
(A + B)T =AT + BT
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.

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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

$\begin{array}{l} Obtain the inverse of the matrix A = [\begin{array}{l} 1 0 4 \\ 1 1 6 \\ - 3 0 - 10 \end{array}] \\ using row operatins . \end{array}$

Ans

$\begin{array}{l} Write A = IA, ie ., \\ [\begin{array}{l} 1 0 4 \\ 1 1 6 \\ - 3 0 - 10 \end{array}] = [\begin{array}{l} 1 0 0 \\ 0 1 0 \\ 0 0 1 \end{array}] A \\ One applying R_{2} \to R_{2} - R_{1} and R_{3} \to R_{3} - R_{1}, we get \\ [\begin{array}{l} 1 0 4 \\ 0 1 2 \\ 0 0 2 \end{array}] = [\begin{array}{l} 1 0 0 \\ - 1 1 0 \\ 3 0 1 \end{array}] A \\ On applying R_{3} \to \frac{1}{2} R_{3}, \\ [\begin{array}{l} 1 0 4 \\ 0 1 2 \\ 0 0 1 \end{array}] = [\begin{array}{l} 1 0 0 \\ - 1 1 0 \\ \frac{3}{2} 0 \frac{1}{2} \end{array}] A \\ On applying R_{1} \to R_{1} - 4 R_{3} and R_{2} \to R_{2} - 2 R_{3}, we get \\ [\begin{array}{l} 1 0 0 \\ 0 1 0 \\ 0 0 1 \end{array}] = [\begin{array}{l} - 5 0 - 2 \\ - 4 1 - 1 \\ \frac{3}{2} 0 \frac{1}{2} \end{array}] A \\ Hence, A^{- 1} = [\begin{array}{l} - 5 0 - 2 \\ - 4 1 - 1 \\ \frac{3}{2} 0 \frac{1}{2} \end{array}] \end{array}$

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

$Find the matrix x so that x [\begin{array}{l} 1 2 3 \\ 4 5 6 \end{array}] = [\begin{array}{l} - 7 - 8 - 9 \\ 2 4 6 \end{array}]$

Ans

$\begin{array}{l} Let x = [\begin{array}{l} a b \\ c d \end{array}] \\ ∴ [\begin{array}{l} a b \\ c d \end{array}] [\begin{array}{l} 1 2 3 \\ 4 5 6 \end{array}] = [\begin{array}{l} - 7 - 8 - 9 \\ 2 4 6 \end{array}] \\ \Rightarrow [\begin{array}{l} a \times 1 + b \times 4 a \times 2 + b \times 5 a \times 3 + b \times 6 \\ c \times 1 + d \times 4 c \times 2 + d \times 5 c \times 3 + d \times 6 \end{array}] = [\begin{array}{l} - 7 - 8 - 9 \\ 2 4 6 \end{array}] \\ \Rightarrow [\begin{array}{l} a + 4 b 2 a + 5 b 3 a + 6 b \\ c + 4 b 2 c + 5 b 3 c + 6 b \end{array}] = [\begin{array}{l} - 7 - 8 - 9 \\ 2 4 6 \end{array}] \\ By using equality and solving \\ \begin{array}{l} a + 4 b = - 7 \\ 2 a + 5 b = -8 \end{array}\} . \dots\dots (1) \begin{array}{l} c + 4 d = 2 \\ 2 c + 5 d = 4 \end{array}\} . \dots\dots (2) \\ Now, by (1) a = 1 b = - 2 \\ by (2) c = 2 d = 0 \\ Hence x = [\begin{array}{l} 1 - 2 \\ 2 0 \end{array}] \end{array}$

Q.4

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

Ans

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

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

$\begin{array}{l} Express A = [\begin{array}{l} 3 3 - 1 \\ - 2 - 2 1 \\ - 4 - 5 2 \end{array}] as the sum of a symmetric \\ and a skew - symmetric matrix . \end{array}$

Ans

$\begin{array}{l} Here A ‘ = [\begin{array}{l} 3 - 2 - 4 \\ 3 - 2 - 5 \\ - 1 1 2 \end{array}] \\ Let P = \frac{1}{2} (A + A ‘) \\ ∴ \frac{1}{2} (A + A ‘) = \frac{1}{2} [\begin{array}{l} 3 3 - 1 \\ - 2 - 2 1 \\ - 4 - 5 2 \end{array}] + [\begin{array}{l} 3 - 2 - 4 \\ 3 - 2 - 5 \\ - 1 1 2 \end{array}] \\ = \frac{1}{2} [\begin{array}{l} 6 1 - 5 \\ 1 - 4 - 4 \\ - 5 - 4 4 \end{array}] = [\begin{array}{l} 3 \frac{1}{2} \frac{- 5}{2} \\ \frac{1}{2} - 2 - 2 \\ \frac{- 5}{2} - 2 2 \end{array}] \\ Now P ‘ = [\begin{array}{l} 3 \frac{1}{2} \frac{- 5}{2} \\ \frac{1}{2} - 2 - 2 \\ \frac{- 5}{2} - 2 2 \end{array}] = P \\ Thus P = \frac{1}{2} [A - A ‘] \\ \frac{1}{2} [A - A ‘] = \frac{1}{2} [[\begin{array}{l} 3 3 - 1 \\ - 2 - 2 1 \\ - 4 - 5 2 \end{array}] - [\begin{array}{l} 3 - 2 - 4 \\ 3 - 2 - 5 \\ - 1 1 2 \end{array}]] \\ = \frac{1}{2} [[\begin{array}{l} 0 5 3 \\ - 2 0 6 \\ - 3 - 6 0 \end{array}] = [\begin{array}{l} 0 \frac{5}{2} \frac{3}{2} \\ - \frac{5}{2} - 3 0 \\ - \frac{3}{2} - 3 0 \end{array}]] \\ Then Q ‘ = [\begin{array}{l} 0 - \frac{5}{2} - \frac{3}{2} \\ \frac{5}{2} 0 - 3 \\ \frac{3}{2} 3 0 \end{array}] = - Q \\ Thus Q = \frac{1}{2} (A - A ‘) is a skew symmetric matrix . \\ Now, P + Q = [\begin{array}{l} 3 \frac{1}{2} \frac{- 5}{2} \\ \frac{1}{2} - 2 - 2 \\ \frac{- 5}{2} - 2 2 \end{array}] + [\begin{array}{l} 0 \frac{5}{2} \frac{3}{2} \\ - \frac{5}{2} 0 3 \\ - \frac{3}{2} - 3 0 \end{array}] = [\begin{array}{l} 3 3 - 1 \\ - 2 - 2 1 \\ - 4 - 5 2 \end{array}] = A \end{array}$

Q.7

$If A = [\begin{array}{l} - 2 \\ 4 \\ 5 \end{array}], B = [1 3 - 6], verify that (AB) ‘ = B ‘ A ‘ .$

Ans

$\begin{array}{l} Here A = [\begin{array}{l} - 2 \\ 4 \\ 5 \end{array}], B = [1 3 - 6] \\ AB = [\begin{array}{l} - 2 \\ 4 \\ 5 \end{array}] [1 3 - 6] = [\begin{array}{l} - 2 - 6 12 \\ 4 12 - 24 \\ 5 15 - 30 \end{array}] \\ (AB ‘) = [\begin{array}{l} - 2 4 5 \\ - 6 12 15 \\ 12 - 24 - 30 \end{array}] \\ Now, A ‘ = [- 2 4 5], B ‘ = [\begin{array}{l} 1 \\ 3 \\ - 6 \end{array}] \\ B ‘ A ‘ = [\begin{array}{l} 1 \\ 3 \\ - 6 \end{array}] [- 2 4 5] = [\begin{array}{l} - 2 4 5 \\ - 6 12 15 \\ 12 - 24 - 30 \end{array}] = (AB) ‘ \\ Hence (AB) ‘ = B ‘ A ‘ \end{array}$

Q.8

$If A = [\begin{array}{l} 1 0 \\ - 1 7 \end{array}] and I = [\begin{array}{l} 1 0 \\ 0 1 \end{array}], then find k so that A^{2} = 8 A + kI .$

Ans

$\begin{array}{l} A^{2} = AA = [\begin{array}{l} 1 0 \\ - 1 7 \end{array}] [\begin{array}{l} 1 0 \\ - 1 7 \end{array}] \\ = [\begin{array}{l} 1 + 0 0 + 0 \\ - 1 - 7 0 + 49 \end{array}] = [\begin{array}{l} 1 0 \\ - 8 49 \end{array}] \\ 8 A + KI = 8 [\begin{array}{l} 1 0 \\ - 1 7 \end{array}] + K [\begin{array}{l} 1 0 \\ 0 1 \end{array}] = [\begin{array}{l} 8 + k 0 \\ - 8 56 + k \end{array}] \\ Given A^{2} = 8 A + kI \\ \Rightarrow [\begin{array}{l} 1 0 \\ - 8 49 \end{array}] = [\begin{array}{l} 8 + k 0 \\ - 8 56 + k \end{array}] \\ \Rightarrow 8 + k = 1 and 56+ k =49 \\ \Rightarrow k = - 7 \end{array}$

Q.9

$Find x and y if x + y = [\begin{array}{l} 5 2 \\ 0 9 \end{array}] and x - y = [\begin{array}{l} 3 6 \\ 0 - 1 \end{array}]$

Ans

$\begin{array}{l} We have \\ (x + y) + (x - y) = [\begin{array}{l} 5 2 \\ 0 9 \end{array}] + [\begin{array}{l} 3 6 \\ 0 - 1 \end{array}] \\ 2 x = [\begin{array}{l} 5 + 3 6 + 2 \\ 0 + 0 9 - 1 \end{array}] = [\begin{array}{l} 8 8 \\ 0 8 \end{array}] \\ \Rightarrow x = \frac{1}{2} [\begin{array}{l} 8 8 \\ 0 8 \end{array}] \\ \Rightarrow x = [\begin{array}{l} 4 4 \\ 0 4 \end{array}] \\ Now (x + y) (x - y) = [\begin{array}{l} 5 2 \\ 0 9 \end{array}] - [\begin{array}{l} 3 6 \\ 0 1 \end{array}] \\ 2 y = [\begin{array}{l} 5 - 3 2 - 6 \\ 0 - 0 9 - 1 \end{array}] \\ \Rightarrow y = \frac{1}{2} [\begin{array}{l} 2 - 4 \\ 0 8 \end{array}] \\ \Rightarrow y = [\begin{array}{l} 1 - 2 \\ 0 4 \end{array}] \\ Hence x = [\begin{array}{l} 4 4 \\ 0 4 \end{array}] y = \frac{1}{2} [\begin{array}{l} 1 - 2 \\ 0 4 \end{array}] \end{array}$

Q.10

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

Ans

$\begin{array}{l} We have A + X = B \\ \Rightarrow [\begin{matrix} \begin{matrix} 8 \\ 4 \\ 3 \end{matrix} & \begin{matrix} 0 \\ - 2 \\ 6 \end{matrix} \end{matrix}] + X = [\begin{matrix} \begin{matrix} 2 \\ 4 \\ - 5 \end{matrix} & \begin{matrix} - 2 \\ 2 \\ 1 \end{matrix} \end{matrix}] \\ \Rightarrow X = [\begin{matrix} \begin{matrix} 2 \\ 4 \\ - 5 \end{matrix} & \begin{matrix} - 2 \\ 2 \\ 1 \end{matrix} \end{matrix}] - [\begin{matrix} \begin{matrix} 8 \\ 4 \\ 3 \end{matrix} & \begin{matrix} 0 \\ - 2 \\ 6 \end{matrix} \end{matrix}] \\ = [\begin{matrix} \begin{matrix} 2 - 8 \\ 4 - 4 \\ - 5 - 3 \end{matrix} & \begin{matrix} - 2 - 0 \\ 2 + 2 \\ 1 - 6 \end{matrix} \end{matrix}] \\ = [\begin{matrix} \begin{matrix} - 6 \\ 0 \\ - 8 \end{matrix} & \begin{matrix} - 2 \\ 4 \\ - 5 \end{matrix} \end{matrix}] \end{array}$

Q.11

$Given A = [\begin{array}{l} 5 - 1 0 \\ 7 3 4 \end{array}] and B = [\begin{array}{l} - 3 2 0 \\ - 5 6 2 \end{array}] . Find A + B, A - B .$

Ans

$\begin{array}{l} Here, A + B = [\begin{array}{l} 5 - 1 0 \\ 7 3 4 \end{array}] + [\begin{array}{l} - 3 2 0 \\ - 5 6 2 \end{array}] \\ = [\begin{array}{l} 2 1 0 \\ 2 9 6 \end{array}] \\ A - B = [\begin{array}{l} 5 - 1 0 \\ 7 3 4 \end{array}] - [\begin{array}{l} - 3 2 0 \\ - 5 6 2 \end{array}] \\ = [\begin{array}{l} 5 + 3 - 1 - 2 0 - 0 \\ 7 + 3 3 - 6 4 - 2 \end{array}] \\ = [\begin{array}{l} 8 - 3 0 \\ 12 - 3 2 \end{array}] \end{array}$

Q.12

$If [\begin{array}{l} x + y 4 \\ 7 xy \end{array}] = [\begin{array}{l} 6 4 \\ 7 8 \end{array}], find the values of x and y .$

Ans

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

Q.13

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

Ans

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

Q.14

$\begin{array}{l} Consider the matrix \\ A = [\begin{array}{l} 2 5 \sqrt{7} 3 \\ 35 6 \frac{5}{3} - 1 \\ \sqrt{3} 1 6 0 \end{array}] \end{array}$

Ans

(i) Order of the matrix = 3

(ii) a₁₃ = 7, a₂₁ = 35

a₂₃ = 5/3, a₃₄ = 0

Q.15

$\begin{array}{l} If A = [\begin{array}{l} 31 \\ 7 5 \end{array}], find x and y such that A^{2} + XI = YA . \\ Hence find A^{- 1} . \end{array}$

Ans

$\begin{array}{l} Given A = [\begin{array}{l} 31 \\ 7 5 \end{array}] \\ ∴ A^{2} = A . A = + [\begin{array}{l} 31 \\ 7 5 \end{array}] [\begin{array}{l} 31 \\ 7 5 \end{array}] \\ \Rightarrow A^{2} = [\begin{array}{l} 9 +7 3+5 \\ 21 + 35 7 + 25 \end{array}] \\ \Rightarrow A^{2} = [\begin{array}{l} 168 \\ 56 32 \end{array}] . \dots.. (1) \\ and xI = x [\begin{array}{l} 10 \\ 0 1 \end{array}] = [\begin{array}{l} x 0 \\ 0 x \end{array}] \\ RHS = yA = y [\begin{array}{l} 31 \\ 7 5 \end{array}] = [\begin{array}{l} 3 y y \\ 7 y 5 y \end{array}] . \dots.. (2) \\ LHS = A^{2} + xI = [\begin{array}{l} 168 \\ 56 32 \end{array}] + [\begin{array}{l} x 0 \\ 0 x \end{array}] \\ By using equality of matrices \\ 16+ x = 3 y \\ \Rightarrow 56 = 7 y \\ \Rightarrow y = 8 \\ And, x = 8 \\ Substituting the values of x and y, we get \\ A^{2} + 8 I = 8 A \\ \Rightarrow 8 I = 8 A - A^{2} \\ \Rightarrow I = \frac{1}{8} [8 A - A^{2}] \\ \Rightarrow {IA}^{- 1} = \frac{1}{8} [8 {AA}^{- 1} - A ({AA}^{- 1})] \\ \Rightarrow {IA}^{- 1} = \frac{1}{8} [8 I - AI] \\ \Rightarrow {IA}^{- 1} = \frac{1}{8} [8 {AA}^{- 1} - A ({AA}^{- 1})] \\ \Rightarrow {IA}^{- 1} = \frac{1}{8} [8 I - AI] \\ \Rightarrow = \frac{1}{8} [8 I - A] \\ \Rightarrow = \frac{1}{8} [[\begin{array}{l} 8 0 \\ 08 \end{array}] - [\begin{array}{l} 31 \\ 75 \end{array}]] \\ \Rightarrow = \frac{1}{8} [\begin{array}{l} 5 -1 \\ - 73 \end{array}] \\ \Rightarrow x = 8, y = 8 and A^{- 1} = \frac{1}{8} [\begin{array}{l} 5 -1 \\ - 73 \end{array}] \end{array}$

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

$\begin{array}{l} If A and B are two equal square matrices such that \\ A = [\begin{array}{l} x y \\ 23 \end{array}] and B = [\begin{array}{l} 14 \\ 2 3 \end{array}], then find the values of x and y . \end{array}$

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

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

Ans

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

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

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

Ans

$\begin{array}{l} A + B = [\begin{array}{l} 10 \\ 24 \end{array}] + [\begin{array}{l} 2 -1 \\ 37 \end{array}] \\ = [\begin{array}{l} 1 +20 - 7 \\ 2 +34 +7 \end{array}] \\ = [\begin{array}{l} 3 - 1 \\ 5 11 \end{array}] \end{array}$

Q.24

$If A = [\begin{array}{l} 1 3 5 \\ 2 4 6 \\ 3 5 7 \end{array}], find the diagonal elements of A .$

Ans

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

Q.25

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

Ans

$\begin{array}{l} AB = [\begin{array}{l} a 0 \\ 0 - a \end{array}] + [\begin{array}{l} a 0 \\ 0 - a \end{array}] = [\begin{array}{l} a \times 0 + 0 \times a a \times a + 0 \times 0 \\ 0 \times 0 + (- a) \times a 0 \times a + (- a) \times 0 \end{array}] \\ = [\begin{array}{l} 0 a^{2} \\ - a^{2} 0 \end{array}] \end{array}$

Q.26

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

Ans

$\begin{array}{l} 2 A + B =2 [\begin{array}{l} 10 \\ 0 -1 \end{array}] + [\begin{array}{l} a 0 \\ 0 - a \end{array}] = [\begin{array}{l} 2 + a 0 \\ 0 -2 - a \end{array}] \\ Given, 2 A + B = 0 \\ = [\begin{array}{l} 2 + a 0 \\ 0 - 2 a \end{array}] \\ \Rightarrow a = - 2 \end{array}$

Q.27

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

Ans

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

Q.28

$\begin{array}{l} Let A = [\begin{array}{l} 0 - \tan \frac{α}{2} \\ \tan \frac{α}{2} 0 \end{array}] and I be the identity matrix of order 2. \\ show that : I + A (I - A) [\begin{array}{l} cosα - sinα \\ sinα cosα \end{array}] \end{array}$

Ans

$\begin{array}{l} I + A = [\begin{array}{l} 1 0 \\ 0 1 \end{array}] + [\begin{array}{l} 0 - \tan \frac{α}{2} \\ \tan \frac{α}{2} 0 \end{array}] = [\begin{array}{l} 1 - \tan \frac{α}{2} \\ \tan \frac{α}{2} 1 \end{array}] \\ I - A = [\begin{array}{l} 1 0 \\ 0 1 \end{array}] - [\begin{array}{l} 0 - \tan \frac{α}{2} \\ \tan \frac{α}{2} 0 \end{array}] = [\begin{array}{l} 1 \tan \frac{α}{2} \\ - \tan \frac{α}{2} 1 \end{array}] \\ ∴ (I - A) [\begin{array}{l} cosα - sinα \\ sinα cosα \end{array}] = [\begin{array}{l} 1 \tan \frac{α}{2} \\ - \tan \frac{α}{2} 1 \end{array}] [\begin{array}{l} cosα - sinα \\ sinα cosα \end{array}] \\ \Rightarrow (I - A) [\begin{array}{l} cosα - sinα \\ sinα cosα \end{array}] = [\begin{array}{l} 1 \tan \frac{α}{2} \\ - \tan \frac{α}{2} 1 \end{array}] [\frac{\frac{1 - \tan^{2} \frac{α}{2}}{1 + \tan^{2} \frac{α}{2}} - \frac{2 \tan \frac{α}{2}}{1 + \tan^{2} \frac{α}{2}}}{\frac{2 \tan \frac{α}{2}}{1 + \tan^{2} \frac{α}{2}} \frac{1 - \tan^{2} \frac{α}{2}}{1 + \tan^{2} \frac{α}{2}}}] \\ \Rightarrow (I - A) [\begin{array}{l} cosα - sinα \\ sinα cosα \end{array}] = [\begin{array}{l} 1 t \\ - t 1 \end{array}] [\frac{\frac{1 - t^{2}}{1 + t^{2}} - \frac{2 t}{1 + t^{2}}}{\frac{2 t}{1 + t^{2}} \frac{1 - t^{2}}{1 + t^{2}}}], where t = \frac{tanα}{2} \\ \Rightarrow (I - A) [\begin{array}{l} cosα - sinα \\ sinα cosα \end{array}] = [\frac{\frac{1 - t^{2} + 2 t^{2}}{1 + t^{2}} \frac{- 2 t + t - t^{3}}{1 + t^{2}}}{\frac{- t + t^{3} + 2 t}{1 + t^{2}} \frac{2 t^{2} + 1 - t^{2}}{1 + t^{2}}}] \\ \Rightarrow (I - A) [\begin{array}{l} cosα - sinα \\ sinα cosα \end{array}] = [\frac{\frac{1 - t^{2}}{1 + t^{2}} \frac{- t + (1 - t^{2})}{1 + t^{2}}}{\frac{t (1 + t^{2})}{1 + t^{2}} \frac{1 + t^{2}}{1 + t^{2}}}] = [\begin{array}{l} 1 - t \\ t 1 \end{array}] \\ \Rightarrow (I - A) [\begin{array}{l} cosα - sinα \\ sinα cosα \end{array}] = [\begin{array}{l} 1 - \tan \frac{α}{2} \\ \tan \frac{α}{2} 1 \end{array}] = I + A \end{array}$

Q.29

$Find the matrix x so that x [\begin{array}{l} 1 2 3 \\ 4 5 6 \end{array}] = [\begin{array}{l} - 7 -8 -9 \\ 4 5 6 \end{array}] .$

Ans

$\begin{array}{l} Let x = [\begin{array}{l} a b \\ c d \end{array}] \\ ∴ [\begin{array}{l} a b \\ c d \end{array}] [\begin{array}{l} 1 2 3 \\ 4 5 6 \end{array}] = [\begin{array}{l} - 7 -8 -9 \\ 4 5 6 \end{array}] \\ \Rightarrow [\begin{array}{l} a \times 1 + b \times 4 a \times 2 + b \times 5 a \times 3 + b \times 6 \\ c \times 1 + d \times 4 c \times 2 + d \times 5 c \times 3 + d \times 6 \end{array}] = [\begin{array}{l} - 7 -8 -9 \\ 4 5 6 \end{array}] \\ \Rightarrow [\begin{array}{l} a + 4 b 2 a + 5 b 3 a + 6 b \\ c + 4 d 2 c + 5 d 3 c + 6 d \end{array}] = [\begin{array}{l} - 7 -8 -9 \\ 2 4 6 \end{array}] \\ By using equality and solving \\ \begin{array}{l} a +4 b = -7 \\ 2 a +5 b = - 8 \end{array}\} . \dots\dots.. (1) \begin{array}{l} c +4 d = 2 \\ 2 c +5 d = 4 \end{array}\} . \dots\dots.. (2) \\ Now, by (1) a = 1 b = - 2 \\ by (2) c = 2 d = 0 \\ Hence x = [\begin{array}{l} 1 - 2 \\ 2 0 \end{array}] \end{array}$

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

$\begin{array}{l} If A and B are two equal square matrices such that \\ A = [\begin{array}{l} x y \\ 2 3 \end{array}] and B = [\begin{array}{l} 1 4 \\ 2 3 \end{array}], then find the values of x and y . \end{array}$

Ans

$x = 1 and y = 4 (Equality of matrices property) .$

Q.33

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

Ans

$A = [\begin{array}{l} \frac{1}{1} \frac{1}{2} \\ \frac{2}{1} \frac{2}{2} \end{array}] = [\begin{array}{l} 1 \frac{1}{2} \\ 2 1 \end{array}]$

Q.34

$\begin{array}{l} If A = [\begin{array}{l} 1 2 3 \\ 3 -2 1 \\ 4 2 1 \end{array}], that show that A^{3} - 23 A - 40 I = 0 . \end{array}$

Ans

$\begin{array}{l} We have A^{2} = A . A = [\begin{array}{l} 1 2 3 \\ 3 -2 1 \\ 4 2 1 \end{array}] [\begin{array}{l} 1 2 3 \\ 3 -2 1 \\ 4 2 1 \end{array}] \\ = [\begin{array}{l} 19 4 8 \\ 1 12 8 \\ 14 6 15 \end{array}] \\ So, A^{3} = A^{2} . A = [\begin{array}{l} 19 4 8 \\ 1 12 8 \\ 14 6 15 \end{array}] [\begin{array}{l} 1 2 3 \\ 3 -2 1 \\ 4 2 1 \end{array}] \\ = [\begin{array}{l} 63 46 69 \\ 69 -6 23 \\ 92 46 63 \end{array}] \\ Now, \\ A^{3} -23 A -40 I = [\begin{array}{l} 63 46 69 \\ 69 -6 23 \\ 92 46 63 \end{array}] - 23 [\begin{array}{l} 1 2 3 \\ 3 -2 1 \\ 4 2 1 \end{array}] - 40 [\begin{array}{l} 1 0 0 \\ 0 1 0 \\ 0 0 1 \end{array}] \\ = [\begin{array}{l} 63 46 69 \\ 69 -6 23 \\ 92 46 63 \end{array}] + [\begin{array}{l} - 23 -46 -69 \\ - 69 46 -23 \\ - 92 -46 -23 \end{array}] + [\begin{array}{l} - 40 0 0 \\ 0 -40 0 \\ 0 0 -40 \end{array}] \\ = [\begin{array}{l} 63 -23 -40 46-46+0 69-69+0 \\ 69 -69 +0 -6+46-40 23-23+0 \\ 92 -92 +0 46-46+0 63-23-40 \end{array}] \\ = [\begin{array}{l} 0 0 0 \\ 0 0 0 \\ 0 0 0 \end{array}] = 0 = RHS \end{array}$

Q.35

$If A = [\begin{array}{l} 1 0 2 \\ 0 2 1 \\ 2 0 3 \end{array}], prove that A^{3} - 6 A^{2} + 7 A + 2 I = 0 .$

Ans

$\begin{array}{l} We have, A = [\begin{array}{l} 1 0 2 \\ 0 2 1 \\ 2 0 3 \end{array}] \\ A^{2} = A . A = [\begin{array}{l} 1 0 2 \\ 0 2 1 \\ 2 0 3 \end{array}] . [\begin{array}{l} 1 0 2 \\ 0 2 1 \\ 2 0 3 \end{array}] = [\begin{array}{l} 5 0 8 \\ 2 4 5 \\ 8 0 13 \end{array}] \\ A^{3} = A^{2} . A = [\begin{array}{l} 5 0 8 \\ 2 4 5 \\ 8 0 13 \end{array}] . [\begin{array}{l} 1 0 2 \\ 0 2 1 \\ 2 0 3 \end{array}] = [\begin{array}{l} 21 0 34 \\ 12 8 23 \\ 34 0 55 \end{array}] \\ LHS : \\ A^{3} - 6 A^{2} + 7 A + 2 I \\ = [\begin{array}{l} 21 0 34 \\ 12 8 23 \\ 34 0 55 \end{array}] - 6 [\begin{array}{l} 5 0 8 \\ 2 4 5 \\ 8 0 13 \end{array}] + 7 [\begin{array}{l} 1 0 2 \\ 0 2 1 \\ 2 0 3 \end{array}] + 2 [\begin{array}{l} 1 0 0 \\ 0 1 0 \\ 0 0 1 \end{array}] \\ = [\begin{array}{l} 21 0 34 \\ 12 8 23 \\ 34 0 55 \end{array}] + [\begin{array}{l} 30 0 -48 \\ - 12 -24 -30 \\ - 48 0 -78 \end{array}] + [\begin{array}{l} 7 0 14 \\ 0 14 7 \\ 14 0 21 \end{array}] + [\begin{array}{l} 2 0 0 \\ 0 2 0 \\ 0 0 2 \end{array}] \\ = [\begin{array}{l} 21 -30+7+2 0+0+0+0 34-48+14+0 \\ 12 -12+0+0 8-24+14+2 23-30+7+0 \\ 34 -48+14+0 0+0+0+0 55-78+21+2 \end{array}] \\ = [\begin{array}{l} 0 0 0 \\ 0 0 0 \\ 0 0 0 \end{array}] \\ = RHS \end{array}$

Q.36

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

Ans

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

Q.37

$\begin{array}{l} Let A = [\begin{array}{l} 0 1 \\ 0 0 \end{array}], show that {(aI + bA)}^{n} = a^{n} I + {na}^{n - 1} bA, where I is the identity \\ matrix of order 2 and n \in N . \end{array}$

Ans

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

Q.38

$\begin{array}{l} Find the inverse of the following matrix, if it exists . \\ [\begin{matrix} \begin{array}{l} 2 - 3 3 \\ 2 2 3 \end{array} \\ 3 - 2 2 \end{matrix}] \end{array}$

Ans

$\begin{array}{l} Let A = [\begin{matrix} \begin{array}{l} 2 - 3 3 \\ 2 2 3 \end{array} \\ 3 - 2 2 \end{matrix}] \\ We know that A = IA \\ ∴ [\begin{matrix} \begin{array}{l} 2 - 3 3 \\ 2 2 3 \end{array} \\ 3 - 2 2 \end{matrix}] = [\begin{matrix} \begin{array}{l} 1 0 0 \\ 0 1 0 \end{array} \\ 0 0 1 \end{matrix}] A \\ \Rightarrow [\begin{matrix} \begin{array}{l} 2 - 3 3 \\ 0 5 0 \end{array} \\ 3 - 2 2 \end{matrix}] = [\begin{matrix} \begin{array}{l} 1 0 0 \\ - 1 1 0 \end{array} \\ 0 0 1 \end{matrix}] A (R_{2} \to R_{2} - R_{1}) \\ \Rightarrow [\begin{matrix} \begin{array}{l} 2 - 3 3 \\ 0 1 0 \end{array} \\ 3 - 2 2 \end{matrix}] = [\begin{matrix} \begin{array}{l} 1 0 0 \\ - \frac{1}{5} \frac{1}{5} 0 \end{array} \\ 0 0 1 \end{matrix}] A (R_{2} \to \frac{1}{5} R_{2}) \\ \Rightarrow [\begin{matrix} \begin{array}{l} - 1 - 1 1 \\ 0 1 0 \end{array} \\ 3 - 2 2 \end{matrix}] = [\begin{matrix} \begin{array}{l} 1 0 - 1 \\ - \frac{1}{5} \frac{1}{5} 0 \end{array} \\ 0 0 1 \end{matrix}] A (R_{1} \to R_{1} - R_{3}) \\ \Rightarrow [\begin{matrix} \begin{array}{l} - 1 0 1 \\ 0 1 0 \end{array} \\ 3 0 2 \end{matrix}] = [\begin{matrix} \begin{array}{l} \frac{4}{5} \frac{1}{5} - 1 \\ - \frac{1}{5} \frac{1}{5} 0 \end{array} \\ - \frac{2}{5} \frac{2}{5} 1 \end{matrix}] A (\begin{array}{l} R_{1} \to R_{1} + R_{2} and \\ R_{3} \to R_{3} + 2 R_{2} \end{array}) \\ \Rightarrow [\begin{matrix} \begin{array}{l} - 1 0 1 \\ 0 1 0 \end{array} \\ 0 0 5 \end{matrix}] = [\begin{matrix} \begin{array}{l} \frac{4}{5} \frac{1}{5} - 1 \\ - \frac{1}{5} \frac{1}{5} 0 \end{array} \\ 2 1 - 2 \end{matrix}] A (R_{3} \to R_{3} + 3 R_{1}) \\ \Rightarrow [\begin{matrix} \begin{array}{l} - 1 0 1 \\ 0 1 0 \end{array} \\ 0 0 1 \end{matrix}] = [\begin{matrix} \begin{array}{l} \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{matrix}] A (R_{3} \to \frac{1}{5} R_{3}) \\ \Rightarrow [\begin{matrix} \begin{array}{l} - 1 0 1 \\ 0 1 0 \end{array} \\ 0 0 1 \end{matrix}] = [\begin{matrix} \begin{array}{l} \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{matrix}] A (R_{3} \to R_{1} - R_{3}) \\ \Rightarrow [\begin{matrix} \begin{array}{l} 1 0 0 \\ 0 1 0 \end{array} \\ 0 0 1 \end{matrix}] = [\begin{matrix} \begin{array}{l} - \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{matrix}] A (R_{1} \to (- 1) R_{1}) \\ A^{- 1} = [\begin{matrix} \begin{array}{l} - \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{matrix}] \end{array}$

Q.39

$\begin{array}{l} Find the values of x, y, z if the matrix \\ A = [\begin{matrix} 0 2 y z \\ x y - z \\ x - y z \end{matrix}] satisfy the equation A ‘ A = I_{3} . \end{array}$

Ans

$\begin{array}{l} A = [\begin{matrix} 0 2 y z \\ x y - z \\ x - y z \end{matrix}] and A ‘ = [\begin{matrix} 0 x x \\ 2 y y - y \\ z - z z \end{matrix}] \\ Now, \\ A ‘ A = I_{3} \Rightarrow [\begin{matrix} 0 x x \\ 2 y y - y \\ z - z z \end{matrix}] . [\begin{matrix} 0 2 y z \\ x y - z \\ x - y z \end{matrix}] = [\begin{matrix} 1 0 0 \\ 0 1 0 \\ 0 0 1 \end{matrix}] \\ [\begin{matrix} 0 + x^{2} + x^{2} 0 + xy - xy 0 - xz + xz \\ 0 + yx - yx 4 y^{2} + y^{2} + y^{2} 2 yz - yz - yz \\ 0 - zx + zx 2 yz - yz - yz z^{2} + z^{2} + z^{2} \end{matrix}] = [\begin{matrix} 1 0 0 \\ 0 1 0 \\ 0 0 1 \end{matrix}] \\ [\begin{matrix} 2 x^{2} 0 0 \\ 0 6 y^{2} 0 \\ 0 0 3 z^{2} \end{matrix}] = [\begin{matrix} 1 0 0 \\ 0 1 0 \\ 0 0 1 \end{matrix}] \\ \Rightarrow 2 x^{2} = 1, 6 y^{2} = 1 and 3 z^{2} = 1 \\ \Rightarrow x = \pm \frac{1}{\sqrt{6}}, y = \pm \frac{1}{\sqrt{6}} and z = \pm \frac{1}{\sqrt{3}} \end{array}$

Q.40

$If A = [\begin{matrix} \begin{array}{l} 1 2 2 \\ 2 1 2 \end{array} \\ 2 2 1 \end{matrix}], prove that A^{2} - 4 A - 5 I = 0. Hence, find A^{- 1} .$

Ans

$\begin{array}{l} Given A = If A = [\begin{matrix} 1 2 2 \\ 2 1 2 \\ 2 2 1 \end{matrix}] \\ \Rightarrow A^{2} = [\begin{matrix} \begin{array}{l} 1 2 2 \\ 2 1 2 \end{array} \\ 2 2 1 \end{matrix}] . [\begin{matrix} \begin{array}{l} 1 2 2 \\ 2 1 2 \end{array} \\ 2 2 1 \end{matrix}] \\ A^{2} = [\begin{array}{l} 1 + 4 + 4 2 + 2 + 4 2 + 4 + 2 \\ 2 + 2 + 4 4 + 1 + 4 4 + 2 + 2 \\ 2 + 4 + 2 4 + 2 + 2 4 + 4 + 1 \end{array}] \\ = [\begin{matrix} \begin{array}{l} 9 8 8 \\ 8 9 8 \end{array} \\ 8 8 9 \end{matrix}] \\ Hence, A^{2} - 4 A - 5 I = [\begin{matrix} \begin{array}{l} 9 8 8 \\ 8 9 8 \end{array} \\ 8 8 9 \end{matrix}] - 4 [\begin{matrix} \begin{array}{l} 1 2 2 \\ 2 1 2 \end{array} \\ 2 2 1 \end{matrix}] - 5 [\begin{matrix} \begin{array}{l} 1 0 0 \\ 0 1 0 \end{array} \\ 0 0 1 \end{matrix}] \\ = [\begin{matrix} \begin{array}{l} 9 8 8 \\ 8 9 8 \end{array} \\ 8 8 9 \end{matrix}] + [\begin{matrix} - 4 & - 8 & - 8 \\ - 8 & - 4 & - 8 \\ - 8 & - 8 & - 4 \end{matrix}] + [\begin{matrix} \begin{array}{l} - 5 0 0 \\ 0 - 5 0 \end{array} \\ 0 0 5 \end{matrix}] \\ \Rightarrow A^{2} - 4 A -5 I = 0 \\ A^{2} - 4 A =5 I \\ A (A - 4 I) = 5 I \\ A^{- 1} = \frac{1}{5} \{[\begin{array}{l} 1 - 4 2 - 02 - 0 \\ 2 - 01 - 42 - 0 \\ 2 - 02 - 01 - 4 \end{array}]\} \\ A^{- 1} = \frac{1}{5} [\begin{matrix} \begin{array}{l} - 3 2 2 \\ 2 - 3 2 \end{array} \\ 2 2 - 3 \end{matrix}] \end{array}$

Q.41

$A = [\begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix}] and f (x) = x^{3} - 6 x^{2} + 7 x + 2, verify that f (A) =0.$

Ans

$\begin{array}{l} We have A = [\begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix}] \\ So, A^{2} = A, A [\begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix}] . [\begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix}] \\ = [\begin{array}{l} 1 + 0 + 4 0 + 0 + 0 2 + 0 + 6 \\ 0 + 0 + 2 0 + 4 + 0 0 + 2 + 3 \\ 2 + 0 + 6 0 + 0 + 0 4 + 0 + 9 \end{array}] \\ A^{2} = [\begin{matrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{matrix}] \\ A^{3} = A^{2}, A [\begin{matrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{matrix}] . [\begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix}] = [\begin{matrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{matrix}] \\ Since, f (x) = x^{3} - 6 x^{2} + 7 x + 2 \\ Putting x = A, we get \\ f (A) = A^{3} - 6 A^{2} + 7 A +2 I \\ = [\begin{matrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{matrix}] - 6 [\begin{matrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{matrix}] + 7 [\begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix}] + 2 [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] \\ = [\begin{matrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{matrix}] + [\begin{matrix} 30 & 0 & - 48 \\ 12 & - 24 & - 30 \\ - 48 & 0 & - 78 \end{matrix}] + [\begin{matrix} 7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21 \end{matrix}] + [\begin{matrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{matrix}] \\ = [\begin{matrix} 21 - 30 + 7 + 2 0 + 0 + 0 + 0 34 - 48 +14+0 \\ 12 - 12 + 0 + 0 8 - 24 + 14 + 2 23 - 30 + 7 + 0 \\ 34 - 48 + 14 + 0 0 + 0 + 0 + 0 55 - 78 + 21 + 2 \end{matrix}] \\ = [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}] \\ ∴ f (A) = 0 \end{array}$

Q.42

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

Ans

$\begin{array}{l} Since A = IA, So \\ [\begin{matrix} 1 & 2 & - 3 \\ 2 & 3 & 2 \\ 3 & - 3 & - 4 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] A \\ [\begin{matrix} 1 & 2 & - 3 \\ 0 & - 1 & 8 \\ 0 & - 9 & 5 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 0 & 1 \end{matrix}] A [\begin{array}{l} R_{2} \to R_{2} + (- 2) R_{1} \\ R_{3} \to R_{3} + (- 3) R_{1} \end{array}] \\ [\begin{matrix} 1 & 2 & - 3 \\ 0 & - 1 & 8 \\ 0 & - 9 & - 5 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 2 & - 1 & 0 \\ 3 & 0 & - 1 \end{matrix}] A [\begin{array}{l} R_{2} \to (- 2) R_{2} \\ R_{3} \to (- 3) R_{3} \end{array}] \\ [\begin{matrix} 1 & 0 & 13 \\ 0 & 1 & - 8 \\ 0 & 0 & 67 \end{matrix}] = [\begin{matrix} - 3 & 2 & 0 \\ 2 & - 1 & 0 \\ - 15 & 9 & - 1 \end{matrix}] A [\begin{array}{l} R_{1} \to R_{1} + (- 2) R_{2} \\ R_{3} \to R_{3} + (- 9) R_{2} \end{array}] \\ [\begin{matrix} 1 & 0 & 13 \\ 0 & 1 & - 8 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} - 3 & 2 & 0 \\ 2 & - 1 & 0 \\ \frac{- 15}{67} & \frac{9}{67} & \frac{- 1}{67} \end{matrix}] A [R_{3} \to \frac{1}{67} R_{3}] \\ [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} \frac{- 3}{67} & \frac{17}{67} & \frac{13}{67} \\ \frac{14}{67} & \frac{5}{67} & \frac{- 8}{67} \\ \frac{- 15}{67} & \frac{9}{67} & \frac{- 1}{67} \end{matrix}] A [\begin{array}{l} R_{1} \to R_{1} + (- 13) R_{3} \\ R_{2} \to R_{2} + 8 R_{3} \end{array}] \\ ∴ A^{- 1} = [\begin{matrix} \frac{- 6}{67} & \frac{17}{67} & \frac{13}{67} \\ \frac{14}{67} & \frac{5}{67} & \frac{- 8}{67} \\ \frac{- 15}{67} & \frac{9}{67} & \frac{- 1}{67} \end{matrix}] \end{array}$ $\begin{array}{l} x +2 y - 3 z =-4 \\ 2 x +3 y +2 z = 2 \\ 3 x -3 y -4 z =11 \\ A = [\begin{matrix} 1 & 2 & - 3 \\ 2 & 3 & 2 \\ 3 & - 3 & - 4 \end{matrix}], x = [\begin{matrix} x \\ y \\ z \end{matrix}] and B [\begin{matrix} - 4 \\ 2 \\ 11 \end{matrix}] \\ So, \\ x = A^{- 1} B = [\begin{matrix} \frac{- 6}{67} & \frac{17}{67} & \frac{13}{67} \\ \frac{14}{67} & \frac{5}{67} & \frac{- 8}{67} \\ \frac{- 15}{67} & \frac{9}{67} & \frac{- 1}{67} \end{matrix}] [\begin{matrix} - 4 \\ 2 \\ 11 \end{matrix}] \\ = \frac{1}{67} [\begin{matrix} - 6 & 17 & 13 \\ 14 & 5 & - 8 \\ - 15 & 9 & - 1 \end{matrix}] [\begin{matrix} - 4 \\ 2 \\ 11 \end{matrix}] \\ = \frac{1}{67} [\begin{array}{l} 201 \\ - 134 \\ 67 \end{array}] [\begin{matrix} \frac{201}{67} \\ \frac{- 134}{67} \\ \frac{67}{67} \end{matrix}] \\ [\begin{matrix} x \\ y \\ z \end{matrix}] = [\begin{matrix} 3 \\ - 2 \\ 1 \end{matrix}] \Rightarrow x = 3, y = - 2 and z = 1 \end{array}$

Q.43

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

Ans

$\begin{array}{l} Since A = IA, So \\ [\begin{matrix} 2 & - 1 & 4 \\ 4 & 0 & 2 \\ 3 & - 2 & 7 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] A \\ [\begin{matrix} 3 & - 2 & 7 \\ 4 & 0 & 2 \\ 2 & - 1 & 4 \end{matrix}] = [\begin{matrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{matrix}] A [R_{1} \leftrightarrow R_{3}] \\ [\begin{matrix} 1 & - 1 & 3 \\ 4 & 0 & 2 \\ 2 & - 1 & 4 \end{matrix}] = [\begin{matrix} - 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{matrix}] A [R_{1} \to R_{3} - R_{3}] \\ [\begin{matrix} 1 & - 1 & 3 \\ 0 & 4 & - 10 \\ 0 & 1 & - 2 \end{matrix}] = [\begin{matrix} - 1 & 0 & 1 \\ 4 & 1 & - 4 \\ 3 & 0 & - 2 \end{matrix}] A [\begin{array}{l} R_{2} \to R_{2} + (- 4) R_{1} \\ R_{3} \to R_{3} + (- 2) R_{1} \end{array}] \\ [\begin{matrix} 1 & - 1 & 3 \\ 0 & 1 & - 2 \\ 0 & 4 & - 10 \end{matrix}] = [\begin{matrix} - 1 & 0 & 1 \\ 3 & 0 & - 2 \\ 4 & 1 & - 4 \end{matrix}] A [R_{2} \leftrightarrow R_{3}] \\ [\begin{matrix} 1 & 0 & 1 \\ 0 & 1 & - 2 \\ 0 & 0 & - 2 \end{matrix}] = [\begin{matrix} 2 & 0 & - 1 \\ 3 & 0 & - 2 \\ - 8 & 1 & 4 \end{matrix}] A [\begin{array}{l} R_{1} \to R_{1} + R_{2} \\ R_{3} \to R_{3} + (- 4) R_{2} \end{array}] \\ [\begin{matrix} 1 & 0 & 1 \\ 0 & 1 & - 2 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 2 & 0 & - 1 \\ 3 & 0 & - 2 \\ 4 & \frac{- 1}{2} & - 2 \end{matrix}] A [R_{3} \to (- \frac{1}{2}) R_{3}] \\ [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} - 2 & \frac{1}{2} & - 1 \\ 11 & - 1 & - 6 \\ 4 & \frac{1}{2} & - 2 \end{matrix}] A [\begin{array}{l} R_{1} \to R_{1} - R_{3} \\ R_{2} \to R_{2} + 2 R_{3} \end{array}] \\ Hence, A^{- 1} = [\begin{matrix} - 2 & \frac{1}{2} & - 1 \\ 11 & - 1 & - 6 \\ 4 & \frac{- 1}{2} & - 2 \end{matrix}] . \end{array}$

Q.44

$If F (x) = [\begin{matrix} cosx & - sinx & 0 \\ sinx & cosx & 0 \\ 0 & 0 & 1 \end{matrix}], show that F (x) F (y) = F (x + y) .$

Ans

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

Q.45

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

Ans

$\begin{array}{l} Given : \\ x [\begin{matrix} 5 \\ 6 \end{matrix}] + y [\begin{matrix} - 1 \\ 1 \end{matrix}] = [\begin{matrix} 15 \\ 7 \end{matrix}] \\ \Rightarrow [\begin{matrix} 5 x \\ 6 x \end{matrix}] + y [\begin{matrix} - y \\ y \end{matrix}] = [\begin{matrix} 15 \\ 7 \end{matrix}] \\ \Rightarrow [\begin{matrix} 5 x - y \\ 6 x + y \end{matrix}] = [\begin{matrix} 15 \\ 7 \end{matrix}] \\ \Rightarrow 5 x - y = 15, 6 x + y = 7 \\ \Rightarrow x = 2 and y = - 5 \end{array}$

Q.46

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

Ans

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

Q.47

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

Ans

$\begin{array}{l} Let A = [\begin{array}{l} \begin{matrix} 1 \end{matrix} - 5 \\ 3 3 \end{array}] \\ Since, A = IA \\ [\begin{matrix} \begin{array}{l} 1 \\ 3 \end{array} & \begin{array}{l} - 5 \\ 3 \end{array} \end{matrix}] = [\begin{matrix} \begin{array}{l} 1 \\ 0 \end{array} & \begin{array}{l} 0 \\ 1 \end{array} \end{matrix}] A \\ Apply R_{2} \to R_{2} - 3 R_{1} \\ [\begin{matrix} \begin{array}{l} 1 \\ 3 \end{array} & \begin{array}{l} - 5 \\ 18 \end{array} \end{matrix}] = [\begin{matrix} \begin{array}{l} 1 \\ - 3 \end{array} & \begin{array}{l} 0 \\ 1 \end{array} \end{matrix}] A \\ R_{2} \to \frac{1}{18} R_{2} \\ ∴ [\begin{matrix} \begin{array}{l} 1 \\ 0 \end{array} & \begin{array}{l} - 5 \\ 1 \end{array} \end{matrix}] = [\begin{matrix} \begin{array}{l} 1 \\ - \frac{1}{6} \end{array} & \begin{array}{l} 0 \\ \frac{1}{18} \end{array} \end{matrix}] A \\ Apply R_{1} \to R_{1} + 5 R_{2} \\ [\begin{matrix} \begin{array}{l} 1 \\ 0 \end{array} & \begin{array}{l} 0 \\ 1 \end{array} \end{matrix}] = [\begin{matrix} \begin{array}{l} \frac{1}{6} \\ - \frac{1}{6} \end{array} & \begin{array}{l} \frac{5}{18} \\ \frac{1}{18} \end{array} \end{matrix}] A \\ A^{- 1} = [\begin{matrix} \begin{array}{l} \frac{1}{6} \\ - \frac{1}{6} \end{array} & \begin{array}{l} \frac{5}{18} \\ \frac{1}{18} \end{array} \end{matrix}] \end{array}$

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CBSE Class 12 Maths Revision Notes Chapter 3

Class 12 Mathematics Chapter 3 Notes

Key Topics Covered In Class 12 Mathematics Chapter 3 Notes

Matrix:

Types of Matrices:

Equality of Matrices:

Negative of a Matrix:

Special Types of Matrices:

Determinant of a Matrix:

Chapter 3 Mathematics class 12 notes: Exercises & Answer Solutions

NCERT Exemplar Class 12 Mathematics

Key Features of Class 12 Mathematics chapter 3 notes

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