If A = cos α sin α - sin α cos α , then verify that A ' A = I

Given: A = cos α sin α - sin α cos α To get transpose interchange rows and columns. ⇒ A ' = cos α - sin α sin α cos α Therefore A ' A = cos α - sin α sin α cos α cos α sin α - sin α cos α = cos 2 α + sin 2 α cos α sin α - cos α sin α cos α sin α - cos α sin α cos 2 α + sin 2 α = 1 0 0 1 = I ∴ A ' A = I Hence, proved

If A = 3 3 2 4 2 0 and B = 2 - 1 2 1 2 4 , then verify that (I) A ' ' = A and (II) ( A + B ) ' = A ' + B '

Given: A = 3 3 2 4 2 0 and B = 2 - 1 2 1 2 4 The transpose of a matrix is found by interchanging its rows into columns or columns into rows Now, A ' = 3 4 3 2 2 0 A ' ' = 3 3 2 4 2 0 = A Now, A + B ' = A ' + B ' LHS : A + B ' = 3 3 2 4 2 0 + 2 - 1 2 1 2 4 ' = 3 + 2 3 + - 1 2 + 2 4 + 1 2 + 2 0 + 4 ' = 5 5 0 . 73 4 4 4 RHS : A ' + B ' = 3 3 2 4 2 0 ' + 2 - 1 2 1 2 4 ' = 3 4 3 2 2 0 + 2 1 - 1 2 2 4 = 3 + 2 4 + 1 3 - 1 2 + 2 2 + 2 0 + 4 = 5 5 0 . 73 4 4 4 So, LHS = RHS .

Find X and Y , if X + Y = 7 0 2 5 and X - Y = 3 0 0 3

Given: X + Y = 7 0 2 5 ⋯ ⋯ ( 1 ) and X - Y = 3 0 0 3 ⋯ ⋯ ( 2 ) Adding equation ( 1 ) and ( 2 ) , we get X + Y + X - Y = 7 0 2 5 + 3 0 0 3 ⇒ 2 X = 7 + 3 0 + 0 2 + 0 5 + 3 2 X = 10 0 2 8 ⇒ X = 1 2 10 0 2 8 ⇒ X = 10 2 0 2 2 2 8 2 ⇒ X = 5 0 1 4 Putting the value of X in equation ( 1 ) , we get 5 0 1 4 + Y = 7 0 2 5 Y = 7 0 2 5 - 5 0 1 4 ⇒ Y = 7 - 5 0 - 0 2 - 1 5 - 4 ⇒ Y = 2 0 1 1 Therefore, X = 5 0 1 4 and Y = 2 0 1 1 .

Compute the indicated products. 1 2 3 2 3 4

To find: 1 2 3 2 3 4 Here order of the first matrix is 3 × 1 and the second matrix is 1 × 3 . As here the number of columns in the first matrix is same as number of rows in the second matrix, by using the multiplication principle of matrices we get, 1 2 3 2 3 4 = 1 × 2 1 × 3 1 × 4 2 × 2 2 × 3 2 × 4 3 × 2 3 × 3 3 × 4 = 2 3 4 4 6 8 6 9 12 Hence, 1 2 3 2 3 4 = 2 3 4 4 6 8 6 9 12

If A = 2 3 1 - 4 and B = 1 - 2 - 1 3 , then verify that ( AB ) - 1 = B - 1 A - 1 .

We have AB = 2 3 1 - 4 1 - 2 - 1 3 AB = - 1 5 5 - 14 Since, AB = - 11 ≠ 0 , ( AB ) - 1 exists and is given by ( AB ) − 1 = 1 | AB | adj ( AB ) = − 1 11 − 14 − 5 − 5 − 1 A B - 1 = 1 11 14 5 5 1 Further , A = - 11 ≠ 0 and B = 1 ≠ 0 .Therefore, A - 1 and B - 1 both exist and are given by A − 1 = − 1 11 − 4 − 3 − 1 2 , B − 1 = 3 2 1 1 B − 1 A − 1 = − 1 11 3 2 1 1 − 4 − 3 − 1 2 B − 1 A − 1 = − 1 11 − 14 − 5 − 5 − 1 = 1 11 14 5 5 1 Hence ( AB ) - 1 = B - 1 A - 1

Express the matrix B = 2 - 2 - 4 - 1 3 4 1 - 2 - 3 as the sum of symmetric and a skew-symmetric matrix.

Given that, B = 2 - 2 - 4 - 1 3 4 1 - 2 - 3 To proof : symmetric and skew symmetric matrix. Let B = P + Q where P = 1 2 B + B ' and Q = 1 2 B - B ' Now, P = 1 2 B + B ' = 1 2 2 - 2 - 4 - 1 3 4 1 - 2 - 3 + 2 - 1 1 - 2 3 - 2 - 4 4 - 3 P = 1 2 B + B ' = 1 2 4 - 3 - 3 - 3 6 2 - 3 2 - 6 P = 1 2 B + B ' = 2 - 3 2 - 3 2 - 3 2 3 1 - 3 2 1 - 3 (symmetric) Q = 1 2 B - B ' = 1 2 2 - 2 - 4 - 1 3 4 1 - 2 - 3 - 2 - 1 1 - 2 3 - 2 - 4 4 - 3 Q = 1 2 B - B ' = 1 2 0 - 1 - 5 1 0 6 5 - 6 0 Q = 1 2 B - B ' = 0 - 1 2 - 5 2 1 2 0 3 5 2 - 3 0 (skew-symmetric) Hence, 2 - 2 - 4 - 1 3 4 1 - 2 - 3 = 2 - 3 2 - 3 2 - 3 2 3 1 - 3 2 1 - 3 + 0 - 1 2 - 5 2 1 2 0 3 5 2 - 3 0 .

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 3 - Matrices>Madhya Pradesh Board-2019

Embibe Experts Solutions for Chapter: Matrices, Exercise 1: Madhya Pradesh Board-2020

Author:Embibe Experts

Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Matrices, Exercise 1: Madhya Pradesh Board-2020

Attempt the free practice questions on Chapter 3: Matrices, Exercise 1: Madhya Pradesh Board-2020 with hints and solutions to strengthen your understanding. EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS solutions are prepared by Experienced Embibe Experts.

Questions from Embibe Experts Solutions for Chapter: Matrices, Exercise 1: Madhya Pradesh Board-2020 with Hints & Solutions

EASY

12th Madhya Pradesh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 3 - Matrices>Madhya Pradesh Board-2019>Q 4

If $A = [\begin{array}{l} \cos α & \sin α \\ - \sin α & \cos α \end{array}]$ , then verify that $A^{'} A = I$

EASY

12th Madhya Pradesh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 3 - Matrices>Madhya Pradesh Board-2019>Q 5

If $A = [\begin{matrix} 3 & \sqrt{3} & 2 \\ 4 & 2 & 0 \end{matrix}]$ and $B = [\begin{matrix} 2 & - 1 & 2 \\ 1 & 2 & 4 \end{matrix}]$ , then verify that (I) ${(A^{'})}^{'} = A$ and (II) $(A + B)^{'} = A^{'} + B^{'}$

EASY

12th Madhya Pradesh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 3 - Matrices>Madhya Pradesh Board-2020>Q 6

FInd the number of all possible matrices of order

$3 \times 3$ with each entry $0$ or $1$ .

EASY

12th Madhya Pradesh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 3 - Matrices>Madhya Pradesh Board-2020>Q 7

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

EASY

12th Madhya Pradesh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 3 - Matrices>Madhya Pradesh Board-2020>Q 8

Compute the indicated products.

$[\begin{array}{l} 1 \\ 2 \\ 3 \end{array}] [\begin{array}{l} 2 & 3 & 4 \end{array}]$

HARD

12th Madhya Pradesh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 3 - Matrices>Madhya Pradesh Board-2020>Q 9

If $A = [\begin{matrix} 2 & 3 \\ 1 & - 4 \end{matrix}]$ and $B = [\begin{matrix} 1 & - 2 \\ - 1 & 3 \end{matrix}],$ then verify that $(AB)^{- 1} = B^{- 1} A^{- 1} .$

MEDIUM

12th Madhya Pradesh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 3 - Matrices>Madhya Pradesh Board-2020>Q 10

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