Prove that x + 4 2 x 2 x 2 x x + 4 2 x 2 x 2 x x + 4 = 5 x + 4 x - 4 2 .

Consider, LHS = x + 4 2 x 2 x 2 x x + 4 2 x 2 x 2 x x + 4 = 5 x + 4 5 x + 4 5 x + 4 2 x x + 4 2 x 2 x 2 x x + 4 applying R 1 → R 1 + R 2 + R 3 = 5 x + 4 1 1 1 2 x x + 4 2 x 2 x 2 x x + 4 taking 5 x + 4 out from R 1 = 5 x + 4 0 0 1 x - 4 - x + 4 2 x 0 x - 4 x + 4 applying C 1 → C 1 - C 2 & C 2 → C 2 - C 3 Now, expand along R 1 = 5 x + 4 0 - 0 + 1 × x - 4 - x + 4 0 x - 4 = ( 5 x + 4 ) ( x - 4 ) 2 = RHS Hence, proved.

If matrix A = cos θ sin θ - sin θ cos θ ,then find A - 1 .

Let A = cos θ - sin θ - sin θ - cos θ = - cos 2 θ - s o n 2 θ = - 1 ≠ 0 therefore A - 1 exists. A 11 = ( - 1 ) 1 + 1 M 11 = M 11 = - cos θ A 12 = ( - 1 ) 1 + 2 M 12 = - M 12 = sin θ A 21 = ( - 1 ) 2 + 1 M 21 = - M 21 = sin θ A 22 = ( - 1 ) 2 + 2 M 22 = M 22 = cos θ Therefore a d j ( A ) = - cos θ sin θ sin θ cos θ T = - cos θ sin θ sin θ cos θ A - 1 = 1 A a d j A A - 1 = cos θ - sin θ - sin θ - cos θ

Prove that 1 + a b c a 1 + b c a b 1 + c = ( 1 + a + b + c ) .

A = 1 + a b c a 1 + b c a b 1 + c Apply the operation c 1 = c 1 + c 2 + c 3 and taking common ( 1 + a + b + c ) A = 1 + a + b + c 1 b c 1 1 + b c 1 b 1 + c Applying C 2 → C 2 - b C 1 and C 3 → C 3 - c C 1 A = ( 1 + a + b + c ) 1 0 0 1 1 0 1 0 1 = ( 1 + a + b + c )

Prove that 1 + a 1 1 1 1 + b 1 1 1 1 + c = a b c 1 + 1 a + 1 b + 1 c

Let △ = 1 + a 1 1 1 1 + b 1 1 1 1 + c Taking a , b and c common from C 1 , C 2 and C 3 respectively, we get: ⇒ Δ = a b c 1 a + 1 1 b 1 c 1 a 1 + 1 b 1 c 1 a 1 b 1 + 1 c ⇒ Δ = a b c 1 + 1 a + 1 b + 1 c 1 b 1 c 1 + 1 a + 1 b + 1 c 1 + 1 b 1 c 1 + 1 a + 1 b + 1 c 1 b 1 + 1 c [Applying C 1 → C 1 + C 2 + C 3 ] ⇒ Δ = a b c 1 + 1 a + 1 b + 1 c 1 1 b 1 c 1 1 + 1 b 1 c 1 1 b 1 + 1 c [Taking 1 + 1 a + 1 b + 1 c common from C 1 ] ⇒ Δ = a b c 1 + 1 a + 1 b + 1 c 1 1 b 1 c 0 1 0 0 0 1 [Apply R 2 → R 2 − R 1 , R 3 → R 3 − R 1 ] ⇒ Δ = a b c ⁡ 1 + 1 a + 1 b + 1 c ⋅ 1 1 0 0 1 [ Expanding along C 1 ] ⇒ Δ = a b c ⁡ 1 + 1 a + 1 b + 1 c Hence proved.

Solve the following system of equations: 3 0 3 2 1 0 4 0 2 x y z = 8 1 4 + 2 y z 3 y

Given 3 0 3 2 1 0 4 0 2 x y z = 8 1 4 + 2 y z 3 y 3 x + 0 y + 3 z 2 x + y + 0 z 4 x + 0 y + 2 z = 8 + 2 y 1 + z 4 + 3 y 3 x - 2 y + 3 z 2 x + y - z 4 x - 3 y + 2 z = 8 1 4 3 - 2 3 2 1 - 1 4 - 3 2 x y z = 8 1 4 Here, A X = B We know that X = A - 1 B A - 1 = 1 | A | adj A | A | = 3 ( 2 - 3 ) + 2 ( 4 + 4 ) + 3 ( - 6 - 4 ) A = - 17 ≠ 0 Now, finding cofactor of A Therefore, A 11 = 2 − 3 = − 1 , A 12 = − ( 4 + 4 ) = − 8 , A 13 = − 6 − 4 = − 10 , A 21 = − ( − 4 + 9 ) = − 5 , A 22 = 6 − 12 = − 6 , A 23 = − ( − 9 + 8 ) = 1 , A 31 = 2 − 3 = − 1 A 32 = − ( − 3 − 6 ) = 9 A 33 = 3 + 4 = 7 We know that, adj A = A 11 A 21 A 31 A 12 A 22 A 32 A 13 A 23 A 34 adj A = − 1 − 5 − 1 − 8 − 6 9 − 10 1 7 X = 1 − 17 − 1 − 5 − 1 − 8 − 6 9 − 10 1 7 8 1 4 x y z = − 1 17 − 8 − 5 − 4 − 64 − 6 + 36 − 80 + 1 + 28 x y z = − 1 17 − 17 − 34 − 51 x y z = 1 2 3 ∴ x = 1 , y = 2 , z = 3 .

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 4 - Determinants>Rajasthan Board-2018

Embibe Experts Solutions for Chapter: Determinants, Exercise 1: Rajasthan Board-2019

Author:Embibe Experts

Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Determinants, Exercise 1: Rajasthan Board-2019

Attempt the free practice questions on Chapter 4: Determinants, Exercise 1: Rajasthan Board-2019 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: Determinants, Exercise 1: Rajasthan Board-2019 with Hints & Solutions

MEDIUM

12th Rajasthan Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 4 - Determinants>Rajasthan Board-2018>Q 2

Prove that $|\begin{matrix} x + 4 & 2 x & 2 x \\ 2 x & x + 4 & 2 x \\ 2 x & 2 x & x + 4 \end{matrix}| = (5 x + 4) {(x - 4)}^{2}$ .

MEDIUM

12th Rajasthan Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 4 - Determinants>Rajasthan Board-2018>Q 3

Solve the following system of equations by using Cramer's rule.

$5 x - 4 y = 7$

$x + 3 y = 9$

EASY

12th Rajasthan Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 4 - Determinants>Rajasthan Board-2019>Q 4

If matrix $A = [\begin{matrix} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{matrix}]$ ,then find $A^{- 1}$ .

HARD

12th Rajasthan Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 4 - Determinants>Rajasthan Board-2019>Q 5

Prove that $|\begin{matrix} 1 + a & b & c \\ a & 1 + b & c \\ a & b & 1 + c \end{matrix}| = (1 + a + b + c)$ .

MEDIUM

12th Rajasthan Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 4 - Determinants>Rajasthan Board-2019>Q 6

Solve the system of linear equations $x + y + 2 z = 0, x + 2 y - z = 9, x - 3 y + 3 z = - 14$ by using matrix method.

EASY

12th Rajasthan Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 4 - Determinants>Rajasthan Board-2020>Q 7

Define singular and Non-singular matrix

MEDIUM

12th Rajasthan Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 4 - Determinants>Rajasthan Board-2020>Q 8

Prove that $|\begin{matrix} 1 + a & 1 & 1 \\ 1 & 1 + b & 1 \\ 1 & 1 & 1 + c \end{matrix}| = a b c (1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c})$

MEDIUM

12th Rajasthan Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 4 - Determinants>Rajasthan Board-2020>Q 9

Solve the following system of equations:

$[\begin{array}{l} 3 & 0 & 3 \\ 2 & 1 & 0 \\ 4 & 0 & 2 \end{array}] [\begin{array}{l} x \\ y \\ z \end{array}] = [\begin{array}{l} 8 \\ 1 \\ 4 \end{array}] + [\begin{matrix} 2 y \\ z \\ 3 y \end{matrix}]$

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