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 .

If A = 3 4 1 2 , then find A - 1 .

Consider, A = 3 4 1 2 We know that, A - 1 = 1 A a d j A Now calculating A A = 3 4 1 2 = 3 × 2 - 4 × 1 = 2 Now finding a d j A a d j A = 3 4 1 2 ' = 3 1 4 2 ∴ A - 1 = 1 2 3 1 4 2

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.

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 .

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

Given x + y + 2 z = 0 x + 2 y − z = 9 x − 3 y + 3 z = − 14 Coefficient matrix is A = 1 1 2 1 2 - 1 1 - 3 3 , B = 0 9 - 14 By matrix method X = A - 1 B We know that, A - 1 = 1 | A | ad j A | A | = 1 ( 6 - 3 ) - 1 ( 3 + 1 ) + 2 ( - 3 - 2 ) = - 11 ≠ 0 Therefore, A - 1 exist a d j A = Transpose of cofactor of A Cofactor of A = A 11 A 12 A 13 A 21 A 22 A 23 A 31 A 32 A 33 , where A i j = ( - 1 ) i + j M i j A 11 = 6 − 3 = 3 A 12 = − ( 3 + 1 ) = − 4 A 13 = − 3 − 2 = − 5 A 21 = − ( 3 + 6 ) = − 9 A 22 = 3 − 2 = 1 A 23 = − ( − 3 − 1 ) = 4 A 31 = − 1 − 4 = − 5 A 32 = − ( − 1 − 2 ) = 3 A 33 = 2 − 1 = 1 Therefore, adj A = 3 − 9 − 5 − 4 1 3 − 5 4 1 ∴ X = 1 - 11 3 - 9 - 5 - 4 1 3 - 5 4 1 0 9 - 14 x = 1 - 11 0 - 81 + 70 0 + 9 - 42 0 - 45 - 14 X = 1 - 11 - 11 - 33 - 55 ∴ x y z = 1 3 5 ∴ x = 1 , y = 3 , z = 5

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 )

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 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 A = 3 4 1 2 , then find A - 1 .

Consider, A = 3 4 1 2 We know that, A - 1 = 1 A a d j A Now calculating A A = 3 4 1 2 = 3 × 2 - 4 × 1 = 2 Now finding a d j A a d j A = 3 4 1 2 ' = 3 1 4 2 ∴ A - 1 = 1 2 3 1 4 2

E M B I B E

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

EASY

12th Rajasthan Board

IMPORTANT

Earn 100

Define singular and Non-singular matrix

Important Questions on Determinants

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-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:

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