## Rajasthan Board RBSE Class 12 Maths Chapter 5 Inverse of a Matrix and Linear Equations Miscellaneous Exercise

Question 1.

If A = , then find A^{-1}.

Solution:

So, A^{-1} exists.

On finding adjoint of A,

F_{11} = 3, F_{12} = – 2, F_{21} = 1, F_{22} = 1

Matrix formed by adjoint A,

Question 2.

If , then find A^{-1}.

Solution:

So, A^{-1}exists.

On finding adjoint of A,

Matrix formed by adjoint of A,

Question 3.

If matrix is a singular matrix, then find x.

Solution:

⇒ (- 6 – 2) + 2(- 3 – x) + 3(2 – 2x) = 0

⇒ – 8 – 6 – 2x + 6 – 6x = 0

⇒ – 8 – 8x = 0

⇒ x =

⇒ x = -1

⇒ Hence, x = -1

Question 4.

Use Cramer’s rule to solve the following system of equations :

(i) 2x – y = 17

3x + 5y = 6

(ii) 3x + ay = 4

2x + ay= 2, a ≠ 0

(iii) x + 2y + 37 = 6

2x + 4y + z = 7

3x + 2y + 9z = 14

Solution:

(i) Given,

2x – y = 17

3x + 5y = 6

∵ ∆ ≠ 0, ∆_{1} ≠ 0 and ∆_{2} ≠ 0

∴ Solution will be unique.

By using Cramer’s rule

Hence, solution is x = 7, y = -3

(ii) Given,

3x + ay = 4

2a + ay = 2,

a ≠ 0

∵ ∆ ≠ 0, ∆_{1} ≠ 0 and ∆_{2} ≠ 0

∴ Solution will be unique.

Using Cramer’s rule

(iii) Given,

x + 2y + 2x = 6

2x + 4y + z = 7

3x + 2y + 92 = 14

= 1(56 – 14) – 2(28 – 21) + 6(4 – 12)

= 42 – 12 – 48

= -20

∵ ∆ ≠ 0, ∆_{1} ≠ 0 , ∆_{2} ≠ 0 and ∆_{3} ≠ 0

∴ Solution will be unique.

Using Cramer’s rule,

Hence, solution is x = 1, y = 1, z = 1.

Question 5.

Use Cramer’s rule and prove that following system of equations are inconsistent :

(i) 2x – y = 5

4x – 2y = 7

(ii) x + y + z = 1

x + 2y + 37 = 2

3x + 4y + 5y = 3

Solution:

(i) Given,

2x – y = 5

4x – 2y = 7

Hence, system of equations is inconsistent.

(ii) Given,

x + y + z = 1

x + 2y + 3z = 2

3x + 4y + 3z = 3

∵ ∆ ≠ 0, ∆_{1} ≠ 0 , ∆_{2} ≠ 0 and ∆_{3} ≠ 0

Hence, system of equations is inconsistent.

Question 6.

Find a matrix A of order 2, where

Solution:

Then,

AB = C

ABB^{-1} = CB^{-1}

AI = CB^{-1} [∵ BB^{-1} = I]

A = CB^{-1} …….(i)

[∵ AI = A]

So, B^{-1} exists.

On finding adjoint of B,

F_{11} = 4, F_{12} = -1, F_{21} = 2, F_{22} = 1

Matrix formed by adjoint of B,

From equation (i),

A = C.B^{-1}

Question 7.

If then prove that A^{2} + 4A – 42I = 0, also find A^{-1}.

Solution:

Question 8.

If then prove that A^{-1} = A.

Solution:

So, A^{-1}exists.

On finding adjoint of A,

F^{11} = – 2, F^{12} = – 5, F^{21} = – 3, F^{22} = 2

Matrix formed by adjoint of A,

Question 9.

If , then find A^{-1} and show that A^{-1}.A= I_{3}.

Solution:

= 1(16 – 9) – 3(4 – 3) + 3(3 – 4)

= 7 – 3 – 3

|A|= 1 ≠ 0

So, A^{-1} exists.

On finding adjoint of A,

Matrix formed by adjoint of A,

Again to prove that

Hence, A^{-1}A = I_{3}.

Question 10.

If , then prove that A^{2} – 4A – 5I = 0, also find A^{-1}.

Solution:

Question 11.

Use matrix method to solve following system of equations :

(i) 5x – 7y = 2

7x – 5y = 3

(ii) 3x + y + z = 3

2x – y – z = 2

– x – y + z = 1

(iii) x + 2y – 2z + 5 = 0

– x + 3y + 4 = 0

– 2y + 7 – 4 = 0

Solution:

(i) Given equations

5x – 7y = 2

7x – 5y = 3

In matrix form AX= B ….(i)…(i)

Determinant of matrix A,

= – 25 + 49 = 24 ≠ 0

So, A^{-1} exists.

On finding adjoint of matrix A

F_{11} = – 5, F_{12} = – 7, F_{21} = 7, F_{22} = 5

Matrix formed by adjoint of A

(ii) Given equations

3x + y + z = 3

2x – y – z = 2

– x – y + z = 1

In matrix form

AX = B ….(i)

= 3(-1 – 1) – 1(2 – 1) + 1(-2 – 1)

= -6 – 1 – 3

= -10 ≠ 0

So, A^{-1} exists.

On finding adjoint of A,

Matrix formed by adjoint of A,

Hence, x = 1, y = -1, z = 1

(iii) Given equations,

x + 2y – 2z + 5 = 0 or x + 2y – 2z = – 5

-x + 3y + 4 = 0 or – x + 3y = – 4

– 2y +z – 4 = 0 or – 2y + z = 4

In matrix form, AX= B ….. (i)

= 1(3 – 0) – 2(-1 – 0) – 2(2 – 0)

= 3 + 2 – 4 = 1 ≠ 0

So, A^{-1}exists.

On finding adjoint of A,

Matrix formed by adjoint of A,

Hence, x = 1, y = – 1, z = 2.

Question 12.

Find area of ∆ABC, if vertices are :

(i) A(- 3, 5), B(3, – 6), C(7, 2)

(ii) A(2, 7), B(2, 2), C(10, 8)

Solution:

(i) Vertices of ∆ABC

A = (-3,5), B = (3, -6), C = (7, 2)

Area of ∆ABC

(ii) Vertices of ∆ABC, A = (2, 7), B = (2, 2), C = (10,8)

Area of ∆ABC

Question 13.

If points (2, – 3), (λ, – 2) and (0, 5) are collinear, then find λ.

Solution:

Given points (2, – 3), (λ, – 2) and (0, 5) are collinear, then

⇒ 2(-2 – 5) + 3(λ – 0).+ 1(5λ – 0) = 0

⇒ – 14 + 3λ + 5λ = 0

⇒ 8λ = 14

⇒ λ = =

Question 14.

Find matrix A, if

Solution:

Given, equation BAC = I ……(i)

On finding adjoint of C,

F_{11} = 5, F_{12} = – 3, F_{21} = – 7, F_{22} = 4

Matrix formed by adjoint of C,

From equation (i)

BAC = I

⇒ B^{-1}(BAC)C^{-1} = B^{-1}C^{-1}

⇒ (B^{-1}B)A(CC^{-1}) = B^{-1}C^{-1}

⇒ IAI = B^{-1}C^{-1}

⇒ (IA)I = B^{-1}C^{-1}

⇒ AI = B^{-1}C^{-1}

⇒ A = B^{-1}C^{-1}

Question 15.

If , then find A^{-1} and solve the following system of equations, using it :

x + y + z = 2,

x + 2y – 37 = 13,

2x – y + 37 = -7.

Solution:

Given

= 1(6 – 3) – 1(3+6) + 1(- 1 – 4)

= 3 – 9 – 5

= -11 ≠ 0

so, A^{-1} exists.

On finding adjoint of A,

Matrix formed by adjoint of A,

Given equations,

x + y + z = 2

x + 2y – 3z = 13

2x – y + 3z = -7

In matrix form,

Question 16.

If , then find A^{-1} and show that aA^{-1} = (a^{2} + bc + 1)I – aA.

Solution:

Given,

So, A^{-1} exists.

On finding adjoint of A,

Question 17.

Using determinant solve the following system of equations:

x + y + z = 1

ax + by + cz = k

a^{2}x + b^{2}y + c^{2}z = k^{2}

Solution:

Given equations,

x + y + z = 1

ax + by + cz = k

a^{2}x + b^{2}y + c^{2}z = k^{2}

Question 18.

If , then find A^{-1} and solve the following system of equations using it :

x + 2y – 3z = – 4,

2x + 3y + 2z = 2,

3x – 3y – 4z = 11

Solution:

= 1(- 12 + 6) – 26 – 8 – 6) – 3(- 6 – 9)

= -6 + 28 + 45

= 67 ≠ 0

So, A^{-1} exists.

On finding adjoint of A,

Given equations,

x + 2y – 32 = -4

2x + 3y + 2z = 2

3x – 3y – 4z = 11

In matrix form,

Question 19.

, then find the value of X.

Solution:

Then, AX = B

X = A^{-1}B …..(i)

Determinants of matrix A,

So, A^{-1} exists.

F_{11} = – 2, F^{12} = – 3, F^{21} = 4, F^{22} = 1

Matrix formed by adjoint of A,

Question 20.

Find the value of a and b, if following system of equations has infinite solutions :

2x + y + az = 4

5x – 2y + z = -2

5x – 5y + z = -2

Solution:

Given equations,

2x + y + az = 4 :

bx – 2y + z = -2

5x – 5y + z = -2

∵ For infinite solution ∆ = 0, ∆_{1} = 0, ∆_{2} = 0 and ∆_{z} = 0.

⇒ 2(- 2 + 5) – 1(b – 5) + d(-5b + 10) = 0

⇒ 6 – b + 5 – 5ab + 10a = 0

⇒ 10a – b – 5ab = -11 …(1)

4(-2 + 5) – 1(-2+2) + a(10 – 4) = 0

⇒ 12 + 6a=0

⇒ 16a = -12

a = = -2

From (i),

10a – b – 5ab = -11

⇒ 10(-2) – b – 5(-2)b = -11

⇒ -20 – b + 10b = -11

⇒ 9a = 9

⇒ b =

Hence, a = -2, b = 1.

##### RBSE Solutions for Class 12 Maths

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