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

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

RBSE Solutions For Class 12 Maths Chapter 5 Miscellaneous Question 1.
If A = , then find A-1.
Solution: So, A-1 exists.
F11 = 3, F12 = – 2, F21 = 1, F22 = 1 RBSE Solutions For Class 12 Maths Chapter 5 Question 2.
If , then find A-1.
Solution: So, A-1exists. Matrix formed by adjoint of A, 12 Maths RBSE Solution 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 = $$\frac { 8 }{ -8 }$$
⇒ x = -1
⇒ Hence, x = -1

Miscellaneous Chapter 5 Class 12 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.

RBSEsolution.Com Class 12 Maths 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.

12th Maths RBSE Solution 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.
F11 = 4, F12 = -1, F21 = 2, F22 = 1
Matrix formed by adjoint of B, From equation (i),
A = C.B-1 Chapter 5 Class 12 Miscellaneous Question 7.
If then prove that A2 + 4A – 42I = 0, also find A-1.
Solution:   Chapter 5 Miscellaneous Class 12 Question 8.
If then prove that A-1 = $$\frac { 1 }{ 19 }$$ A.
Solution: So, A-1exists.
F11 = – 2, F12 = – 5, F21 = – 3, F22 = 2
Matrix formed by adjoint of A, Class 12 Maths Chapter 5 Miscellaneous Exercise Solutions Question 9.
If , then find A-1 and show that A-1.A= I3.
Solution: = 1(16 – 9) – 3(4 – 3) + 3(3 – 4)
= 7 – 3 – 3
|A|= 1 ≠ 0
So, A-1 exists. Matrix formed by adjoint of A, Again to prove that Hence, A-1A = I3.

Ch 5 Miscellaneous Class 12 Question 10.
If , then prove that A2 – 4A – 5I = 0, also find A-1.
Solution:  Class 12 Maths Chapter 5 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
F11 = – 5, F12 = – 7, F21 = 7, F22 = 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. 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-1exists. Matrix formed by adjoint of A, Hence, x = 1, y = – 1, z = 2.

5 Miscellaneous Class 12 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  Miscellaneous Exercise On Chapter 5 Class 12 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 = $$\frac { 14 }{ 8 }$$ = $$\frac { 7 }{ 4 }$$

Chapter 5 Maths Class 12 Question 14.
Find matrix A, if Solution: Given, equation BAC = I ……(i) F11 = 5, F12 = – 3, F21 = – 7, F22 = 4
Matrix formed by adjoint of C, From equation (i)
BAC = I
⇒ B-1(BAC)C-1 = B-1C-1
⇒ (B-1B)A(CC-1) = B-1C-1
⇒ IAI = B-1C-1
⇒ (IA)I = B-1C-1
⇒ AI = B-1C-1
⇒ A = B-1C-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. 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 = (a2 + bc + 1)I – aA.
Solution:
Given, So, A-1 exists.  Question 17.
Using determinant solve the following system of equations:
x + y + z = 1
ax + by + cz = k
a2x + b2y + c2z = k2
Solution:
Given equations,
x + y + z = 1
ax + by + cz = k
a2x + b2y + c2z = k2    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.  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-1B …..(i)
Determinants of matrix A, So, A-1 exists.
F11 = – 2, F12 = – 3, F21 = 4, F22 = 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 = $$\frac { -12 }{ 6 }$$ = -2
From (i),
10a – b – 5ab = -11
⇒ 10(-2) – b – 5(-2)b = -11
⇒ -20 – b + 10b = -11
⇒ 9a = 9
⇒ b = $$\frac { 9 }{ 9 }$$
Hence, a = -2, b = 1.