RBSE Solutions for Class 12 Maths Chapter 5 Inverse of a Matrix and Linear Equations Ex 5.2

Rajasthan Board RBSE Class 12 Maths Chapter 5 Inverse of a Matrix and Linear Equations Ex 5.2

Question 1.
Find area of triangle, whose vertices are:
(i) (2, 5), (- 2, – 3) and (6,0)
(ii) (3, 8), (2, 7) and (5, – 1)
(iii) (0, 0), (5, 0) and (3, 4)
Solution:
(i) Area of triangle

(ii) Area of triangle

(iii) Area of triangle

Question 2.
Use determinant to find the area of triangle whose vertices are (1, 4), (2, 3) and (-5, – 3), Are these points collinear ?
Solution:
Area of triangle

These points are not collinear because area of triangle is not equal to zero.

Question 3.
Find the value of k, if area of triangle is 35 sq. unit and vertices of triangle are (k, 4), (2, -6) and (5, 4).
Solution:
Given points (k, 4), (2, -6) and (5, 4) and area of triangle = 35 sq. unit

taking +ve sign
⇒ – 10k + 50 = 70
⇒ – 10k = 70 – 50
⇒ – 10k = 20
⇒ k = -2
taking -ve sign
– 10k + 50 = – 70
⇒ 10k = – 70 – 50
⇒ – 10k = – 120
⇒ k = 12
⇒ k= -2, 12.

Question 4.
Use determinant to find k, if points (k, 2 – 2k), (-k + 1, 2k) and (-4 – k, 6 – 2k) and collinear.
Solution:
Given points (k, 2 – 2k), (-k + 1, 2k) and (-4 – k, 6 – 2k) are collinear.

⇒ k [2k – ( 6 – 2k)] – (2 – 2k) [(-k + 1) – (-4 – k)] + 1 [(-k + 1) (6 – 2k) – (-4 – k) (2k)] = 0
⇒ k [2k – 6 + 2k] – (2 – 2k) [-k + 1 + 4 + k ] + 1[- 6k + 2k² + 6 – 2k + 8k + 2k²] = 0
⇒ K(4k – 6) – (2 – 2k) (5) + 1(4k² + 6) = 0
⇒ 4k² – 6k – 10 + 10k + 4k² + 6 = 0
⇒ 8k² + 4k – 4 = 0
⇒ 2k² + k – 1 = 0
⇒ 2k ² + (2 – 1)k – 1 = 0
⇒ 2k(k + 1) – 1(k + 1) = 0
⇒ (k + 1) (2k -1)=0
Hence, k = -1, 1/2.

Question 5.
If point (3,-2), (x, 2) and (8, 8) are collinear, then find the value of x, using determinant.
Solution:
Given (3,-2), (x, 2) and (8, 8) are collinear.

⇒ 3(2 – 8) + 2(x – 8) + 1(8x -16) = 0
⇒ – 18 + 2x – 16 + 8x – 16=0
⇒ 10x – 50 = 0
⇒ 10x = 50
⇒ x = 50/10
Hence, x = 5

Question 6.
Find the equation of line passing through the two points (3, 1) and (9, 3), using determinant, also find the area of traingle if third point is (– 2, – 4).
Solution:
Equation of line passing through (3, 1) and (93),

⇒ x(1 – 3) – (3 – 9) + 1(9 – 9) = 0
⇒ – 2x + 6 + 0 = 0
⇒ – 2(x – 3y) = 0
⇒ x – 3y = 0
Area of triangle

Area of triangle ∆ = 10 sq. unit (leaving -ive sign)

Question 7.
Solve the following system of equations by Cramer’s rule :
(i) 2x + 3y = 9, 3x – 2y = 7
(ii) 2x – 7y – 13 = 0, 5x + 6y – 9 = 0
Solution:
Given equations
2x + 3y = 9
3x – 2y = 7
Here,

Solution of equation be x = 3, y = -1.

Question 8.
Prove that following system of equation are inconsistent :
(i) 3x + y + 2z = 3
2x + y + 3z = 5
x – 2y – z = 1

(ii) x + 6y + 11 = 0
3x + 20y – 67 + 3 = 0
6y – 187 + 1 = 0
Solution:
Given equations
3x + y + 2z = 3
2x + y + 3z = 5
x – 2y – z= 1

= 3(- 1 + 6) – 1(-2 – 3) + 2 (- 4 – 1)
= 15 + 5 – 10 = 10 ≠ 0
∵ ∆ ≠ 0
∴ Solution is not possible.
Hence, system of equation are in consistent.
Hence proved.

(ii) Given equation
x + 6y + 11 = 0
or
x + 6y = – 11

3x + 20y – 6z + 3 = 0
or
3x + 20y – 6z = -3

6y – 18z = -1
Here

∵ ∆ = 0, ∆₁ ≠ 0, ∆₂ ≠ 0 and ∆₃ ≠ 0
∴ Solution of equations is not possible.
Hence, system of equation are inconsistent.
Hence proved.

Question 9.
Solve the following system of equations by Cramer’s rule :
(i) x + 2y + 4z = 16
4x + 3y – 2z = 5
3x – 5y + z = 4

(ii) 2x + y – z = 0
x – y + z = 6
x + 2y + z = 3
Solution:
(i) Given equations
x + 2y + 4z = 16
4x + 3y – 2z= 5
3x – 5y + z = 4

(ii) Given equations
2x + y – z = 0
x – y + z = 6
x + 2y + z = 3

= 2(-1 – 2) – 1(1 – 1) – 1(2 + 1)
= -6 – 0 – 3 = -9

= 0(- 1 – 2) – 1(16 – 3) – 1(12 + 3)
= 0 – 3 – 15
= -18

= 2(6 – 3) – 0(1 – 1) – 1(3 – 6)
= 6 – 0 + 3
= 9

= 2(-3 – 12) – 1(3 – 6) + 0(2 + 1)
= -30 + 3 + 0
= -27
Using Cramer’s rule,

Hence, x = 2, y = -1, z = 3.

Question 10.
Solve the following system of equation using determinants :
(i) 6x + y – 3z = 5
x + 3y – 2z = 5
2x + y + 4z = 8

Solution:
(i) Given equations
6x + y – 3z = 5
x + 3y – 2z = 5
2x + y + 4z = 8

= 6(12 + 2) – 1(4 + 4) – 3(1 – 6)
= 84 – 8 + 15
= 91

= 5(12 + 2) – 1(20 + 16) – 3(5 – 24)
= 70 – 36 + 57
= 91

= 6(20 + 16) – 5(4 + 4) – 3(8 – 10)
= 216 – 40 + 6
= 182

= 6(24 – 5) – 1(8 – 10) + 5(1 – 6)
= 114 + 2 – 25
= 91

Using Cramer’s rule

Question 11.
Use matrix method to solve following system of equations :
(i) 2x – y = -2
3x + 4y = 3

(ii) 5x + 7y + 2 = 0
4x + 6y + 3 = 0

(iii) x + y – 7 = 1
3x + y – 2z = 3
x – y – z = -1

(iv) 6x – 12y + 25z = 4
4x + 15y – 20z = 3
2x + 18y + 15z = 10
Solution:
(i) Given equations
2x – y = -2
3x + 4y = 3
Let AX = B

So, A^{– 1} exists.
On finding adjoint of matrix A
F₁₁ = 4, F₁₂ = -3, F₂₁ = 1, F₂₂ = 2
Matrix formed by adjoint of A,

(ii) Given equations
5x + 7y + 2 = 0 or 5x + 7y = -2
4x + 6y + 3 = 0 or 4x + 6y = – 3
⇒ AX = B …(i)

So, A^{– 1} exists.
On finding adjoint of matrix A,
F₁₁ = 6, F₁₂ = – 4, F₂₁ = – 7, F₂₂ = 5
Matrix formed by adjoint of A,

(iii) Given equations,
x + y – z = 1
3x + y – 2z = 3
x – y – z = -1
Let AX = B ……(i)

(iv) Given equations
6x – 12y + 25z = 4
4x + 15y + 15z = 3
2x + 18y + 15z = 10
AX = B ……….(i)

= 6(225 + 360) + 12(60 + 40) + 25(72 – 30)
= 3510 + 1200 + 1050 = 5760 = 0
So, A^{– 1} exists.
On finding adjoint of matrix A,

Question 12.
If img then, find A^{– 1} and solve the following system of linear equations :
x – 2y = 10,
2x + y + 3x = 8,
– 2y + z = 7
Solutin:
Given

= 1(1 + 6) + 2(2 – 0) + 0(- 4 – 0)
= 7 + 4 + 0
|A| = 11 ≠ 0
So, A^{– 1} exists.
On finding adjoint of A,

Matrix formed by adjoint of A,

Now, given equations
x – 2y = 10
2x + y + 3z = 8
– 2y + z = 7
In matrix form

Question 13.
Find the product of matrices
with the help of it, solve the following system of linear equations :
x – y + z = 4,
x – 2y – 2z = 9,
2x + y + 3z = 1
Solution:
Let

Matrix form of equations

Question 14.
Find the inverse matrix of matrix
and with the help of it, solve the following system of equations :

Solution:

= 1(1 + 3) + 1(2 + 3) + 1(2 – 1)
= 4 + 5 + 1 = 10 ≠ 0
So, A^{– 1} exists.
On finding adjoint of A,

Matrix formed by adjoint of A,

Matrix form of equations