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

RBSE Solutions For Class 12 Maths Chapter 5 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
RBSE Solutions For Class 12 Maths Chapter 5 Inverse Of A Matrix And Linear Equations

(ii) Area of triangle
RBSE Solutions For Class 12 Maths Chapter 5.2 Inverse Of A Matrix And Linear Equations

(iii) Area of triangle
Ex 5.2 Class 12 RBSE Inverse Of A Matrix And Linear Equations

RBSE Solutions For Class 12 Maths Chapter 5.2 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
RBSE Solutions For Class 12 Maths Chapter 5 Miscellaneous Inverse Of A Matrix And Linear Equations

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

Ex 5.2 Class 12 RBSE 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
Exercise 5.2 Class 12 Solutions Inverse Of A Matrix And Linear Equations
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.

RBSE Solutions For Class 12 Maths Chapter 5 Miscellaneous 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.
5.2 Class 12 Inverse Of A Matrix And Linear Equations

⇒ 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.

Exercise 5.2 Class 12 Solutions 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.
RBSE Solutions For Class 10 Maths Chapter 5.2 Inverse Of A Matrix And Linear Equations
⇒ 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

5.2 Class 12 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),
Ex 5.2 Class 10 RBSE Inverse Of A Matrix And Linear Equations Exercise 5.2 Class 10 RBSE Inverse Of A Matrix And Linear Equations
⇒ x(1 – 3) – (3 – 9) + 1(9 – 9) = 0
⇒ – 2x + 6 + 0 = 0
⇒ – 2(x – 3y) = 0
⇒ x – 3y = 0
Area of triangle
Ex 5.2 Class 12 Inverse Of A Matrix And Linear Equations
Area of triangle ∆ = 10 sq. unit (leaving -ive sign)

RBSE Solutions For Class 10 Maths Chapter 5.2 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,
RBSE Solutions For Class 12 Maths Chapter 5.2 Inverse Of A Matrix And Linear Equations
RBSE Class 10 Maths Chapter 5.2 Inverse Of A Matrix And Linear Equations
Solution of equation be x = 3, y = -1.

Cramer’s Rule Class 12 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
Exercise 5.2 Class 12 Inverse Of A Matrix And Linear Equations

= 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
Class 12 Maths Ex 5.2 Inverse Of A Matrix And Linear Equations
5.2 12th Maths Inverse Of A Matrix And Linear Equations
∵ ∆ = 0, ∆1 ≠ 0, ∆2 ≠ 0 and ∆3 ≠ 0
∴ Solution of equations is not possible.
Hence, system of equation are inconsistent.
Hence proved.

Ex 5.2 Class 10 RBSE 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
Class 12 Maths Exercise 5.2 Inverse Of A Matrix And Linear Equations
Maths Class 12 Chapter 5 Exercise 5.2 Inverse Of A Matrix And Linear Equations

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

Math Ch 5 Class 12 Inverse Of A Matrix And Linear Equations
= 2(-1 – 2) – 1(1 – 1) – 1(2 + 1)
= -6 – 0 – 3 = -9

RBSE Solutions For Class 12 Maths Chapter 5 Pdf Inverse Of A Matrix And Linear Equations
= 0(- 1 – 2) – 1(16 – 3) – 1(12 + 3)
= 0 – 3 – 15
= -18

Exercise 5.2 Maths Class 12 Inverse Of A Matrix And Linear Equations
= 2(6 – 3) – 0(1 – 1) – 1(3 – 6)
= 6 – 0 + 3
= 9

RBSE Solutions For Class 12 Maths Chapter 5 Inverse Of A Matrix And Linear Equations
= 2(-3 – 12) – 1(3 – 6) + 0(2 + 1)
= -30 + 3 + 0
= -27
Using Cramer’s rule,
RBSE Solutions Class 8 Maths Chapter 5 Exercise 5.2 Inverse Of A Matrix And Linear Equations
Hence, x = 2, y = -1, z = 3.

Exercise 5.2 Class 10 RBSE Question 10.
Solve the following system of equation using determinants :
(i) 6x + y – 3z = 5
x + 3y – 2z = 5
2x + y + 4z = 8

Inverse Matrix  RBSE Solutions Class 8 Maths Chapter 5 Exercise 5.2
Solution:
(i) Given equations
6x + y – 3z = 5
x + 3y – 2z = 5
2x + y + 4z = 8
Matrices Class 8 Solutions RBSE
= 6(12 + 2) – 1(4 + 4) – 3(1 – 6)
= 84 – 8 + 15
= 91

12th Maths Exercise 5.2
= 5(12 + 2) – 1(20 + 16) – 3(5 – 24)
= 70 – 36 + 57
= 91

12th Math 5.2 Solution Inverse Of A Matrix And Linear Equations RBSE
= 6(20 + 16) – 5(4 + 4) – 3(8 – 10)
= 216 – 40 + 6
= 182

Ex5 5 Class 12 Inverse Of A Matrix And Linear Equations RBSE
= 6(24 – 5) – 1(8 – 10) + 5(1 – 6)
= 114 + 2 – 25
= 91

Using Cramer’s rule
5.2 Class 12 Maths Inverse Of A Matrix And Linear Equations RBSE
Exercise 5.2 Class 12 Maths Solutions RBSE
RBSE Maths Class 12 Chapter 5 Exercise 5.2 Inverse Of A Matrix And Linear Equations

Ex 5.2 Class 12 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
Exercise 5.2 Class 12 Maths Inverse Of A Matrix And Linear Equations RBSE
So, A– 1 exists.
On finding adjoint of matrix A
F11 = 4, F12 = -3, F21 = 1, F22 = 2
Matrix formed by adjoint of A,
Class 12 Math Chapter 5 Inverse Of A Matrix And Linear Equations RBSE

(ii) Given equations
5x + 7y + 2 = 0 or 5x + 7y = -2
4x + 6y + 3 = 0 or 4x + 6y = – 3
⇒ AX = B …(i)
Class 12 Maths RBSE Chapter 5 Solutions Inverse Of A Matrix And Linear Equations
So, A– 1 exists.
On finding adjoint of matrix A,
F11 = 6, F12 = – 4, F21 = – 7, F22 = 5
Matrix formed by adjoint of A,
Chapter 5 Maths Class 12 Inverse Of A Matrix And Linear Equations
Exercise 5.2 Class 12 Maths RBSE Solutions Inverse Of A Matrix And Linear Equations

(iii) Given equations,
x + y – z = 1
3x + y – 2z = 3
x – y – z = -1
Let AX = B ……(i)
Exercise 5.2 Question Number 12 Inverse Of A Matrix And Linear Equations
RBSE Solutions for Class 12 Maths Chapter 5 Inverse of a Matrix and Linear Equations
RBSE Solutions for Class 12 Maths Chapter 5 Inverse of a Matrix and Linear Equations
RBSE Solutions for Class 12 Maths Chapter 5 Inverse of a Matrix and Linear Equations

(iv) Given equations
6x – 12y + 25z = 4
4x + 15y + 15z = 3
2x + 18y + 15z = 10
AX = B ……….(i)
RBSE Solutions for Class 12 Maths Chapter 5 Inverse of a Matrix and Linear Equations
= 6(225 + 360) + 12(60 + 40) + 25(72 – 30)
= 3510 + 1200 + 1050 = 5760 = 0
So, A– 1 exists.
On finding adjoint of matrix A,
RBSE Solutions for Class 12 Maths Chapter 5 Inverse of a Matrix and Linear Equations
RBSE Solutions for Class 12 Maths Chapter 5 Inverse of a Matrix and Linear Equations
RBSE Solutions for Class 12 Maths Chapter 5 Inverse of a Matrix and Linear Equations

RBSE Solutions For Class 12 Maths Chapter 5.2 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
RBSE Solutions for Class 12 Maths Chapter 5 Inverse of a Matrix and Linear Equations
RBSE Solutions for Class 12 Maths Chapter 5 Inverse of a Matrix and Linear Equations
= 1(1 + 6) + 2(2 – 0) + 0(- 4 – 0)
= 7 + 4 + 0
|A| = 11 ≠ 0
So, A– 1 exists.
On finding adjoint of A,
RBSE Solutions for Class 12 Maths Chapter 5 Inverse of a Matrix and Linear Equations
Matrix formed by adjoint of A,
RBSE Solutions for Class 12 Maths Chapter 5 Inverse of a Matrix and Linear Equations
Now, given equations
x – 2y = 10
2x + y + 3z = 8
– 2y + z = 7
In matrix form
RBSE Solutions for Class 12 Maths Chapter 5 Inverse of a Matrix and Linear Equations

RBSE Class 10 Maths Chapter 5.2 Question 13.
Find the product of matrices
RBSE Solutions for Class 12 Maths Chapter 5 Inverse of a Matrix and Linear Equations RBSE Solutions for Class 12 Maths Chapter 5 Inverse of a Matrix and Linear Equations 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
RBSE Solutions for Class 12 Maths Chapter 5 Inverse of a Matrix and Linear Equations
RBSE Solutions for Class 12 Maths Chapter 5 Inverse of a Matrix and Linear Equations
Matrix form of equations
RBSE Solutions for Class 12 Maths Chapter 5 Inverse of a Matrix and Linear Equations

Exercise 5.2 Class 12 Question 14.
Find the inverse matrix of matrix
RBSE Solutions for Class 12 Maths Chapter 5 Inverse of a Matrix and Linear Equations and with the help of it, solve the following system of equations :
RBSE Solutions for Class 12 Maths Chapter 5 Inverse of a Matrix and Linear Equations
Solution:
RBSE Solutions for Class 12 Maths Chapter 5 Inverse of a Matrix and Linear Equations
= 1(1 + 3) + 1(2 + 3) + 1(2 – 1)
= 4 + 5 + 1 = 10 ≠ 0
So, A– 1 exists.
On finding adjoint of A,
RBSE Solutions for Class 12 Maths Chapter 5 Inverse of a Matrix and Linear Equations
Matrix formed by adjoint of A,
RBSE Solutions for Class 12 Maths Chapter 5 Inverse of a Matrix and Linear Equations
RBSE Solutions for Class 12 Maths Chapter 5 Inverse of a Matrix and Linear Equations
Matrix form of equations
RBSE Solutions for Class 12 Maths Chapter 5 Inverse of a Matrix and Linear Equations

Class 12 Maths Ex 5.2 Question 15.
If (x1, y1), (x2, y2), (x3, y3) are vertices and a is side of an equilateral triangle respectively then prove that
RBSE Solutions for Class 12 Maths Chapter 5 Inverse of a Matrix and Linear Equations
Solution:
RBSE Solutions for Class 12 Maths Chapter 5 Inverse of a Matrix and Linear Equations
RBSE Solutions for Class 12 Maths Chapter 5 Inverse of a Matrix and Linear Equations

RBSE Solutions for Class 12 Maths