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

May 27, 2019 by Fazal Leave a Comment

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
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 Inverse of a Matrix and Linear Equations

(iii) Area of triangle
RBSE Solutions for Class 12 Maths Chapter 5 Inverse of a Matrix and Linear Equations

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 Inverse of a Matrix and Linear Equations

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
RBSE Solutions for Class 12 Maths Chapter 5 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.

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.
RBSE Solutions for Class 12 Maths Chapter 5 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.

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 12 Maths Chapter 5 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

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),
RBSE Solutions for Class 12 Maths Chapter 5 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
RBSE Solutions for Class 12 Maths Chapter 5 Inverse of a Matrix and Linear Equations
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,
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
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
RBSE Solutions for Class 12 Maths Chapter 5 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
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
∵ ∆ = 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
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

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

RBSE Solutions for Class 12 Maths Chapter 5 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 Inverse of a Matrix and Linear Equations
= 0(- 1 – 2) – 1(16 – 3) – 1(12 + 3)
= 0 – 3 – 15
= -18

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

RBSE Solutions for Class 12 Maths Chapter 5 Inverse of a Matrix and Linear Equations
Solution:
(i) Given equations
6x + y – 3z = 5
x + 3y – 2z = 5
2x + y + 4z = 8
RBSE Solutions for Class 12 Maths Chapter 5 Inverse of a Matrix and Linear Equations
= 6(12 + 2) – 1(4 + 4) – 3(1 – 6)
= 84 – 8 + 15
= 91

RBSE Solutions for Class 12 Maths Chapter 5 Inverse of a Matrix and Linear Equations Ex 5.2
= 5(12 + 2) – 1(20 + 16) – 3(5 – 24)
= 70 – 36 + 57
= 91

RBSE Solutions for Class 12 Maths Chapter 5 Inverse of a Matrix and Linear Equations
= 6(20 + 16) – 5(4 + 4) – 3(8 – 10)
= 216 – 40 + 6
= 182

RBSE Solutions for Class 12 Maths Chapter 5 Inverse of a Matrix and Linear Equations
= 6(24 – 5) – 1(8 – 10) + 5(1 – 6)
= 114 + 2 – 25
= 91

Using Cramer’s rule
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

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
RBSE Solutions for Class 12 Maths Chapter 5 Inverse of a Matrix and Linear Equations
So, A– 1 exists.
On finding adjoint of matrix A
F11 = 4, F12 = -3, F21 = 1, F22 = 2
Matrix formed by adjoint of A,
RBSE Solutions for Class 12 Maths Chapter 5 Inverse of a Matrix and Linear Equations

(ii) Given equations
5x + 7y + 2 = 0 or 5x + 7y = -2
4x + 6y + 3 = 0 or 4x + 6y = – 3
⇒ AX = B …(i)
RBSE Solutions for Class 12 Maths Chapter 5 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,
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

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

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

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

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

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

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