RBSE Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6

Rajasthan Board RBSE Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 Textbook Exercise Questions and Answers.

Rajasthan Board RBSE Solutions for Class 12 Maths in Hindi Medium & English Medium are part of RBSE Solutions for Class 12. Students can also read RBSE Class 12 Maths Important Questions for exam preparation. Students can also go through RBSE Class 12 Maths Notes to understand and remember the concepts easily.

RBSE Class 12 Maths Solutions Chapter 4 Determinants Ex 4.6

Question 1.
x + 2y = 2
2x + 3 y = 3
Answer:
Given system of equations is :
x + 2y = 2
2x + 3y = 3
Given system of equations can be written in matrix form as:
RBSE Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 1
Determinant of matrix A is
|A| = \(\left|\begin{array}{ll} 1 & 2 \\ 2 & 3 \end{array}\right|\)
= 1 × 3 - 2 × 2
= 3 - 4 = - 1
∴ |A| = - 1 ≠ 0
Thus, given system of equations is consistent.

RBSE Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6

Question 2.
2x - y = 5
x + y = 4
Answer:
Given system of equations is :
2x - y = 5
x + y = 4
In matrix form, the given system of equations can be written as
AX = B
RBSE Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 2
Determinant of matrix A is
|A| = \(\left|\begin{array}{rr} 2 & -1 \\ 1 & 1 \end{array}\right|\) = 2 × 1 - 1 × (- 1) = 2 + 1 = 3
i.e., |A| ≠ 0
Thus, given system of equations is consistent.

Question 3.
x + 3y = 5
2x + 6y = 8
Answer:
Given system of equations is :
x + 3y = 5
2x + 6y = 8
In matrix form, the given system of equations can be written as:
AX = B
RBSE Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 3
Determinant of matrix A
|A| = \(\left|\begin{array}{ll} 1 & 3 \\ 2 & 6 \end{array}\right|\) = (1 × 6) - 2 × (3) = 6 - 6 = 0
∴ |A| = 0
i.e., matrix A is singular.
Thus, the given system of equations is in-consistent.

RBSE Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6

Question 4.
x + y + z = 1
2x + 3y + 2z = 2
ax + ay + 2az = 4
Answer:
Given system of equations is:
x + y + z = 1
2x + 3y + 2z = 2
ax + ay + 2az = 4
In matrix form, this can be written as:
AX = B
RBSE Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 4
It means A is non-singular so given system of equations is consistent.

Question 5.
3x - y - 2z = 2
2y - z = - 1
3x - 5y = 3
Answer:
Given system of equations is :
3x - y - 2z = 2
0x + 2y - z = - 1
3x - 5y + 0z = 3
Above system can be expressed in matrix form as
AX = B
RBSE Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 5
= 3{2 × 0 - (- 5) × (- 1)} + 1{0 × 0 - 3 × (- 1)} - 2{0 × (- 5) - 3 × 2}
= 3(- 5) + 1(3) - 2(- 6)
= - 15 + 3 + 12 = 0
⇒ |A| = 0
So, matrix A is singular.
Thus, given system of equations is in-consistent.

RBSE Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6

Question 6.
5x - y + 4z = 5
2x + 3y + 5z = 2
5x - 2y + 6z = - 1
Answer:
Given system of equations is:
5x - y + 4z = 5
2x + 3y + 5z = 2
5x - 2y + 6z = - 1
We can write the above system in matrix form as :
AX = B
RBSE Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 6
= 5{3 × 6 - (- 2) × 5} + 1(2 × 6 - 5 × 5) + 4{2 × (- 2) - 5 × 3}
= 5(18 + 10) + 1(12 - 25) + 4(- 4 - 15)
= 5 × 28 - 13 + 4 × (- 19)
= 140 -13 - 76 = 140 - 89 = 51
∴ | A | = 51 ≠ 0
Thus, given system of equations is consistent.

Solve system of linear equations, using matrix method in questions 7 to 14:

Question 7.
5x + 2y = 4
7x + 3y = 5
Answer:
Given system of equations is :
5x + 2y = 4
7x + 3y = 5
Writing the given system of equations in matrix form
AX = B
RBSE Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 7
Determinant of matrix |A| is
|A| = \(\left|\begin{array}{ll} 5 & 2 \\ 7 & 3 \end{array}\right|\)
= 5 × 3 - 7 × 2 = 15 - 14 = 1
∴ |A| = 1 ≠ 0
Since, matrix A is non-singular so A-1 exists and system is consistent.
If Aij is cofactor of aij in A, then
A11 = (- 1)1 + 1(3) = 3
A12 = (- 1)1 + 2 = 7 = - 7
A21 = (- 1)2 + 12 = - 2
A22 = (- 1)2 + 25 = 5
Matrix formed by the cofactors of |A| is:
RBSE Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 8

RBSE Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6

Question 8.
2x - y = - 2
3x + 4y = 3
Answer:
Given system of equations is:
2x - y = - 2
3x + 4y = 3
Writing the given system of equations in matrix form,
AX = B ........ (1)
RBSE Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 9
Determinant of matrix A is
|A| = \(\left|\begin{array}{rr} 2 & -1 \\ 3 & 4 \end{array}\right|\)
= 2 × 4 - 3 × (- 1) = 8 + 3 = 11
∴ |A| = 1 ≠ 0
Since, matrix A is non-singular so A-1 exists, and system is consistent.
If Aij is cofactor of aij in A
A11 = (- 1)1 + 1 4 = 4
A12 = (- 1)1 + 2 3 = - 3
A21 = (- 1)2 + 1 (- 1) = (- 1) (- 1) = 1
A22 = (- 1)2 + 2 2 = 2
Matrix formed by the cofactor of |A| is:
RBSE Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 10

Question 9.
4x - 3y = 3
3x - 5y = 7
Answer:
Given system of equations is:
4y - 3 y = 3
3x - 5y = 7
Writing the given system of equations in matrix form
AX = B .......... (1)
RBSE Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 11
Determinant of matrix A,
|A| = \(\left|\begin{array}{rr} 4 & -3 \\ 3 & -5 \end{array}\right|\)
= 4 × (- 5) - 3 × (- 3) = - 20 + 9 = - 11
∴ |A| = - 11 ≠ 0
Since, matrix A is non-singular so A-1 exists and given system is consistent.
If Aij is cofactor of aij in A, then
A11 = (- 1)1 + 1 (- 5) = - 5
A12 = (- 1)1 + 2 (3) = (- 1) × 3 = - 3
A21 = (- 1)2 + 1 (- 3) = (- 1) (- 3) = 3
A22 = (- 1)2 + 2 4 = 4
Matrix formed by the cofactor of |A| is:
RBSE Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 12

RBSE Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6

Question 10.
5x + 2y = 3
3x + 2y = 5
Answer:
Given system of equations is:
5x + 2y = 3
3x + 2y = 5
Writing the given system of equations in matrix form
AX = B .......... (1)
RBSE Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 13
Determinant of matrix A,
|A| = \(\left|\begin{array}{ll} 5 & 2 \\ 3 & 2 \end{array}\right|\)
= 5 × 2 - 3 × 2 = 10 - 6 = 4
∴ |A| = 4 ≠ 0
Since, matrix A is non-singular so A-1 exists and given, system is consistent.
If Aij is cofactor of aij in A, then
A11 = (- 1)1 + 1 (2) = 2
A12 = (- 1)1 + 2 (3) = - 3
A21 = (- 1)2 + 1 (2) = - 2
A22 = (- 1)2 + 2 (5) = 5
RBSE Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 14

Question 11.
2x + y + 2 = 1
x - 2y - z = 3/2
3y - 5z = 9
Answer:
Given system of equations is :
2x + y + z = 1
x - 2y - z = 3/2
0x + 3y - 5z = 9
Writing the given system of equations in matrix form
AX = B ...(1)
RBSE Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 15
Determinant of matrix A,
|A| = \(\left|\begin{array}{rrr} 2 & 1 & 1 \\ 1 & -2 & -1 \\ 0 & 3 & -5 \end{array}\right|\)
= 2{- 2 × (- 5) - 3 × (- 1)} - 1(1 × (- 5) - 0 × (-1)} + 1{1 × 3 - 0 × (- 2)}
= 2{(10 + 3) -1(- 5 + 0) + 1(3 + 0)}
= 2 × 13 - 1 × (- 5) + 1 × 3 = 26 + 5 + 3 = 34
∴ |A| = 34 ≠ 0
Since, matrix A is non-singular so A-1 exists and given system of equations is consistent.
If Aij is cofactor of aij in A, then
A11 = + [2 × (- 5) - 3 × ( - 1)] = 10 + 3 = 13
A12 = (- 1) [(- 5 + 0)] = ( - 1) (- 5) = 5
A13 = [1 × 3 - 0 × (- 2)] = 3 + 0 = 3
A21 = (- 1) [- 5- 3] = (- 1) (- 8) = 8
A22 = + [2 × ( - 5) - 0 × 1] = - 10 - 0 = - 10
A23 = (- 1) [6 - 0] = - 6
A31 = +[1 × (- 1) - (- 2) × 1] = - 1 + 2 = 1
A32 = (- 1) [- 2 - 1] = (- 1) (- 3) = 3
A33 = + [2 × (- 2) - 1 × 1] = - 4 - 1 = - 5
RBSE Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 16

RBSE Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6

Question 12.
x - y + z = 4
2x + y - 3z = 0
x + y + z = 2
Answer:
Given system of equations is:
x - y + z = 4
2x + y - 3z = 0
x + y + z = 2
Writing it in matrix form
AX = B ...... (1)
RBSE Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 17
= (1 + 3) + (2 + 3) + (2 - 1) = 4 + 5 + 1 = 10
∴ |A| = 10 ≠ 0
Since, matrix A is non-singular so A-1 exists and given tern is consistent.
If Aij is cofactor of in aij, then
A11 = +[1 × 1 - 1 × (- 3)] = 1 + 3 = 4
A12 = (- 1) [2 + 3] = (- 1)5 = - 5
A13 = +[2 × 1 - 1 × 1] = 2 - 1 = 1
A21 = (- 1) [ - 1 - 1] = (- 1) (- 2) = 2
A22 = + [1 × 1 - 1 × 1] = 1 - 1 = 0
A23 = - 1[1 + 1] = - 2
A31 = +[(- 1) × (- 3) - 1 × 1] = 3 - 1 = 2
A32 = (- 1) [- 3 - 2] = (- 1) (- 5) = 5
A33 = +[1 × 1 - 2 × (- 1)] = 1 + 2 = 3
RBSE Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 18

Question 13.
2x + 3y + 3z = 5
x - 2y + z = - 4
3x - y - 2z = 3
Answer:
Given system of equations is:
2x + 3y + 3z = 5
x - 2y + z = - 4
3x - y - 2z = 3
Writing it in matrix form
AX = B ........ (1)
RBSE Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 19
Determinant of matrix A is
|A| = \(\left|\begin{array}{rrr} 2 & 3 & 3 \\ 1 & -2 & 1 \\ 3 & -1 & -2 \end{array}\right|\)
= 2(4 + 1) - 3 (- 2 - 3) + 3(- 1 + 6)
= 2 × 5 - 3 × (- 5) + 3 × 5 = 10 + 15 + 15 = 40
∴ |A| = 40 ≠ 0
Since, matrix A is non-singular so A-1 exists and system of equations is consistent.
If Aij is cofactor of aij in A, then
A11 = + [- 2 × (- 2) - (- 1) × 1] = 4 + 1 = 5
A12 = (- 1) [- 2 - 3] = (- 1) (- 5) = 5
A13 = + [1 × (- 1) - 3 × (- 2)] = - 1 + 6 = 5
A21 = (- 1) [- 6 + 3] = (- 1) (- 3) = 3
A22 = + [2 × (- 2) - 3 × 3] = - 4 - 9 = - 13
A23 = (- 1) [- 2 - 9] = (- 1) (- 11) = 11
A31 = + [3 × 1 - (- 2) × 3] = 3 + 6 = 9
A32 = (- 1) [2 - 3] = - 1(- 1) = 1
A33 = + [2 × (- 2) - 1 × 3] = - 4 - 3 = - 7
RBSE Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 20

RBSE Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6

Question 14.
x - y + 2z = 7
3x + 4y - 5z = - 5
2x - y + 3z = 12
Answer:
Given system of equations is:
x - y + 2z = 7
3x + 4y - 5z = - 5
2x - y,+ 3z = 12
Writing it in matrix form
AX = B
RBSE Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 21
Determinant of matrix A,
|A| = \(\left|\begin{array}{rrr} 1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3 \end{array}\right|\)
= (12 - 5) + (9 + 10) + 2(- 3 - 8)
= 7 + 19 + 2(- 11) = 26 - 22 = 4
∴ |A| = 4 ≠ 0
Since, matrix A is non-singular so A-1 exists and system of equations is consistent.
If Aij is cofactor of aij in A, then
A11 = + [4 × 3 - (- 1) × (- 5)] = 12 - 5 = 7
A12 = (- 1) [9 + 10] = (- 1) (19) = -19
A13 = + [3 × (- 1) - 2 × 4] = - 3 - 8 = - 11
A21 = (- 1) [- 3 + 2] = (-1) (- 1) = 1
A22 = + [1 × 3 - 2 × 2] = 3 - 4 = - 1
A23 = (-1)[- 1 + 2] = (- 1)(1) = - 1
A31 = + [(- 1) × (- 5) - 4 × 2] = 5 - 8 = - 3
A32 = (- 1) [- 5 - 6] = (- 1) (- 11) = 11
A33 = + [1 × 4 - 3 × (- 1)] = 4 + 3 = 7
RBSE Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 22

Question 15.
If A = \(\left[\begin{array}{rrr} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{array}\right]\), find A-1. Using A-1 solve the following system of equations:
2x - 3y + 5z = 11
3x + 2y - 4z = - 5
x + y - 2z = - 3
Answer:
Given, A = \(\left[\begin{array}{rrr} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{array}\right]\)
Determinant of matrix A is
|A| = \(\left|\begin{array}{rrr} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{array}\right|\)
⇒ |A| - 2(- 4 + 4) + 3(- 6 + 4) + 5(3 - 2)
= 2 × 0 + 3(- 2) + 5 × 1 = - 6 + 5 = - 1
∴ |A| = 4 ≠ 0
So, matrix A is non-singular so A-1 exist.
If Aij is cofactor of aij then
A11 = + [2 × (- 2) -1 × (- 4)] = - 4 + 4 = 0
A12 = (-1) [- 6 + 4] = (-1) (- 2) = 2
A13 = + [3 × 1 -1 × 2] - 3 - 2 = 1
A21 = (- 1) [6 - 5] = - 1
A22 = + [2 × (- 2) - 1 × 5] = - 4 - 5 = - 9
A23 = (- 1) [2 + 3] = - 5
A31 = + [- 3 × (- 4) - 2 × 5] = 12 - 10 = 2
A32 = (-1) [- 8 - 15] = (- 1) (- 23) = 23
A33 = + [2 × 2 - 3 × (- 3)] = 4 + 9 = 13
RBSE Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 23
Writing the given system of equations in matrix form, we get
RBSE Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 24
Since, matrix A is non-singular so A-1 exists.
since, A × B ⇒ x = A-1B
RBSE Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 25

RBSE Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6

Question 16.
The cost of 4 kg onion, 3 kg wheat and 2 kg rice is ₹ 60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is ₹ 90. The cost of 6 kg onion, 2 kg wheat and 3 kg rice is ₹ 70. Find the cost of each item (per kg) by matrix method
Answer:
Let cost of 1 kg onion = ₹ x
cost of 1 kg wheat = ₹ y
And cost of 1 kg rice = ₹ z
We have,
cost of 4 kg onion = ₹ 4x
cost of 3 wheat = ₹ 3y
And cost of 2 kg rice = ₹ 2z
∴ 4x + 3y + 2z = 60 .......... (1)
Again cost of 2 kg onion = ₹ 2x
cost of 4 kg wheat = ₹ 4y
and cost of 6 kg rice = ₹6z
∴ 2x + 4y + 6z = 90 ........... (2)
Now cost of 6 kg onion = ₹ 6x
cost of 2 kg wheat = ₹2y
and cost of 3 kg rice = ₹3z
∴ 6x + 2y + 3z = 70 ........... (3)
Thus, system of equations is :
4x + 3y + 2z = 60
2x + 4y + 6z = 90
6x + 2y + 3z = 70
or
4x + 3y + 2z = 60
x + 2y + 3z = 45
6x + 2y + 3z = 70
Writing in matrix form
AX = B ......... (4)
RBSE Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 26
Determinant of matrix A is
|A| = \(\left|\begin{array}{lll} 4 & 3 & 2 \\ 1 & 2 & 3 \\ 6 & 2 & 3 \end{array}\right|\)
= 4(2 × 3 - 2 × 3) - 3(1 × 3 - 6 × 3) + 2(1 × 2 - 6 × 2)
= 4(6 - 6) - 3(3 - 18) + 2(2 - 12)
= 4 × 0 - 3(- 15) + 2(- 10)
= 0 + 45 - 20 = 25
∴ |A| = 25 ≠ 0
If Aij is cofactor of aij in A, then
A11 = + [2 × 3 - 2 × 3] = 0
A12 = (- 1) [1 × 3 - 6 × 3] = (- 1) (3 - 18) = (- 1) (- 15) = 15
A13 = + [1 × 2 - 6 × 2] = 2 - 12 = - 10
A21 = (- 1) [3 × 3 - 2 × 2] = (- 1) (9 - 4) = (- 1)5 = - 5
A22 = [4 × 3 - 6 × 2] = 12 - 12 = 0
A23 = (- 1) [4 × 2 - 6 × 3] = (-1) (8 - 18) = (-1) (-10) = 10
A31 = +[3 × 3 - 2 × 2] = 9 - 4 = 5
A32 = (- 1) [12 - 2] = - 10
A33 = + [4 × 2 - 1 × 3] = 8 - 3 = 5
RBSE Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 27
⇒ x = 5, y = 8, z = 8
Thus, Onion = ₹ 5, Wheat = ₹ 8, Rice = ₹ 8

Bhagya
Last Updated on Nov. 2, 2023, 9:25 a.m.
Published Nov. 1, 2023