# NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6

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## NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6

Get Free NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 PDF in Hindi and English Medium. Sets Class 12 Maths NCERT Solutions are extremely helpful while doing your homework. Determinants Exercise 4.6 Class 12 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 4 Class 12 Determinants Ex 4.6 provided in NCERT Textbook.

Free download NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, BIE, Intermediate and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20.

The topics and sub-topics included in the Determinants chapter are the following:

 Section Name Topic Name 4 Determinants 4.1 Introduction 4.2 Determinant 4.3 Properties of Determinants 4.4 Area of a Triangle 4.5 Adjoint and Inverse of a Matrix 4.6 Applications of Determinants and Matrices 4.7 Summary

### NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6

Examine the consistency of the system of equations in Questions 1 to 6:

Ex 4.6 Class 12 Maths Question 1.
x + 2y = 2
2x + 3y = 3
Solution:
x + 2y = 2,
2x + 3y = 3
=> (begin{bmatrix} 1 & 2 \ 2 & 3 end{bmatrix}left[ begin{matrix} x \ y end{matrix} right] =left[ begin{matrix} 2 \ 3 end{matrix} right] )
=> AX = B
Now |A| = (begin{vmatrix} 1 & 2 \ 2 & 3 end{vmatrix})
= 3 – 4
= – 1 ≠ 0.
Hence, equations are consistent.

Ex 4.6 Class 12 Maths Question 2.
2x – y = 5
x + y = 4
Solution:
2x – y = 5,
x + y = 4
=> (begin{bmatrix} 2 & -1 \ 1 & 1 end{bmatrix}left[ begin{matrix} x \ y end{matrix} right] =left[ begin{matrix} 5 \ 4 end{matrix} right] )
=> AX = B
Now |A| = (begin{vmatrix} 2 & -1 \ 1 & 1 end{vmatrix})
= 2 + 1
= 3 ≠ 0.
Hence, equations are consistent.

Ex 4.6 Class 12 Maths Question 3.
x + 3y = 5,
2x + 6y = 8
Solution:
x + 3y = 5,
2x + 6y = 8
=> (begin{bmatrix} 1 & 3 \ 2 & 6 end{bmatrix}left[ begin{matrix} x \ y end{matrix} right] =left[ begin{matrix} 5 \ 8 end{matrix} right] )
=> AX = B
Now |A| = (begin{vmatrix} 1 & 3 \ 2 & 6 end{vmatrix})
= 6 – 6
= 0. Hence, equations are consistent with no solution

Ex 4.6 Class 12 Maths Question 4.
x + y + z = 1
2x + 3y + 2z = 2
ax + ay + 2az = 4
Solution:
x + y + z = 1
2x + 3y + 2z = 2
x + y + z = (\ frac { 4 }{ a } ) Ex 4.6 Class 12 Maths Question 5.
3x – y – 2z = 2
2y – z = – 1
3x – 5y = 3
Solution:
(left[ begin{matrix} 3 & -1 & -2 \ 0 & 2 & -1 \ 3 & -5 & 0 end{matrix} right] left[ begin{matrix} x \ y \ z end{matrix} right] =left[ begin{matrix} 2 \ -1 \ 3 end{matrix} right] )
=> AX = B  Ex 4.6 Class 12 Maths Question 6.
5x – y + 4z = 5
2x + 3y + 5z = 2
5x – 2y + 6z = -1
Solution:
Given
5x – y + 4z = 5
2x + 3y + 5z = 2
5x – 2y + 6z = -1
(left[ begin{matrix} 5 & -1 & 4 \ 2 & 3 & 5 \ 5 & -2 & 6 end{matrix} right] left[ begin{matrix} x \ y \ z end{matrix} right] =left[ begin{matrix} 5 \ 2 \ -1 end{matrix} right] )
(AX=B|A|=left[ begin{matrix} 5 & -1 & 4 \ 2 & 3 & 5 \ 5 & -2 & 6 end{matrix} right] )
= 5(18 + 10)+1(12 – 25)+4(-4-15)
= 140-13-76
= 51 ≠ 0
Hence equations are consistent with a unique
solution.

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

Ex 4.6 Class 12 Maths Question 7.
5x + 2y = 4
7x + 3y = 5
Solution:
The given system of equations can be written as  Ex 4.6 Class 12 Maths Question 8.
2x – y = – 2
3x + 3y = 3
Solution:
The given system of equations can be written Ex 4.6 Class 12 Maths Question 9.
4x – 3y = 3
3x – 5y = 7
Solution:
The given system of equations can be written as
(begin{bmatrix} 4 & -3 \ 3 & -5 end{bmatrix}left[ begin{matrix} x \ y end{matrix} right] =left[ begin{matrix} 3 \ 7 end{matrix} right] i.e,,AX=B)
where (A=begin{bmatrix} 4 & -3 \ 3 & -5 end{bmatrix}) Ex 4.6 Class 12 Maths Question 10.
5x + 2y = 3
3x + 2y = 5
Solution:
The given system of equations can be written as
(begin{bmatrix} 5 & 2 \ 3 & 2 end{bmatrix}left[ begin{matrix} x \ y end{matrix} right] =left[ begin{matrix} 3 \ 5 end{matrix} right] i.e,,AX=B)
where (A=begin{bmatrix} 5 & 2 \ 3 & 2 end{bmatrix}) Ex 4.6 Class 12 Maths Question 11.
2x + y + z = 1,
x – 2y – z = 3/2
3y – 5z = 9
Solution:
The given system of equations are
2x + y + z = 1,
x – 2y – z = 3/2,
3y – 5z = 9
We know AX = B => X = A-1B Ex 4.6 Class 12 Maths Question 12.
x – y + z = 4
2x + y – 3z = 0
x + y + z = 2.
Solution:
The given system of equations can be written
(left[ begin{matrix} 1 & -1 & 1 \ 2 & 1 & -3 \ 1 & 1 & 1 end{matrix} right] left[ begin{matrix} x \ y \ z end{matrix} right] =left[ begin{matrix} 4 \ 0 \ 2 end{matrix} right] i.e,,AX=B)  Ex 4.6 Class 12 Maths Question 13.
2x + 3y + 3z = 5
x – 2y + z = – 4
3x – y – 2z = 3
Solution:
The given system of equations can be written as:
(left[ begin{matrix} 2 & 3 & 3 \ 1 & -2 & 1 \ 3 & -1 & -2 end{matrix} right] left[ begin{matrix} x \ y \ z end{matrix} right] =left[ begin{matrix} 5 \ -4 \ 3 end{matrix} right] i.e,,AX=B )  Ex 4.6 Class 12 Maths Question 14.
x – y + 2z = 7
3x + 4y – 5z = – 5
2x – y + 3z = 12.
Solution:
The given system of equations can be written
(left[ begin{matrix} 1 & -1 & 2 \ 3 & 4 & -5 \ 2 & -1 & 3 end{matrix} right] left[ begin{matrix} x \ y \ z end{matrix} right] =left[ begin{matrix} 7 \ -5 \ 12 end{matrix} right] i.e,,AX=B )  Ex 4.6 Class 12 Maths Question 15.
If A = (left[ begin{matrix} 2 & -3 & 5 \ 3 & 2 & -4 \ 1 & 1 & -2 end{matrix} right] ) Find A-1. Using A-1Solve the following system of linear equations 2x – 3y + 5z = 11,3x + 2y – 4z = – 5, x + y – 2z = – 3
Solution:
We have AX = B
where (A=left[ begin{matrix} 2 & -3 & 5 \ 3 & 2 & -4 \ 1 & 1 & -2 end{matrix} right] ,X=left[ begin{matrix} x \ y \ z end{matrix} right] ) Ex 4.6 Class 12 Maths Question 16.
The cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs. 69. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is Rs. 90. The cost of 6 kg onion, 2 kg wheat and 3 kg rice is Rs. 70. Find the cost of each item per kg by matrix method
Solution:
Let cost of 1 kg onion = Rs x
and cost of 1 kg wheat = Rs y
and cost of 1 kg rice = Rs z
4x+3y+2z=60
2x+4y+6z=90
6x+2y+3z=70 ### NCERT Solutions for Class 12 Maths Chapter 4 Determinants Hindi Medium Ex 4.6                  More Resources for NCERT Solutions Class 12: