NCERT Solutions - Determinants, Exercise 4.6 JEE Notes | EduRev

Mathematics (Maths) Class 12

JEE : NCERT Solutions - Determinants, Exercise 4.6 JEE Notes | EduRev

The document NCERT Solutions - Determinants, Exercise 4.6 JEE Notes | EduRev is a part of the JEE Course Mathematics (Maths) Class 12.
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Determinants

Exercise 4.6

Question 1: Examine the consistency of the system of equations.
 x + 2y = 2
 2x + 3y = 3

Answer
The given system of equations is:
x + 2y = 2
2x + 3y = 3
The given system of equations can be written in the form of AX = B, where

NCERT Solutions - Determinants, Exercise 4.6 JEE Notes | EduRev
NCERT Solutions - Determinants, Exercise 4.6 JEE Notes | EduRev A is non-singular.
Therefore, A−1 exists.
Hence, the given system of equations is consistent.

Question 2: Examine the consistency of the system of equations.
 2x − y = 5 x
 + y = 4

Answer
The given system of equations is:
2x − y = 5 x
+ y = 4
The given system of equations can be written in the form of AX = B, where
NCERT Solutions - Determinants, Exercise 4.6 JEE Notes | EduRev
NCERT Solutions - Determinants, Exercise 4.6 JEE Notes | EduRev A is non-singular.
Therefore, A−1 exists.
Hence, the given system of equations is consistent.

Question 3: Examine the consistency of the system of equations.
 x + 3y = 5
 2x + 6y = 8

Answer
The given system of equations is:
x + 3y = 5
2x + 6y = 8
The given system of equations can be written in the form of AX = B, where
NCERT Solutions - Determinants, Exercise 4.6 JEE Notes | EduRev
Thus, the solution of the given system of equations does not exist. Hence, the system of equations is inconsistent.

Question 4: Examine the consistency of the system of equations.
 x + y + z = 1 

2x + 3y + 2z = 2 

ax + ay + 2az = 4

Answer
The given system of equations is:
x + y + z = 1 2x
+ 3y + 2z = 2 ax
+ ay + 2az = 4
This system of equations can be written in the form AX = B, where

NCERT Solutions - Determinants, Exercise 4.6 JEE Notes | EduRev
NCERT Solutions - Determinants, Exercise 4.6 JEE Notes | EduRev A is non-singular.
Therefore, A−1 exists.
Hence, the given system of equations is consistent.

Question 5: Examine the consistency of the system of equations.
 3x − y − 2z = 2
 2y − z = −1
 3x − 5y = 3

Answer
The given system of equations is:
3x − y − 2z = 2
2y − z = −1
3x − 5y = 3
This system of equations can be written in the form of AX = B, where

NCERT Solutions - Determinants, Exercise 4.6 JEE Notes | EduRev

A is a singular matrix. 

NCERT Solutions - Determinants, Exercise 4.6 JEE Notes | EduRev

NCERT Solutions - Determinants, Exercise 4.6 JEE Notes | EduRev

NCERT Solutions - Determinants, Exercise 4.6 JEE Notes | EduRev

Thus, the solution of the given system of equations does not exist. Hence, the system of equations is inconsistent.

Question 6: Examine the consistency of the system of equations.
 5x − y + 4z = 5
 2x + 3y + 5z = 2
 5x − 2y + 6z = −1

Answer
The given system of equations is:
5x − y + 4z = 5
2x + 3y + 5z = 2
5x − 2y + 6z = −1
This system of equations can be written in the form of AX = B, where

NCERT Solutions - Determinants, Exercise 4.6 JEE Notes | EduRev
Therefore, A−1 exists.
Hence, the given system of equations is consistent.

Question 7: Solve system of linear equations, using matrix method.

NCERT Solutions - Determinants, Exercise 4.6 JEE Notes | EduRev

Answer
The given system of equations can be written in the form of AX = B, where

NCERT Solutions - Determinants, Exercise 4.6 JEE Notes | EduRev
Thus, A is non-singular. Therefore, its inverse exists.

NCERT Solutions - Determinants, Exercise 4.6 JEE Notes | EduRev


Question 8: Solve system of linear equations, using matrix method.

NCERT Solutions - Determinants, Exercise 4.6 JEE Notes | EduRev

Answer :The given system of equations can be written in the form of AX = B, where

NCERT Solutions - Determinants, Exercise 4.6 JEE Notes | EduRev
Thus, A is non-singular. Therefore, its inverse exists.

NCERT Solutions - Determinants, Exercise 4.6 JEE Notes | EduRev

Question 9: Solve system of linear equations, using matrix method.

NCERT Solutions - Determinants, Exercise 4.6 JEE Notes | EduRev

Answer The given system of equations can be written in the form of AX = B, where

NCERT Solutions - Determinants, Exercise 4.6 JEE Notes | EduRev
Thus, A is non-singular. Therefore, its inverse exists.

NCERT Solutions - Determinants, Exercise 4.6 JEE Notes | EduRev

NCERT Solutions - Determinants, Exercise 4.6 JEE Notes | EduRev

NCERT Solutions - Determinants, Exercise 4.6 JEE Notes | EduRev

NCERT Solutions - Determinants, Exercise 4.6 JEE Notes | EduRev

Question 10: Solve system of linear equations, using matrix method.
 5x + 2y = 3
 3x + 2y = 5


Answer: The given system of equations can be written in the form of AX = B, where

NCERT Solutions - Determinants, Exercise 4.6 JEE Notes | EduRev
Thus, A is non-singular. Therefore, its inverse exists.

Question 11: Solve system of linear equations, using matrix method.

NCERT Solutions - Determinants, Exercise 4.6 JEE Notes | EduRev


Answer: The given system of equations can be written in the form of AX = B, where
NCERT Solutions - Determinants, Exercise 4.6 JEE Notes | EduRev
Thus, A is non-singular. Therefore, its inverse exists.

NCERT Solutions - Determinants, Exercise 4.6 JEE Notes | EduRev


Question 12: Solve system of linear equations, using matrix method.
 x − y + z = 4 

2x + y − 3z = 0 

x + y + z = 2

Answer: The given system of equations can be written in the form of AX = B, where

NCERT Solutions - Determinants, Exercise 4.6 JEE Notes | EduRev
Thus, A is non-singular. Therefore, its inverse exists.

NCERT Solutions - Determinants, Exercise 4.6 JEE Notes | EduRev


Question 13: Solve system of linear equations, using matrix method.
 2x + 3y + 3z = 5
 x − 2y + z = −4
 3x − y − 2z = 3

Answer: The given system of equations can be written in the form AX = B, where

NCERT Solutions - Determinants, Exercise 4.6 JEE Notes | EduRev
Thus, A is non-singular. Therefore, its inverse exists.

NCERT Solutions - Determinants, Exercise 4.6 JEE Notes | EduRev


Question 14: Solve system of linear equations, using matrix method.

x − y + 2z = 7
 3x + 4y − 5z = −5
 2x − y + 3z = 12


Answer: The given system of equations can be written in the form of AX = B, where

NCERT Solutions - Determinants, Exercise 4.6 JEE Notes | EduRev

Thus, A is non-singular. Therefore, its inverse exists.

NCERT Solutions - Determinants, Exercise 4.6 JEE Notes | EduRev
Question 15:

If  NCERT Solutions - Determinants, Exercise 4.6 JEE Notes | EduRev
find A−1. Using A−1 solve the system of equations


Answer

NCERT Solutions - Determinants, Exercise 4.6 JEE Notes | EduRev
Now, the given system of equations can be written in the form of AX = B, where

NCERT Solutions - Determinants, Exercise 4.6 JEE Notes | EduRev

NCERT Solutions - Determinants, Exercise 4.6 JEE Notes | EduRev

Question 16: The cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs 60. 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 cost of each item per kg by matrix method.


Answer
Let the cost of onions, wheat, and rice per kg be Rs x, Rs y,and Rs z respectively.
Then, the given situation can be represented by a system of equations as:

NCERT Solutions - Determinants, Exercise 4.6 JEE Notes | EduRev
This system of equations can be written in the form of AX = B, where

NCERT Solutions - Determinants, Exercise 4.6 JEE Notes | EduRev
Now,
X = A−1 B

NCERT Solutions - Determinants, Exercise 4.6 JEE Notes | EduRev
Hence, the cost of onions is Rs 5 per kg, the cost of wheat is Rs 8 per kg, and the cost of rice is Rs 8 per kg.

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