You cannot copy content of this page

NCERT solution class 12 chapter 4 Determinants exercise 4.6 mathematics part 1

EXERCISE 4.6


Page No 136:

Question 1:

Examine the consistency of the system of equations.

+ 2= 2

2x + 3= 3

Answer:

The given system of equations is:

+ 2= 2

2x + 3= 3

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

∴ 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.

2− y = 5

x + = 4

Answer:

The given system of equations is:

2− y = 5

x + = 4

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

∴ 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

∴ A is a singular matrix.

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

∴ 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

∴ A is a singular matrix.

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

∴ A is non-singular.

Therefore, A−1 exists.

Hence, the given system of equations is consistent.

Question 7:

Solve system of linear equations, using matrix method.

Answer:

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

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

Question 8:

Solve system of linear equations, using matrix method.

Answer:

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

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

Question 9:

Solve system of linear equations, using matrix method.

Answer:

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

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

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

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

Question 11:

Solve system of linear equations, using matrix method.

Answer:

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

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

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

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

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

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

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

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

Page No 137:

Question 15:

If, find A−1. Using A−1 solve the system of equations

Answer:

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

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:

This system of equations can be written in the form of AX = B, where

Now,

X = A−1 B

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.


 

Leave a Comment

Your email address will not be published. Required fields are marked *

error: Content is protected !!
Free Web Hosting