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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

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

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

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

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.

**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.

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