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

Now,

|A| = 1(3) -2(2) = 3 – 4 = -1 ≠ 0

 

∴ 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

Now,

|A| = 2(1) – (-1)(1) = 2 + 1 = 3 ≠ 0

 

∴ 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

Now,

|A| = 1(6) – 3(2) = 6 – 6 = 0

∴ A is a singular matrix.

Now,

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

Now,

|A| = 16a – 2a) -1 (4a – 2a) +1 (2a – 3a)

= 4a – 2a – a = 4a – 3a = a ≠ 0

∴ 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

Now,

|A| = 3(0 -5) -0 + 3 (1 + 4) = -15 + 15 = 0

∴ A is singular matrix.

Now,

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

Now,

|A| = 5(18 + 10) + 1 (12 - 25) + 4 (-4 – 15)

= 5(28) +1(-13) +4(-19)

= 140 – 13 – 76

= 51 ≠ 0

∴ 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

5x + 2y = 4

7x + 3y = 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.

Now,

A^-1 = (1 / |A|) (adjA)

Hence, x = 2 and y = -3.

 

Question 8:

Solve system of linear equations, using matrix method.

2x – y = -2

3x + 4y = 3

Answer

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

Now,

|A| = 8 + 3 = 11 ≠ 0

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

Now,

Hence, x = (-5/11)  and y = (12/11).

 

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

4x – 3y = 3

3x + 5y = 7

Answer

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

Now,

|A| = -20 + 9 = -11 ≠ 0

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

Now,

Hence, x = (-6/11) and y = (-19/11).

 

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

Now,

|A| = 10 - 6 = 4 ≠ 0

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

 

Question 11:

Solve system of linear equations, using matrix method.

2x + y + z = 1

x – 2y – z = 3/2

3y – 5z = 9

Answer

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

Now,

|A| = 2(10 + 3) – 1 (-5 - 3) + 0 = 2(13) -1 (-8) = 26 + 8 = 34 ≠ 0 

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

Now, A₁₁ = 13, A₁₂ = 5, A₁₃ = 3

A₂₁ = 1, A₃₂ = 3, A₃₃ = -5

Hence, x = 1, y = ½ , and z = 3/2.

 

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

Now,

|A| = 1(1 + 3) + 1(2 + 3) + 1(2 - 1) = 4 + 5 + 1 ≠ 0

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

Now, A₁₁ = 4, A₁₂ = -5, A₁₃ = 1

A₂₁ = 2, A₂₂ = 0, A₂₃ = -2

A₃₁ = 2, A₃₂ = 5, A₃₃ = 3

Hence, x = 2, y = -1, and z = 1.

 

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

Now,

|A| = 2(4 + 1) - 3(-2 - 3) + 3(-1 + 6) = 2(5) -3(5) + 3(5) = 10 + 15 + 15 = 40 ≠ 0

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

Now, A₁₁ = 5, A₁₂ = 5, A₁₃ = 5

A₂₁ = 3, A₂₂ = -13, A₂₃ = 11

A₃₁ = 9, A₃₂ = 1, A₃₃ = -7

Hence, x = 1, y = 2, and z = 1.

 

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

Now,

|A| = 1(12 - 5) + 1(9 + 10) + 2(-3 - 8) = 7 + 19 - 22 = 4 ≠ 0

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

Now, A₁₁ = 7, A₁₂ = -19, A₁₃ = -11

A₂₁ = 1, A₂₂ = -1, A₂₃ = -1

A₃₁ = -3, A₃₂ = 11, A₃₃ = 7

Hence, x = 2, y = 1, and z = 3.

 

Question 15:

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

Answer

∴ |A| = 2(-4 + 4) + 3(-6 + 4) + 5(3 - 2) = 0 - 6 + 5 = -1 ≠ 0

Now, A₁₁ = 0, A₁₂ = 2, A₁₃ = 1

A₂₁ = -1, A₂₂ = -9, A₂₃ = -5

A₃₁ = 2, A₃₂ = 23, A₃₃ = 13

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

The solution of the system of equations is given by X = A^-1 B.

X = A^-1B

Hence, x = 1, y = 2, and z = 3.

 

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:

4x + 3y + 2z = 60

2x + 4y + 6z = 90

6x + 2y + 3z = 70

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

|A| = 4(12 - 12) - 3(6 - 36) + 2(4 - 24) = 0 + 90 - 40 = 50 ≠ 0

Now, A₁₁ = 0, A₁₂ = 30, A₁₃ = -20

A₂₁ = -5, A₂₂ = 0, A₂₃ = 10

A₃₁ = 10, A₃₂ = -20, A₃₃ = 10

Now,

X = A^-1 B