**Question **1. State, whether the following statements are true or false. If false, give a reason.

(i) If A and B are two matrices of orders 3 × 2 and 2 × 3 respectively; then their sum A + B is possible.

(ii) The matrices A2 × 3 and B2 × 3 are conformable for subtraction.

(iii) Transpose of a 2 × 1 matrix is a 2 × 1 matrix.

(iv) Transpose of a square matrix is a square matrix.

Solution:

(i) False, When the order of both matrices A and B is the same, the total A + B is achievable.

(ii) True

(iii) False, A 1×2 matrix is the transpose of a 2×1 matrix.

(iv) True

(v) False, There is only one column in a column matrix, yet there are numerous rows.

**Question **2. Given:

Here matrices are equal then corresponding elements are equal.

𝑥 = 3________(𝑖)

𝑦 + 2 = 1

𝑦 = −1________(𝑖𝑖)

𝑧 – 1 = 2

𝑧 = 3________(𝑖𝑖𝑖)

Hence, the value of 𝑥, 𝑦 and 𝑧 is 3, -1 and 3.

**Question **3. Solve for a, b and c if

Matrices are equal then corresponding elements are equal.

𝑎 + 5 = 2_______(𝑖)

𝑎 = −3

−4 = 𝑏 + 4

𝑏 = −8__________(𝑖𝑖)

2 = 𝑐– 1

𝑐 = 3__________(𝑖𝑖𝑖)

Hence, the value of 𝑥, 𝑦 and 𝑧 is -3, -8 and 3 respectively.

(ii) We have,

Matrices are equal then corresponding elements are equal.

𝑎 = 3____________(𝑖)

𝑎 − 𝑏 = −1____________(𝑖𝑖)

𝑏 + 𝑐 = 2____________(𝑖𝑖𝑖)

From equation (ii) we get,

𝑎 − 𝑏 = −1

3 − 𝑏 = −1

−𝑏 = −1 − 3

−𝑏 = −4

𝑏 = 4

From equation (iii) we get,

𝑏 + 𝑐 = 2

4 + 𝑐 = 2

𝑐 = 2 − 4

𝑐 = −2

Hence, the value of a, b and c is 3, 4 and -2 respectively.

**Question **4. If 𝐴 = [8 −3] and 𝐵 = [4 −5]; find:

(𝑖)𝐴 + 𝐵

(𝑖𝑖) 𝐵– 𝐴

Solution:

(i) 𝐴 + 𝐵

[8 −3] + [4 −5]

[8 + 4 −3 − 5]

[12 −8]

Hence, the value of 𝐴 + 𝐵 is [12 −8].

(ii) 𝐵 − 𝐴

[4 −5] − [8 −3]

[4 − 8 −5 − (−3)]

[−4 −5 + 3]

[−4 −2]

Hence, the value of 𝐵 − 𝐴 is [−4 −2].

**Question **5. If

**(𝑖) 𝐵 + 𝐶(𝑖𝑖) 𝐴 − 𝐶(𝑖𝑖𝑖) 𝐴 + 𝐵 − 𝐶(𝑖𝑣) 𝐴 − 𝐵 + 𝐶Solution:**

(i) It is given that,

**Question **6. Wherever possible, write each of the following as a single matrix.

**Question **7. Find x and y from the following equations:

Matrices are equal then corresponding elements are equal.

3 − 𝑥 = 7____________(i)

𝑦 − 4 = 2____________(ii)

From equation (i) we get,

−𝑥 = 7 − 3

−𝑥 = 4

𝑥 = −4

From equation (ii) we get,

𝑦 − 4 = 2

𝑦 = 2 + 4

𝑦 = 6

Hence, the value of x and y is -4 and 6 respectively.

**Question **8. Given:

**Question **9. Write the additive inverse of matrices A, B and C:

**Question **10. Given A = [2 −3], B = [0 2] and C = [−1 4]; find the matrix X in each of the following:

(𝑖) 𝑋 + 𝐵 = 𝐶 − 𝐴

(𝑖𝑖) 𝐴 − 𝑋 = 𝐵 + 𝐶

Solution:

It is given that,

A = [2 −3], B = [0 2] and C = [−1 4]

(i) X + B = C − A

Put the value of A, B and C.

**Question **11. Given A =

### Exercise 9B

**Question **1. Evaluate

**Question **2. Find x and y if:

**Solution:**

(i) 3[4 𝑥] + 2[𝑦 −3] = [10 0]

[12 3𝑥] + [2𝑦 −6] = [10 0]

[12 + 2𝑦 3𝑥 + (−6)] = [10 0]

By comparing,

12 + 2𝑦 = 10_____________(𝑖)

3𝑥 − 6 = 0_____________(𝑖𝑖)

From equation (i) we get,

12 + 2𝑦 = 10

2𝑦 = 10 − 12

2𝑦 = −2

𝑦 = −2/2

𝑦 = −1

From equation (ii) we get,

3𝑥 = 6

𝑥 = 6/3

𝑥 = 2

Hence, the value of x and y is 2 and -1.

**Question **3. Given

**Question **4. If

**Question **5. Given

**Question **6. If

By comparing both the matrix is we get,

2𝑥 + 9 = −7_______________(𝑖)

3𝑦 = 15_______________(𝑖𝑖)

𝑧 = 9_______________(𝑖𝑖𝑖)

From equation (i) we get,

2𝑥 = −7 − 9

2𝑥 = −16

𝑥 = −8

From equation (ii) we get,

3𝑦 = 15

𝑦 = 15/3

𝑦 = 5

Hence, the value of 𝑥, 𝑦 and 𝑧 are -8, 5 and 9 respectively.

**Question **7. Given

**Question **8. Given

**Question **9. If

**Question **10. If I is**the unit matrix of order 2×2; find the matrix M, such that:**

**Question **11. If

### Exercise 9C

**Question **1. Evaluate: If possible:

In the given matrix number of columns does not equal the number of rows in the second matrix. Hence, product can’t be possible.

**Question **2. If

**Question **3. If

4𝑥 = 16____________(𝑖)

1 = −𝑦**__**(𝑖𝑖)

From equation (i) we get,

4𝑥 = 16

𝑥 = 16

4

𝑥 = 4

From equation (ii) we get,

1 = −𝑦

𝑦 = −1

Hence, the value of x and y is 4 and -1 respectively.

**Question **4. Find x and y, if:

By comparing both the sides,

20 + 3𝑥 = 𝑦(𝑖)

5𝑥 − 2 = 8___________(𝑖𝑖)

From equation (𝑖𝑖) get the value of 𝑥

5𝑥 − 2 = 8

5𝑥 = 8 + 2

5𝑥 = 10

𝑥 = 10/5

𝑥 = 2

Put the value of x in equation (i)

20 + 3(2) = 𝑦

20 + 6 = 𝑦

26 = 𝑦

𝑦 = 26

Hence, the value of x and y is 2 and 26 respectively.

**Question **5. If

**Question **6. Given

(iii) Because the number of columns in matrix A is not equal to the number of rows, the product AA=A^2 is not possible.

**Question **7. Let

**Question **8. If

**Question **9. If

**Question **10. Given

**Question **11. If

**Question **12. Find the matrix A, if

**Question **13. If

By comparing,

1 + 𝑎 = 1_____________(𝑖)

𝑎 + 𝑏^{2} = 1___________(𝑖𝑖)

From equation (i),

1 + 𝑎 = 1

𝑎 = 1 − 1

𝑎 = 0

Put the value of a in equation (𝑖𝑖)

𝑎 + 𝑏^{2} = 1

0 + 𝑏^{2} = 1

𝑏^{2} = 1

𝑏 = 1

**Question **14. If

**Question **15. If

**Question **16 (i). Solve for x and

**Question **16 (ii). Solve for x and y:

**Question **16 (iii). Solve for x and y:

By comparing both matrix,

−4 = 2𝑦**___**(𝑖)

2𝑥 = 6_______________(𝑖𝑖)

From equation (i) we get, the value of 𝑦

2𝑦 = −4

𝑦 = −4/2

𝑦 = −2

From equation (ii) we get, the value of 𝑥

2𝑥 = 6

𝑥 = 6/2

𝑥 = 3

Hence, the value of x and y is 3 and -2.

**Question **17. In each case given below, find:

(a) The order of matrix M.

(b) The matrix M.

[𝑎 + 0 𝑎 + 2𝑏] = [1 2]

By comparing

𝑎 + 0 = 1__________(𝑖)

𝑎 + 2𝑏 = 2__________(𝑖𝑖)

From equation (i) we get,

𝑎 = 1

Put the value of a in equation (ii),

1 + 2𝑏 = 2

2𝑏 = 2 − 1

2𝑏 = 1

𝑏 = 1/2

Hence, the value of a and b is 1 and 1/2.

By comparing both the matrix,

𝑎 + 4𝑏 = 13___________________(𝑖)

2𝑎 + 𝑏 = 5 **_______**(𝑖𝑖)

From equation (i) we get,

𝑎 + 4𝑏 = 13

𝑎 = 13 − 4𝑏

**(𝑖𝑖𝑖)**

*_______*Value of a put in equation (ii)

2(13 − 4𝑏) + 𝑏 = 5

26 − 8𝑏 + 𝑏 = 5

26 − 7𝑏 = 5

−7𝑏 = 5 − 26

−7𝑏 = −21

𝑏 = −21/−7

𝑏 = 3

Put the value of b in equation (iii)

𝑎 = 13 − 4(3)

𝑎 = 13 − 12

𝑎 = 1

Hence, the value of a and b is 1 and 3.

**Question **18. If

**Question **19. If

**Question **20. If A and B are any two 2 × 2 matrices such that 𝐴𝐵 = 𝐵𝐴 = 𝐵 and B is not a zero matrix, what can you say about the matrix A?

Solution:

It is given that,

𝐴𝐵 = 𝐵𝐴 = 𝐵

We know that, 𝐴𝐼 = 𝐼𝐴, with I being the identified matrix.

As a result, B stands for the identified matrix.

**Question **21. Given

By comparing,

3𝑎 = 3 + 𝑎**__**(𝑖)

3𝑏 + 0 = 𝑏

**(𝑖𝑖)**

*__*4𝑐 = 4 + 𝑐

**(𝑖𝑖𝑖)**

*__*From equation (i) we get,

3𝑎 − 𝑎 = 3

2𝑎 = 3

𝑎 = 3/2

From equation (ii) we get,

3𝑏 + 0 = 𝑏

3𝑏 − 𝑏 = 0

𝑏 = 0

From equation (iii) we get,

4𝑐 = 4 + 𝑐

4𝑐 − 𝑐 = 4

3𝑐 = 4

𝑐 = 4/3

Hence, the value of a, b and c is 3/2, 0 and 4/3.

**Question **22. If

**Question **23. Given the matrices:

**Question **24. If

**Question **25. If

**Question **26. If

**Question **27. If

**Question **28. If

**Question **29. If

By comparing,

6+6y=0____________(i)

2x+12=0____________(ii)

From equation (i) we get,

6+6y=0

6y=-6

y=-6/6

y=-1

From equation (ii) we get,

2x+12=0

2x=-12

x=-12/2

x=-6

Hence, the value x and y is -1 and -6.

**Question **30. Evaluation without using tables:

**Question **31. State, with reason, whether the following are true or false. A, B and C are matrices of order 2 × 2.**(𝑖) 𝐴 + 𝐵 = 𝐵 + 𝐴****(𝑖𝑖) 𝐴 – 𝐵 = 𝐵 – 𝐴****(𝑖𝑖𝑖) (𝐵. 𝐶). 𝐴 = 𝐵. (𝐶. 𝐴)****(𝑖𝑣) (𝐴 + 𝐵). 𝐶 = 𝐴. 𝐶 + 𝐵. 𝐶****(𝑣) (𝐵 – 𝐶) = 𝐴. 𝐵 – 𝐴. 𝐶****(𝑣𝑖) (𝐴 – 𝐵). 𝐶 = 𝐴. 𝐶 – 𝐵. 𝐶****(𝑣𝑖𝑖) 𝐴² – 𝐵² = (𝐴 + 𝐵) (𝐴 – 𝐵)****(𝑣𝑖𝑖𝑖) (𝐴 – 𝐵)² = 𝐴² – 2𝐴. 𝐵 + 𝐵²****Solution:**

(i) True matrices addition is commutative.

(ii) False matrices subtraction is commutative.

(iii) True matrices multiplication is associative.

(iv) True matrices multiplication is distributive over addition.

(v) True matrices multiplication is distributive over subtraction.

(vi) Matrix multiplication is distributive when compared to subtraction.

(vii) For factorization and expansion, the False Laws of Algebra do not apply to matrices.

(viii) For factorization and expansion, the False Laws of Algebra do not apply to matrices.

### Exercise 9D

**Question **1. Find x and y if:

By comparing,

6𝑥 − 10 = 8____________(𝑖)

−2𝑥 + 14 = 4𝑦**_**(𝑖𝑖)

From equation (i) we get,

6𝑥 − 10 = 8

6𝑥 = 8 + 10

6𝑥 = 18

𝑥 = 18/6

𝑥 = 3

Put the value of x in equation (𝑖𝑖)

−2𝑥 + 14 = 4𝑦

−2(3) + 14 = 4𝑦

−6 + 14 = 4𝑦

−8 = 4𝑦/

8/4 = 𝑦

𝑦 = 2

**Question **2. Find x and 𝑦, if:

[3𝑥 + 24 12𝑥 + 56] − [6 −21] = [15 10𝑦]

[3𝑥 + 24 − 6 12𝑥 + 56 − (−21)] = [15 10𝑦]

[3𝑥 + 18 12𝑥 + 77] = [15 10𝑦]

By comparing,

3𝑥 + 18 = 15_________(𝑖)

12𝑥 + 77 = 10𝑦** _**(𝑖)

From equation (i) we get,

3𝑥 + 18 = 15

3𝑥 = 15 − 18

3𝑥 = −3

𝑥 = − 3/3

𝑥 = −1

**Question **3. If

(i) 𝑥, 𝑦 ∈ 𝑊 (whole number)

It can be observed that the above two equations are satisfied when 𝑥 = 3 and 𝑦 = 4.

(ii) 𝑥, 𝑦 ∈ 𝑍 (integers)

It can be observed that the above two equations are satisfied when 𝑥 = ±3 and 𝑦 = ±4.

**Question **4. Given

**Question **5. Evaluate:

**Question **6. If

**Question **7. If

By comparing,

𝑎 + 1 = 5______________(𝑖)

2 + 𝑏 = 0______________(𝑖𝑖)

−1 − 𝑐 = 3____________(𝑖𝑖𝑖)

Form equation (i) we get,

𝑎 = 5 − 1

𝑎 = 4

Form equation (ii) we get,

2 + 𝑏 = 0

𝑏 = −2

Form equation (iii) we get,

−𝑐 = 3 + 1

−𝑐 = 4

𝑐 = −4

Hence, the value of a, b and c is 4, -2 and -4.

**Question **8. If 𝐴 =

**Question **9. Find x and y, if:

**Question **10. If matrix

**Question **11. Given

**Question **12. Find the value of x, given that A^{2}=B,

**Question **13. If

**Question **14. Given

**Question **15. Let

**Question **16. Let

**Question **17. If

**Question **18. Given

**Question **19. Given

**Question **20. Evaluate: