# Class 12 RD Sharma Solutions – Chapter 5 Algebra of Matrices – Exercise 5.2 | Set 1

### Question 1(i): Compute the following sum:.

**Solution:**

As the matrices are of the same dimensions, we can add them to get a matrix of the same dimensions which is 2×2.

=>

=>

### Question 1(ii): Compute the following sum:.

**Solution:**

As the matrices are of the same dimensions, we can add them to get a matrix of the same dimensions which is 3×3.

=>

=>

### Question 2: Let A =, B =and C =. Find each of the following:

### (i): 2A – 3B

**Solution:**

Both the matrices A and B are of the same order which is 2×2, hence the operation can be performed.

=> 2A =

=> 3B =

=> 2A – 3B =

=> 2A – 3B=

### (ii): B – 4C

**Solution:**

Both the matrices B and C are of the same order which is 2×2, hence the operation can be performed.

=> B =

=> 4C =

=> B – 4C =

=> B – 4C=

### (iii): 3A – C

**Solution:**

Both the matrices A and C are of the same order which is 2×2, hence the operation can be performed.

=> 3A =

=> C =

=> 3A – C =

=> 3A – C=

### (iv): 3A -2B + 3C

**Solution:**

The matrices A, B and C are of the same order which is 2×2, hence the operation can be performed.

=> 3A =

=> 2B =

=> 3C =

=> 3A – 2B + 3C =

=> 3A – 2B + 3C=

### Question 3: If A =, B =, C =, find:

### (i): A + B and B + C

**Solution:**

A and B can not be added since A’s order is 2×2 which is different from B’s order which is 2×3.

B+C can be computed and is solved as follows:

=> B + C =

=> B + C=

### (ii): 2B + 3A and 3C – 4B

**Solution:**

A and B can not be added since A’s order is 2×2 which is different from B’s order which is 2×3, and thus 2B + 3A can not be calculated.

3C – 4B can be computed and is solved as follows:

=> 3C =

=> 4B =

=> 3C – 4B =

=> 3C – 4B=

### Question 4: Let A =, B =and C =. Compute 2A – 3B + 4C.

**Solution:**

The result can be computed since A, B and C are of the same order which is 2×3.

=> 2A =

=> 3B =

=> 4C =

=> 2A – 3B + 4C =

=> 2A – 3B + 4C=

### Question 5: If A =diag(2, -5, 9), B = diag(1, 1, -4) and C = diag(-6, 3, 4), find:

### (i): A – 2B

**Solution:**

In the given question A and B are diagonal matrices of the order 3×3, thus the only non-zero elements are present in the diagonal.

=> A = diag(2, -5, 9)

=> 2B = 2. diag(1, 1, -4) = diag(2, 2, -8)

=> A – 2B = diag(2-2, -5-2, 9+8)

=> A – 2B= diag(0, -7, 17)

### (ii): B + C – 2A

**Solution:**

In the given question A, B and C are diagonal matrices of the order 3×3, thus the only non-zero elements are present in the diagonal.

=> B = diag(1, 1, -4)

=> C = diag(-6, 3, 4)

=> 2A = 2. diag(2, -5, 9) = diag(4, -10, 18)

=> B + C – 2A = diag(1-6-4, 1+3+10, -4+4-18)

=> B + C – 2A= diag(-9, 14, -18)

### (iii): 2A + 3B – 5C

**Solution:**

In the given question A, B and C are diagonal matrices of the order 3×3, thus the only non-zero elements are present in the diagonal.

=> 2A = 2. diag(2, -5, 9) = diag(4, -10, 18)

=> 3B = 3. diag(1, 1, -4) = diag(3, 3, -12)

=> 5C = 5. diag(-6, 3, 4) = diag(-30, 15, 20)

=> 2A + 3B – 5C = diag(4+3+30, -10+3-15, 18-12-20)

=> 2A + 3B – 5C= diag(37, -22, -14)

### Question 6: Given the matrices A =, B =and C =. Verify that (A + B) + C = A + (B + C).

**Solution:**

Given L.H.S :

=> (A + B) =

=> (A + B) =

=> (A + B) =

=> (A + B) + C =

=> (A + B) + C =

=> (A + B) + C=Given R.H.S :

=> (B + C) =

=> (B + C) =

=> (B + C) =

=> A + (B + C) =

=> A + (B + C) =

=> A + (B + C)=

Hence R.H.S = L.H.S has been verified.

### Question 7: Find the matrices X and Y, if X + Y =and X – Y =.

**Solution:**

We know that (X + Y) + (X – Y) = 2X.

=> (X + Y) + (X – Y) =

=> 2X =

=> 2X =

=> X =

=> X =Now Y = (X + Y) – X

=> Y =

=> Y =

=> Y =

### Question 8: Find X, if Y =and 2X + Y =.

**Solution:**

Given 2X + Y =

=> 2X +=

=> 2X =

=> 2X =

=> 2X =

=> X =

=> X =

### Question 9: Find matrices X and Y, if 2X – Y =and X + 2Y =.

**Solution:**

We know that 2 (2X – Y) + (X + 2Y) = 4X – 2Y + X + 2Y = 5X .

=> 2 (2X – Y) =

=> 2 (2X -Y) =

=> 2 (2X – Y) + (X + 2Y) =

=> 5X =

=> 5X =

=> X =

=> X =As (X + 2Y) =

=>

=> 2Y =

=> 2Y =

=> Y =

=> Y =

### Question 10: If X – Y =and X + Y =, find X and Y.

**Solution:**

We know that (X + Y) + (X – Y) = 2X.

=> 2X =

=> 2X =

=> 2X =

=> X =

=> X =Also (X + Y) – (X -Y) = 2Y.

=> 2Y =

=> 2Y =

=> 2Y =

=> Y =

=> Y =