# Class 12 NCERT Solutions – Mathematics Part I – Chapter 3 Matrices – Exercise 3.3

### Question 1. Find the transpose of each of the following matrices:

**(i) **** **

**(ii) **

**(iii) **

**Solution:**

(i)Let A =∴Transpose of A = A’ = A

^{T }=

(ii)Let A =∴Transpose of A = A’ = AT =

(iii)Let A =∴Transpose of A = A’ = AT =

### Question 2. If A = and B = then verify that:

**(i) (A+B)’ = A’+B’**

**(ii) (A-B)’ = A’- B’**

**Solution:**

(i)A+B =L.H.S. = (A+B)’ =

R.H.S. = A’+B’ =

∴L.H.S = R.H.S.

Hence, proved.

(ii)A-B =L.H.S. = (A-B)’

R.H.S. = A’-B’ =

∴ L.H.S. = R.H.S.

Hence, proved.

### Question 3. If A’ = and B = , then verify that:

**(i) (A+B)’=A’+B’**

**(ii) (A-B)’=A’-B’**

**Solution:**

Given A’=and B=

then, (A’)’ = A =

(i)A+B =∴ L.H.S. = (A+B)’=

R.H.S.= A’+B’ =

∴ L.H.S. = R.H.S.

Hence, proved.

(ii)A-B =∴ L.H.S. = (A-B)’=

R.H.S.= A’-B’ =

∴ L.H.S. = R.H.S.

Hence, proved.

### Question 4. If A’ = and B = then find (A+2B)’.

**Solution:**

Given: A’ =and B =

then (A’)’ =A=

Now, A+2B =

∴(A+2B)’ =

### Question 5. For the matrices A and B, verify that (AB)′ = B′A′, where

**(i) A =**** and B = **

**(ii) A =**** and B =**

**Solution:**

(i)AB = =∴ L.H.S. = (AB)′ =

R.H.S.= B′A’ =

∴ L.H.S. = R.H.S.

Hence, proved.

(ii)AB =∴ L.H.S. = (AB)′ =

Now, R.H.S.=B’A’ =

∴ L.H.S. = R.H.S.

Hence, proved.

### Question 6. If (i) A = , then verify that A′ A = I.

### (ii) A = ,then verify that A′ A = I.

**Solution:**

(i)= I = R.H.S.

∴ L.H.S. = R.H.S.

(ii)= I = R.H.S.

∴ L.H.S. = R.H.S.

### Question 7. (i) Show that the matrix A = is a symmetric matrix.

### (ii) Show that the matrix A = is a symmetric matrix.

(i)Given: A =Now, A’=

∵ A = A’

∴ A is a symmetric matrix.

(ii)Given: A =Now, A’=

∵ A = A’

∴ A is a symmetric matrix.

### Question 8. For the matrix A =, verify that:

**(i) (A + A′) is a symmetric matrix**

**(ii) (A – A′) is a skew symmetric matrix**

**Solution:**

(i)Given: A =Let B = (A+A’) =

Now, B’ = (A+A’)’ =

∵ B = B’

∴ B=(A+A’) is a symmetric matrix.

(ii)Given: A =Let B = (A-A’) =

Now, B’ = (A-A’)’ =

∵ -B = B’

∴ B=(A-A’) is a skew symmetric matrix.

### Question 9. Find 1/2(A+A’) and 1/2(A-A’) ,when A =.

**Solution:**

Given: A =

∴ A’ =

Now, A+A’ = +

Now, A-A’ =

### Question 10. Express the following matrices as the sum of a symmetric and a skew symmetric matrix:

**(i) **

**(ii) **

**(iii) **

**(iv) **

**Solution:**

(i)Given : A =⇒ A’=

Let P =

and Q =

Now, P =…..(1)

& P’ =

∵ P=P’

∴ P is a symmetric matrix.

Now, Q =…..(2)

& Q’ =

∵ -Q=Q’

∴ Q is a skew symmetric matrix.

By adding (1) and (2), we get,

Therefore, A =P + Q

(ii)Given :⇒ A’=

P =

…..(1)

Q =

……(2)

By adding (1) and (2), we get,

\begin{bmatrix}0 & 0 & 0\\0 & 0 & 0\\0 & 0 & 0\end{bmatrix}

Therefore, A =P + Q

(iii)Given: A =⇒ A’=

P = }…..(1)

Q = ……(2)

By adding (1) and (2), we get

}

Therefore, A =P + Q

(iv)Given: A =⇒ A’=

P =

…..(1)

Q =

…..(2)

By adding (1) and (2), we get

Therefore, A =P + Q

### Question 11. If A, B are symmetric matrices of same order, then AB – BA is a

**(A) Skew symmetric matrix (B) Symmetric matrix**

**(C) Zero matrix (D) Identity matrix **

**Solution:**

Given: A and B are symmetric matrices.

⇒ A=A’

⇒ B=B’

Now, ( AB – BA)’ =(AB)’-(BA)’ [∵ (X-Y)’=X’-Y’]

=B’A’-A’B’ [∵ (XY)’=Y’X’]

=BA-AB [∵ Given]

= -(AB-BA)

∴(AB-BA) is a skew symmetric matrix.

∴ The option (A) is correct.

### Question 12. If A =, and A + A′ = I, then the value of α is

**(A)π/6 (B) π/3**

**(C) π (D)3π/2**

**Solution:**

On comparing both sides, we get

2cosα = 1

⇒ cosα =

⇒ cosα = cos

⇒ α =

∴ The option (B) is correct.