NCERT Solutions Class 12 maths Chapter-3 (matrices)Exercise 3.3
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.
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’ = AT = 
(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.
Solution:
(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 =
,, 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,
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.
Exercise 3.3){kind=link}