# Exercise 3.3 Solutions

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Exercise 3.3 Solutions
1. Find the transpose of each of the following matrices:
(i) {\left[\begin{array}{c} 5 \\[5pt] \dfrac{1}{2} \\[10pt] -1 \end{array}\right]} (ii) {\left[\begin{array}{cc} 1 & -1 \\[5pt] 2 & 3 \end{array}\right]} (iii) {\left[\begin{array}{ccc} -1 & 5 & 6 \\[5pt] \sqrt{3} & 5 & 6 \\[5pt] 2 & 3 & -1 \end{array}\right]}
(i) To find the transpose of {\left[\begin{array}{c} 5 \\[5pt] \dfrac{1}{2} \\[10pt] -1 \end{array}\right]}
Transpose of {\left[\begin{array}{c} 5 \\[5pt] \dfrac{1}{2} \\[10pt] -1 \end{array}\right]} = {\left[\begin{array}{ccc} 5 & \dfrac{1}{2} & -1 \end{array}\right]}
(ii) To find the transpose of {\left[\begin{array}{cc} 1 & -1 \\[5pt] 2 & 3 \end{array}\right]}
Transpose of {\left[\begin{array}{cc} 1 & -1 \\[5pt] 2 & 3 \end{array}\right]} = {\left[\begin{array}{ccc} 1 & 2 \\[5pt] -1 & 3 \end{array}\right]}
(iii) To find the transpose of {\left[\begin{array}{ccc} -1 & 5 & 6 \\[5pt] \sqrt{3} & 5 & 6 \\[5pt] 2 & 3 & -1 \end{array}\right]}
Transpose of {\left[\begin{array}{ccc} -1 & 5 & 6 \\[5pt] \sqrt{3} & 5 & 6 \\[5pt] 2 & 3 & -1 \end{array}\right]} = {\left[\begin{array}{ccc} -1 & \sqrt{3} & 2 \\[5pt] 5 & 5 & 3 \\[5pt] 6 & 6 & -1 \end{array}\right]}

2. If {\text{A} = \left[\begin{array}{cc} -1 & 2 & 3 \\[5pt] 5 & 7 & 9 \\[5pt] -2 & 1 & 1 \end{array}\right]} and {\text{B} = \left[\begin{array}{cc} -4 & 1 & -5 \\[5pt] 1 & 2 & 0 \\[5pt] 1 & 3 & 1 \end{array}\right],} then verify that
(i) (A + B) = A + B,
(ii) (A – B) = A + B

Given that {\text{A} = \left[\begin{array}{cc} -1 & 2 & 3 \\[5pt] 5 & 7 & 9 \\[5pt] -2 & 1 & 1 \end{array}\right]} and {\text{B} = \left[\begin{array}{cc} -4 & 1 & -5 \\[5pt] 1 & 2 & 0 \\[5pt] 1 & 3 & 1 \end{array}\right].} So,
A
{= \left[\begin{array}{cc} -1 & 5 & -2 \\[5pt] 2 & 7 & 1 \\[5pt] 3 & 9 & 1 \end{array}\right]}
B
{= \left[\begin{array}{cc} -4 & 1 & 1 \\[5pt] 1 & 2 & 3 \\[5pt] -5 & 0 & 1 \end{array}\right]}
(i) To verify that (A + B) = A + B
A + B
{= \left[\begin{array}{cc} -1 & 2 & 3 \\[5pt] 5 & 7 & 9 \\[5pt] -2 & 1 & 1 \end{array}\right] + \left[\begin{array}{cc} -4 & 1 & -5 \\[5pt] 1 & 2 & 0 \\[5pt] 1 & 3 & 1 \end{array}\right]}
{= \left[\begin{array}{cc} (-1) + (-4) & 2 + 1 & 3 + (-5) \\[5pt] 5 + 1 & 7 + 2 & 9 + 0 \\[5pt] -2 + 1 & 1 + 3 & 1 + 1 \end{array}\right]}
{= \left[\begin{array}{cc} -5 & 3 & -2 \\[5pt] 6 & 9 & 9 \\[5pt] -1 & 4 & 2 \end{array}\right]}
Now,
(A + B)
{= \left[\begin{array}{cc} -5 & 6 & -1 \\[5pt] 3 & 9 & 4 \\[5pt] -2 & 9 & 2 \end{array}\right]}
A + B
{= \left[\begin{array}{cc} -1 & 5 & -2 \\[5pt] 2 & 7 & 1 \\[5pt] 3 & 9 & 1 \end{array}\right] + \left[\begin{array}{cc} -4 & 1 & 1 \\[5pt] 1 & 2 & 3 \\[5pt] -5 & 0 & 1 \end{array}\right]}
{= \left[\begin{array}{cc} (-1) + (-4) & 5 + 1 & -2 + 1 \\[5pt] 2 + 1 & 7 + 2 & 1 + 3 \\[5pt] 3 + (-5) & 9 + 0 & 1 + 1 \end{array}\right]}
{= \left[\begin{array}{cc} -5 & 6 & -1 \\[5pt] 3 & 9 & 4 \\[5pt] -2 & 9 & 2 \end{array}\right]}
(A + B) = A + B
(ii) To verify that (A – B) = A – B
A – B
{= \left[\begin{array}{cc} -1 & 2 & 3 \\[5pt] 5 & 7 & 9 \\[5pt] -2 & 1 & 1 \end{array}\right] - \left[\begin{array}{cc} -4 & 1 & -5 \\[5pt] 1 & 2 & 0 \\[5pt] 1 & 3 & 1 \end{array}\right]}
{= \left[\begin{array}{cc} (-1) - (-4) & 2 - 1 & 3 - (-5) \\[5pt] 5 - 1 & 7 - 2 & 9 - 0 \\[5pt] -2 - 1 & 1 - 3 & 1 - 1 \end{array}\right]}
{= \left[\begin{array}{cc} -1 + 4 & 2 - 1 & 3 + 5 \\[5pt] 5 - 1 & 7 - 2 & 9 - 0 \\[5pt] -2 - 1 & 1 - 3 & 1 - 1 \end{array}\right]}
{= \left[\begin{array}{cc} 3 & 1 & 8 \\[5pt] 4 & 5 & 9 \\[5pt] -3 & -2 & 0 \end{array}\right]}
Now,
(A – B)
{= \left[\begin{array}{cc} 3 & 4 & -3 \\[5pt] 1 & 5 & -2 \\[5pt] 8 & 9 & 0 \end{array}\right]}
A – B
{= \left[\begin{array}{cc} -1 & 5 & -2 \\[5pt] 2 & 7 & 1 \\[5pt] 3 & 9 & 1 \end{array}\right] - \left[\begin{array}{cc} -4 & 1 & 1 \\[5pt] 1 & 2 & 3 \\[5pt] -5 & 0 & 1 \end{array}\right]}
{= \left[\begin{array}{cc} (-1) - (-4) & 5 - 1 & -2 - 1 \\[5pt] 2 - 1 & 7 - 2 & 1 - 3 \\[5pt] 3 - (-5) & 9 - 0 & 1 - 1 \end{array}\right]}
{= \left[\begin{array}{cc} 3 & 4 & -3 \\[5pt] 1 & 5 & -2 \\[5pt] 8 & 9 & 1 \end{array}\right]}
(A – B) = A – B

3. If A = {\left[\begin{array}{cc} 3 & 4 \\[5pt] -1 & 2 \\[5pt] 0 & 1 \end{array}\right]} and {\text{B} = \left[\begin{array}{ccc} - 1 & 2 & 1 \\[5pt] 1 & 2 & 3 \end{array}\right],} then verify that
(i) (A + B) = A + B
(ii) (A – B) = A – B

Given that A = {\left[\begin{array}{cc} 3 & 4 \\[5pt] -1 & 2 \\[5pt] 0 & 1 \end{array}\right]}
{⇒ \text{A} = \left[\begin{array}{ccc} 3 & -1 & 0 \\[5pt] 4 & 2 & 1 \end{array}\right]}
It is also given that {\text{B} = \left[\begin{array}{ccc} -1 & 2 & 1 \\[5pt] 1 & 2 & 3 \end{array}\right]}
⇒ B = {\left[\begin{array}{cc} -1 & 1 \\[5pt] 2 & 2 \\[5pt] 1 & 3 \end{array}\right]}
(i) To verify that (A + B) = A + B
A + B
{= \left[\begin{array}{ccc} 3 & -1 & 0 \\[5pt] 4 & 2 & 1 \end{array}\right] + \left[\begin{array}{ccc} -1 & 2 & 1 \\[5pt] 1 & 2 & 3 \end{array}\right]}
{= \left[\begin{array}{ccc} 3 + (-1) & -1 + 2 & 0 + 1 \\[5pt] 4 + 1 & 2 + 2 & 1 + 3 \end{array}\right]}
{= \left[\begin{array}{ccc} 2 & 1 & 1 \\[5pt] 5 & 4 & 4 \end{array}\right]}
L.H.S.
= (A + B)
{= \left[\begin{array}{cc} 2 & 5 \\[5pt] 1 & 4 \\[5pt] 1 & 4 \end{array}\right]}
R.H.S.
= A + B
{= \left[\begin{array}{cc} 3 & 4 \\[5pt] -1 & 2 \\[5pt] 0 & 1 \end{array}\right] + \left[\begin{array}{cc} -1 & 1 \\[5pt] 2 & 2 \\[5pt] 1 & 3 \end{array}\right]}
{= \left[\begin{array}{cc} 3 + (-1) & 4 + 1 \\[5pt] -1 + 2 & 2 + 2 \\[5pt] 0 + 1 & 1 + 3 \end{array}\right]}
{= \left[\begin{array}{cc} 2 & 5 \\[5pt] 1 & 4 \\[5pt] 1 & 4 \end{array}\right]}
As L.H.S. and R.H.S. are equal
(A + B) = A + B
(ii) To verify that (A – B) = A – B
A – B
{= \left[\begin{array}{ccc} 3 & -1 & 0 \\[5pt] 4 & 2 & 1 \end{array}\right] - \left[\begin{array}{ccc} -1 & 2 & 1 \\[5pt] 1 & 2 & 3 \end{array}\right]}
{= \left[\begin{array}{ccc} 3 - (-1) & -1 - 2 & 0 - 1 \\[5pt] 4 - 1 & 2 - 2 & 1 - 3 \end{array}\right]}
{= \left[\begin{array}{ccc} 4 & -3 & -1 \\[5pt] 3 & 0 & -2 \end{array}\right]}
L.H.S.
= (A – B)
{= \left[\begin{array}{cc} 4 & 3 \\[5pt] -3 & 0 \\[5pt] -1 & -2 \end{array}\right]}
R.H.S.
= A – B
{= \left[\begin{array}{cc} 3 & 4 \\[5pt] -1 & 2 \\[5pt] 0 & 1 \end{array}\right] - \left[\begin{array}{cc} -1 & 1 \\[5pt] 2 & 2 \\[5pt] 1 & 3 \end{array}\right]}
{= \left[\begin{array}{cc} 3 - (-1) & 4 - 1 \\[5pt] -1 - 2 & 2 - 2 \\[5pt] 0 - 1 & 1 - 3 \end{array}\right]}
{= \left[\begin{array}{cc} 4 & 3 \\[5pt] -3 & 0 \\[5pt] -1 & -2 \end{array}\right]}
As L.H.S. and R.H.S. are equal
(A – B) = A – B

4. If A = {\left[\begin{array}{cc} -2 & 3 \\[5pt] 1 & 2 \end{array}\right]} and {\text{B} = \left[\begin{array}{cc} -1 & 0 \\[5pt] 1 & 2 \end{array}\right],} then find (A + 2B)
Given that A = {\left[\begin{array}{cc} -2 & 3 \\[5pt] 1 & 2 \end{array}\right]} and {\text{B} = \left[\begin{array}{cc} -1 & 0 \\[5pt] 1 & 2 \end{array}\right]}
⇒ A
= (A)
{= \left[\begin{array}{cc} -2 & 1 \\[5pt] 3 & 2 \end{array}\right]}
A + 2B
{= \left[\begin{array}{cc} -2 & 1 \\[5pt] 3 & 2 \end{array}\right] + 2\left[\begin{array}{cc} -1 & 0 \\[5pt] 1 & 2 \end{array}\right]}
{= \left[\begin{array}{cc} -2 & 1 \\[5pt] 3 & 2 \end{array}\right] + \left[\begin{array}{cc} 2 × (-1) & 2 × 0 \\[5pt] 2 × 1 & 2 × 2 \end{array}\right]}
{= \left[\begin{array}{cc} -2 & 1 \\[5pt] 3 & 2 \end{array}\right] + \left[\begin{array}{cc} -2 & 0 \\[5pt] 2 & 4 \end{array}\right]}
{= \left[\begin{array}{cc} -2 + (-2) & 1 + 0 \\[5pt] 3 + 2 & 2 + 4 \end{array}\right]}
{= \left[\begin{array}{cc} -4 & 1 \\[5pt] 5 & 6 \end{array}\right]}
So,
(A + 2B)
= {\left[\begin{array}{cc} -4 & 5 \\[5pt] 1 & 6 \end{array}\right]}

5. For the matrices A and B, verify that (AB) = BA, where (i) {\text{A} = \left[\begin{array}{c} 1 \\[5pt] -4 \\[5pt] 3 \end{array}\right], \text{B} = \left[\begin{array}{ccc} -1 & 2 & 1 \end{array}\right]} (ii) {\text{A} = \left[\begin{array}{c} 0 \\[5pt] 1 \\[5pt] 2 \end{array}\right], \text{B} = \left[\begin{array}{ccc} 1 & 5 & 7 \end{array}\right]}
(i) To verify that (AB) = BA where A = {\left[\begin{array}{c} 1 \\[5pt] -4 \\[5pt] 3 \end{array}\right]} and B = {\left[\begin{array}{ccc} -1 & 2 & 1 \end{array}\right]}
Given that A = {\left[\begin{array}{c} 1 \\[5pt] -4 \\[5pt] 3 \end{array}\right]} and B = {\left[\begin{array}{ccc} -1 & 2 & 1 \end{array}\right]}
AB
{= \left[\begin{array}{c} 1 \\[5pt] -4 \\[5pt] 3 \end{array}\right] \left[\begin{array}{ccc} -1 & 2 & 1 \end{array}\right]}
{= \left[\begin{array}{c} 1 × (-1) & 1 × 2 & 1 × 1 \\[5pt] (-4) × (-1) & -4 × 2 & -4 × 1 \\[5pt] 3 × (-1) & 3 × 2 & 3 × 1 \end{array}\right]}
{= \left[\begin{array}{c} -1 & 2 & 1 \\[5pt] 4 & -8 & -4 \\[5pt] -3 & 6 & 3 \end{array}\right]}
L.H.S.
= (AB)
{= \left[\begin{array}{ccc} -1 & 4 & -3 \\[5pt] 2 & -8 & 6 \\[5pt] 1 & -4 & 3 \end{array}\right]}
B
{= \left[\begin{array}{c} -1 \\[5pt] 2 \\[5pt] 1 \end{array}\right]}
A
{= \left[\begin{array}{c} 1 & -4 & 3 \end{array}\right]}
R.H.S.
= BA
{= \left[\begin{array}{c} -1 \\[5pt] 2 \\[5pt] 1 \end{array}\right] \left[\begin{array}{c} 1 & -4 & 3 \end{array}\right]}
{= \left[\begin{array}{ccc} -1 × 1 & (-1) × (-4) & -1 × 3 \\[5pt] 2 × 1 & 2 × (-4) & 2 × 3 \\[5pt] 1 × 1 & 1 × (-4) & 1 × 3 \end{array}\right]}
{= \left[\begin{array}{ccc} 1 & 4 & -3 \\[5pt] 2 & -8 & 6 \\[5pt] 1 & -4 & 3 \end{array}\right]}
∴ L.H.S. = R.H.S.
So, it is verified that (AB) = BA
(ii) To verify that (AB) = BA where A = {\left[\begin{array}{c} 0 \\[5pt] 1 \\[5pt] 2 \end{array}\right]} and B = {\left[\begin{array}{ccc} 1 & 5 & 7 \end{array}\right]}
Given that A = {\left[\begin{array}{c} 0 \\[5pt] 1 \\[5pt] 2 \end{array}\right]} and B = {\left[\begin{array}{ccc} 1 & 5 & 7 \end{array}\right]}
AB
{= \left[\begin{array}{c} 0 \\[5pt] 1 \\[5pt] 2 \end{array}\right] \left[\begin{array}{ccc} 1 & 5 & 7 \end{array}\right]}
{= \left[\begin{array}{c} 0 × 1 & 0 × 5 & 0 × 7 \\[5pt] 1 × 1 & 1 × 5 & 1 × 7 \\[5pt] 2 × 1 & 2 × 5 & 2 × 7 \end{array}\right]}
{= \left[\begin{array}{c} 0 & 0 & 0 \\[5pt] 1 & 5 & 7 \\[5pt] 2 & 10 & 14 \end{array}\right]}
L.H.S.
= (AB)
{= \left[\begin{array}{ccc} 0 & 1 & 2 \\[5pt] 0 & 5 & 10 \\[5pt] 0 & 7 & 14 \end{array}\right]}
B
{= \left[\begin{array}{c} 1 \\[5pt] 5 \\[5pt] 7 \end{array}\right]}
A
{= \left[\begin{array}{c} 0 & 1 & 2 \end{array}\right]}
R.H.S.
= BA
{= \left[\begin{array}{c} 1 \\[5pt] 5 \\[5pt] 7 \end{array}\right] \left[\begin{array}{c} 0 & 1 & 2 \end{array}\right]}
{= \left[\begin{array}{ccc} 1 × 0 & 1 × 1 & 1 × 2 \\[5pt] 5 × 0 & 5 × 1 & 5 × 2 \\[5pt] 7 × 0 & 7 × 1 & 7 × 2 \end{array}\right]}
{= \left[\begin{array}{ccc} 0 & 1 & 2 \\[5pt] 0 & 5 & 10 \\[5pt] 0 & 7 & 14 \end{array}\right]}
∴ L.H.S. = R.H.S.
So, it is verified that (AB) = BA

6. If (i) A = {\left[\begin{array}{cc} \cos α & \sin α \\[5pt] -\sin α & \cos α \end{array}\right],} then verify that AA = I
(ii) A = {\left[\begin{array}{cc} \sin α & \cos α \\[5pt] -\cos α & \sin α \end{array}\right],} then verify that AA = I
(i) To verify that AA = I when {\text{A} = \left[\begin{array}{cc} \cos α & \sin α \\[5pt] -\sin α & \cos α \end{array}\right]}
Given that
A
{= \left[\begin{array}{cc} \cos α & \sin α \\[5pt] -\sin α & \cos α \end{array}\right]}
⇒ A
{= \left[\begin{array}{cc} \cos α & -\sin α \\[5pt] \sin α & \cos α \end{array}\right]}
Now,
L.H.S.
= AA
{= \left[\begin{array}{cc} \cos α & -\sin α \\[5pt] \sin α & \cos α \end{array}\right] \left[\begin{array}{cc} \cos α & \sin α \\[5pt] -\sin α & \cos α \end{array}\right]}
{= \left[\begin{array}{cc} (\cos α) (\cos α) + (-\sin α) (-\sin α) & (\cos α) (\sin α) + (-\sin α) (\cos α) \\[5pt] (\sin α) (\cos α) + (\cos α) (-\sin α) & (\sin α) (\sin α) + (\cos α)(\cos α) \end{array}\right]}
{= \left[\begin{array}{cc} \cos^2 α + \sin^2 α & \cos α \sin α - \sin α \cos α \\[5pt] \sin α \cos α - \cos α \sin α & \sin^2 α + \cos^2 α \end{array}\right]}
{= \left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]}
= I
= R.H.S.
As L.H.S. = R.H.S., it is verified that AA = I
(ii) To verify that AA = I when {\text{A} = \left[\begin{array}{cc} \sin α & \cos α \\[5pt] -\cos α & \sin α \end{array}\right]}
Given that
A
{= \left[\begin{array}{cc} \sin α & \cos α \\[5pt] -\cos α & \sin α \end{array}\right]}
⇒ A
{= \left[\begin{array}{cc} \sin α & -\cos α \\[5pt] \cos α & \sin α \end{array}\right]}
Now,
L.H.S.
= AA
{= \left[\begin{array}{cc} \sin α & -\cos α \\[5pt] \cos α & \sin α \end{array}\right] \left[\begin{array}{cc} \sin α & \cos α \\[5pt] -\cos α & \sin α \end{array}\right]}
{= \left[\begin{array}{cc} (\sin α) (\sin α) + (-\cos α) (-\cos α) & (\sin α) (\cos α) + (-\cos α) (\sin α) \\[5pt] (\cos α) (\sin α) + (\sin α) (-\cos α) & (\cos α) (\cos α) + (\sin α)(\sin α) \end{array}\right]}
{= \left[\begin{array}{cc} \sin^2 α + \cos^2 α & \sin α \cos α - \cos α \sin α \\[5pt] \cos α \sin α - \sin α \cos α & \cos^2 α + \sin^2 α \end{array}\right]}
{= \left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]}
= I
= R.H.S.
As L.H.S. = R.H.S., it is verified that AA = I

7. (i) Show that the matrix {\text{A} = \left[\begin{array}{cc} 1 & -1 & 5 \\[5pt] -1 & 2 & 1 \\[5pt] 5 & 1 & 3 \end{array}\right]} is a symmetric matrix.
(ii) Show that the matrix {\text{A} = \left[\begin{array}{cc} 0 & 1 & -1 \\[5pt] -1 & 0 & 1 \\[5pt] 1 & -1 & 0 \end{array}\right]} is a skew symmetric matrix.
(i) To show that the matrix {\text{A} = \left[\begin{array}{cc} 1 & -1 & 5 \\[5pt] -1 & 2 & 1 \\[5pt] 5 & 1 & 3 \end{array}\right]} is a symmetric matrix.
Given that
A
{= \left[\begin{array}{cc} 1 & -1 & 5 \\[5pt] -1 & 2 & 1 \\[5pt] 5 & 1 & 3 \end{array}\right]}
⇒ A
{= \left[\begin{array}{cc} 1 & -1 & 5 \\[5pt] -1 & 2 & 1 \\[5pt] 5 & 1 & 3 \end{array}\right]}
= A
∴ As A = A, A is a symmetric matrix.
(ii) To show that the matrix {\text{A} = \left[\begin{array}{cc} 0 & 1 & -1 \\[5pt] -1 & 0 & 1 \\[5pt] 1 & -1 & 0 \end{array}\right]} is a skew symmetric matrix.
Given that
A
{= \left[\begin{array}{cc} 0 & 1 & -1 \\[5pt] -1 & 0 & 1 \\[5pt] 1 & -1 & 0 \end{array}\right]}
⇒ A
{= \left[\begin{array}{cc} 0 & -1 & 1 \\[5pt] 1 & 0 & -1 \\[5pt] -1 & 1 & 0 \end{array}\right]}
{= -\left[\begin{array}{cc} 0 & 1 & -1 \\[5pt] -1 & 0 & 1 \\[5pt] 1 & -1 & 0 \end{array}\right]}
= -A
∴ As A = -A, A is a skew symmetric matrix.

8. For the matrix {\text{A} = \left[\begin{array}{cc} 1 & 5 \\[5pt] 6 & 7 \end{array}\right],} verify that
(i)
(A + A) is a symmetric matrix
(ii)
(A – A) is a skew symmetric matrix
Given that
A
{= \left[\begin{array}{cc} 1 & 5 \\[5pt] 6 & 7 \end{array}\right]}
⇒ A
{= \left[\begin{array}{cc} 1 & 6 \\[5pt] 5 & 7 \end{array}\right]}
(i) To verify that (A + A) is a symmetric matrix.
We have,
A + A
{= \left[\begin{array}{cc} 1 & 5 \\[5pt] 6 & 7 \end{array}\right] + \left[\begin{array}{cc} 1 & 6 \\[5pt] 5 & 7 \end{array}\right]}
{= \left[\begin{array}{cc} 1 + 1 & 5 + 6 \\[5pt] 6 + 5 & 7 + 7 \end{array}\right]}
{= \left[\begin{array}{cc} 2 & 11 \\[5pt] 11 & 14 \end{array}\right]}
So,
(A + A)
{= \left[\begin{array}{cc} 2 & 11 \\[5pt] 11 & 14 \end{array}\right]}
= A + A
∴ As (A + A) = (A + A), (A + A) is a symmetric matrix.
(ii) To verify that (A – A) is a skew symmetric matrix.
We have,
A – A
{= \left[\begin{array}{cc} 1 & 5 \\[5pt] 6 & 7 \end{array}\right] - \left[\begin{array}{cc} 1 & 6 \\[5pt] 5 & 7 \end{array}\right]}
{= \left[\begin{array}{cc} 1 - 1 & 5 - 6 \\[5pt] 6 - 5 & 7 - 7 \end{array}\right]}
{= \left[\begin{array}{cc} 0 & -1 \\[5pt] 1 & 0 \end{array}\right]}
So,
(A – A)
{= \left[\begin{array}{cc} 0 & 1 \\[5pt] -1 & 0 \end{array}\right]}
{= -\left[\begin{array}{cc} 0 & -1 \\[5pt] 1 & 0 \end{array}\right]}
= -(A – A)
∴ As (A – A) = -(A – A), (A – A) is a skew symmetric matrix.

9. Find \dfrac{1}{2}(A + A) and \dfrac{1}{2}(A – A), when {\text{A} = \left[\begin{array}{cc} 0 & a & b \\[5pt] -a & 0 & c \\[5pt] -b & -c & 0 \end{array}\right]}
Given that
A
{= \left[\begin{array}{cc} 0 & a & b \\[5pt] -a & 0 & c \\[5pt] -b & -c & 0 \end{array}\right]}
⇒ A
{= \left[\begin{array}{cc} 0 & -a & -b \\[5pt] a & 0 & -c \\[5pt] b & c & 0 \end{array}\right]}
{= -\left[\begin{array}{cc} 0 & a & b \\[5pt] -a & 0 & c \\[5pt] -b & -c & 0 \end{array}\right]}
= -A
So,
\dfrac{1}{2}(A + A)
= \dfrac{1}{2}(A + (-A))
= \dfrac{1}{2}(A-A)
= \dfrac{1}{2}O
= O
\dfrac{1}{2}(A – A)
= \dfrac{1}{2}(A + A)
= \dfrac{1}{2}(2A)
= A

10. Express the following matrices as the sum of a symetric and a skew symmetric matrix:
(i) {\left[\begin{array}{cc} 3 & 5 \\[5pt] 1 & -1 \end{array}\right]}
(ii) {\left[\begin{array}{ccc} 6 & -2 & 2 \\[5pt] -2 & 3 & -1 \\[5pt] 2 & -1 & 3 \end{array}\right]}
(iii) {\left[\begin{array}{ccc} 3 & 3 & -1 \\[5pt] -2 & -2 & 1 \\[5pt] -4 & -5 & 2 \end{array}\right]}
(iv) {\left[\begin{array}{cc} 1 & 5 \\[5pt] -1 & 2 \end{array}\right]}
To express each of the given matrices as the sum of a symmetric and a skew symmetric matrices, we use the following theorem.
Theorem: Any square matrix can be expressed as the sum of a symmetric and a skew symmetric matrix.
So, any square matrix A can be represented as
A = \dfrac12(A + A) + \dfrac12(A – A)
We also know that (A + A) is a symmetric matrix and (A – A) is a skew symmetric matrix. We also know that, for any given matrix A, (kA) = kA, it follows that \dfrac12(A + A) is symmetric matrix and \dfrac12(A – A) is a skew symmetric matrix. Thus, any square matrix can be expressed as the sum of a symmetric and a skew symmetric matrix. We use this theorem to express the given matrices as the sum of a symmetric and a skew symmetric matrix.
(i) To express {\left[\begin{array}{cc} 3 & 5 \\[5pt] 1 & -1 \end{array}\right]} as the sum of a symmetric and a skew symmetric matrix:
Let the given metrix be A
So,
A
{= \left[\begin{array}{cc} 3 & 5 \\[5pt] 1 & -1 \end{array}\right]}
⇒ A
{= \left[\begin{array}{cc} 3 & 1 \\[5pt] 5 & -1 \end{array}\right]}
Now,
A + A
{= \left[\begin{array}{cc} 3 & 5 \\[5pt] 1 & -1 \end{array}\right] + \left[\begin{array}{cc} 3 & 1 \\[5pt] 5 & -1 \end{array}\right]}
{= \left[\begin{array}{cc} 3 + 3 & 5 + 1 \\[5pt] 1 + 5 & -1 + (-1) \end{array}\right]}
{= \left[\begin{array}{cc} 6 & 6 \\[5pt] 6 & -2 \end{array}\right]}
\dfrac12(A + A)
{= \dfrac12 × \left[\begin{array}{cc} 6 & 6 \\[5pt] 6 & -2 \end{array}\right]}
{= \left[\begin{array}{cc} \dfrac12 × 6 & \dfrac12 × 6 \\[10pt] \dfrac12 × 6 & \dfrac12 × -2 \end{array}\right]}
{= \left[\begin{array}{cc} 3 & 3 \\[5pt] 3 & -1 \end{array}\right]}
Similarly,
A – A
{= \left[\begin{array}{cc} 3 & 5 \\[5pt] 1 & -1 \end{array}\right] - \left[\begin{array}{cc} 3 & 1 \\[5pt] 5 & -1 \end{array}\right]}
{= \left[\begin{array}{cc} 3 - 3 & 5 - 1 \\[5pt] 1 - 5 & -1 - (-1) \end{array}\right]}
{= \left[\begin{array}{cc} 0 & 4 \\[5pt] -4 & 0 \end{array}\right]}
\dfrac12(A – A)
{= \dfrac12 × \left[\begin{array}{cc} 0 & 4 \\[5pt] -4 & 0 \end{array}\right]}
{= \left[\begin{array}{cc} \dfrac12 × 0 & \dfrac12 × 4 \\[10pt] \dfrac12 × (-4) & \dfrac12 × 0 \end{array}\right]}
{= \left[\begin{array}{cc} 0 & 2 \\[5pt] -2 & 0 \end{array}\right]}
Thus the given matrix A can be represented as the sum of a symmetric and skew symmetric matrices as
A = \dfrac12(A + A) + \dfrac12(A – A)
{\left[\begin{array}{cc} 3 & 5 \\[5pt] 1 & 0 \end{array}\right] = \left[\begin{array}{cc} 3 & 3 \\[5pt] 3 & -1 \end{array}\right] + \left[\begin{array}{cc} 0 & 2 \\[5pt] -2 & 0 \end{array}\right]}
(ii) To express {\left[\begin{array}{ccc} 6 & -2 & -2 \\[5pt] -2 & 3 & -1 \\[5pt] 2 & -1 & 3 \end{array}\right]} as the sum of a symmetric and a skew symmetric matrix:
Let the given metrix be A
So,
A
{= \left[\begin{array}{ccc} 6 & -2 & -2 \\[5pt] -2 & 3 & -1 \\[5pt] 2 & -1 & 3 \end{array}\right]}
⇒ A
{= \left[\begin{array}{cc} 6 & -2 & -2 \\[5pt] -2 & 3 & -1 \\[5pt] 2 & -1 & 3 \end{array}\right]}
= A
Now,
A + A
= A + A
= 2A
\dfrac12(A + A)
= \dfrac12 × 2A
= A
Similarly,
A – A
= A – A
= O
\dfrac12(A – A)
= \dfrac12 × O
= O
Thus the given matrix A can be represented as the sum of a symmetric and skew symmetric matrices as
A = \dfrac12(A + A) + \dfrac12(A – A)
{\left[\begin{array}{ccc} 6 & -2 & -2 \\[5pt] -2 & 3 & -1 \\[5pt] 2 & -1 & 3 \end{array}\right] = \left[\begin{array}{ccc} 6 & -2 & -2 \\[5pt] -2 & 3 & -1 \\[5pt] 2 & -1 & 3 \end{array}\right] + \left[\begin{array}{ccc} 0 & 0 & 0 \\[5pt] 0 & 0 & 0 \\[5pt] 0 & 0 & 0 \end{array}\right]}
(iii) To express {\left[\begin{array}{ccc} 3 & 3 & -1 \\[5pt] -2 & -2 & 1 \\[5pt] -4 & -5 & 2 \end{array}\right]} as the sum of a symmetric and a skew symmetric matrix:
Let the given metrix be A
So,
A
{= \left[\begin{array}{ccc} 3 & 3 & -1 \\[5pt] -2 & -2 & 1 \\[5pt] -4 & -5 & 2 \end{array}\right]}
⇒ A
{= \left[\begin{array}{ccc} 3 & -2 & -4 \\[5pt] 3 & -2 & -5 \\[5pt] -1 & 1 & 2 \end{array}\right]}
Now,
A + A
{= \left[\begin{array}{ccc} 3 & 3 & -1 \\[5pt] -2 & -2 & 1 \\[5pt] -4 & -5 & 2 \end{array}\right] + \left[\begin{array}{ccc} 3 & -2 & -4 \\[5pt] 3 & -2 & -5 \\[5pt] -1 & 1 & 2 \end{array}\right]}
{= \left[\begin{array}{ccc} 3 + 3 & 3 + (-2) & -1 + (-4) \\[5pt] -2 + 3 & -2 + (-2) & 1 + (-5) \\[5pt] -4 + (-1) & -5 + 1 & 2 + 2 \end{array}\right]}
{= \left[\begin{array}{ccc} 6 & 1 & -5 \\[5pt] 1 & -4 & -4 \\[5pt] -5 & -4 & 4 \end{array}\right]}
\dfrac12(A + A)
{= \dfrac12 × \left[\begin{array}{cc} 6 & 1 & -5 \\[5pt] 1 & -4 & -4 \\[5pt] -5 & -4 & 4 \end{array}\right]}
{= \left[\begin{array}{ccc} \dfrac12 × 6 & \dfrac12 × 1 & \dfrac12 × -5 \\[10pt] \dfrac12 × 1 & \dfrac12 × -4 & \dfrac12 × -4 \\[10pt] \dfrac12 × -5 & \dfrac12 × -4 & \dfrac12 × 4 \end{array}\right]}
{= \left[\begin{array}{ccc} 3 & \dfrac12 & \dfrac{-5}{2} \\[10pt] \dfrac12 & -2 & -2 \\[10ptx] \dfrac{-5}{2} & -2 & 2 \end{array}\right]}
Similarly,
A – A
{= \left[\begin{array}{ccc} 3 & 3 & -1 \\[5pt] -2 & -2 & 1 \\[5pt] -4 & -5 & 2 \end{array}\right] - \left[\begin{array}{ccc} 3 & -2 & -4 \\[5pt] 3 & -2 & -5 \\[5pt] -1 & 1 & 2 \end{array}\right]}
{= \left[\begin{array}{ccc} 3 - 3 & 3 - (-2) & -1 - (-4) \\[5pt] -2 - 3 & -2 - (-2) & 1 - (-5) \\[5pt] -4 - (-1) & -5 - 1 & 2 - 2 \end{array}\right]}
{= \left[\begin{array}{ccc} 0 & 5 & 3 \\[5pt] -5 & 0 & 6 \\[5pt] -3 & -6 & 0 \end{array}\right]}
\dfrac12(A – A)
{= \dfrac12 × \left[\begin{array}{ccc} 0 & 5 & 3 \\[5pt] -5 & 0 & 6 \\[5pt] -3 & -6 & 0 \end{array}\right]}
{= \left[\begin{array}{ccc} \dfrac12 × 0 & \dfrac12 × 5 & \dfrac12 × 3 \\[10pt] \dfrac12 × -5 & \dfrac12 × 0 & \dfrac12 × 6 \\[10pt] \dfrac12 × -3 & \dfrac12 × -6 & \dfrac12 × 0 \end{array}\right]}
{= \left[\begin{array}{ccc} 0 & \dfrac52 & \dfrac32 \\[10pt] \dfrac{-5}{2} & 0 & 3 \\[10pt] \dfrac{-3}{2} & -3 & 0 \end{array}\right]}
Thus the given matrix A can be represented as the sum of a symmetric and skew symmetric matrices as
A = \dfrac12(A + A) + \dfrac12(A – A)
{\left[\begin{array}{ccc} 3 & 3 & -1 \\[5pt] -2 & -2 & 1 \\[5pt] -4 & -5 & 2 \end{array}\right] = \left[\begin{array}{ccc} 3 & \dfrac12 & \dfrac{-5}{2} \\[10pt] \dfrac12 & -2 & -2 \\[10pt] \dfrac{-5}{2} & -2 & 2 \end{array}\right] + \left[\begin{array}{ccc} 0 & \dfrac52 & \dfrac32 \\[10pt] \dfrac{-5}{2} & 0 & 3 \\[10pt] \dfrac{-3}{2} & -3 & 0 \end{array}\right]}
(iv) To express {\left[\begin{array}{cc} 1 & 5 \\[5pt] -1 & 2 \end{array}\right]} as the sum of a symmetric and a skew symmetric matrix:
Let the given metrix be A
So,
A
{= \left[\begin{array}{cc} 1 & 5 \\[5pt] -1 & 2 \end{array}\right]}
⇒ A
{= \left[\begin{array}{cc} 1 & -1 \\[5pt] 5 & 2 \end{array}\right]}
Now,
A + A
{= \left[\begin{array}{cc} 1 & 5 \\[5pt] -1 & 2 \end{array}\right] + \left[\begin{array}{cc} 1 & -1 \\[5pt] 5 & 2 \end{array}\right]}
{= \left[\begin{array}{cc} 1 + 1 & 5 + (-1) \\[5pt] -1 + 5 & 2 + 2 \end{array}\right]}
{= \left[\begin{array}{cc} 2 & 4 \\[5pt] 4 & 4 \end{array}\right]}
\dfrac12 × (A + A)
{= \dfrac12 × \left[\begin{array}{cc} 2 & 4 \\[5pt] 4 & 4 \end{array}\right]}
{= \left[\begin{array}{cc} \dfrac12 × 2 & \dfrac12 × 4 \\[10pt] \dfrac12 × 4 & \dfrac12 × 4 \end{array}\right]}
{= \left[\begin{array}{cc} 1 & 2 \\[5pt] 2 & 2 \end{array}\right]}
Similarly,
A – A
{= \left[\begin{array}{cc} 1 & 5 \\[5pt] -1 & 2 \end{array}\right] - \left[\begin{array}{cc} 1 & -1 \\[5pt] 5 & 2 \end{array}\right]}
{= \left[\begin{array}{cc} 1 - 1 & 5 - (-1) \\[5pt] -1 - 5 & 2 - 2 \end{array}\right]}
{= \left[\begin{array}{cc} 0 & 6 \\[5pt] -6 & 0 \end{array}\right]}
\dfrac12(A – A)
{= \dfrac12 × \left[\begin{array}{cc} 0 & 6 \\[5pt] -6 & 0 \end{array}\right]}
{= \left[\begin{array}{cc} \dfrac12 × 0 & \dfrac12 × 6 \\[10pt] \dfrac12 × -6 & \dfrac12 × 0 \end{array}\right]}
{= \left[\begin{array}{cc} 0 & 3 \\[5pt] -3 & 0 \end{array}\right]}
Thus the given matrix A can be represented as the sum of a symmetric and skew symmetric matrices as
A = \dfrac12(A + A) + \dfrac12(A – A)
{\left[\begin{array}{cc} 1 & 5 \\[5pt] 5 & 2 \end{array}\right] = \left[\begin{array}{cc} 1 & 2 \\[5pt] 2 & 2 \end{array}\right] + \left[\begin{array}{cc} 0 & 3 \\[5pt] -3 & 0 \end{array}\right]}

Choose the correct answer in the Exercises 11 and 12.
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
As it is given that both A and B are symmetric matrices, it implies that
A = A
B = B
Now,
(AB – BA)
= (AB) – (BA) (∵ (A – B) = A – B)
= BA – AB (∵ (AB) = BA)
= BA – AB (∵ A = A and B = B)
= -(AB – BA)
As (AB – BA) = – (AB – BA), AB – BA is a skew symmetric matrix.
option A is the correct answer.
12. If {\text{A} = \left[\begin{array}{cc} \cos α & -\sin α \\[5pt] \sin α & \cos α \end{array}\right],} and A + A = I, then the value of α is
(A) \dfracπ6
(B) \dfracπ3
(C) π
(D) \dfrac{3π}{2}
Given that
A
{= \left[\begin{array}{cc} \cos α & -\sin α \\[5pt] \sin α & \cos α \end{array}\right]}
⇒ A
{= \left[\begin{array}{cc} \cos α & \sin α \\[5pt] -\sin α & \cos α \end{array}\right]}
Now,
A + A
{= \left[\begin{array}{cc} \cos α & -\sin α \\[5pt] \sin α & \cos α \end{array}\right] + \left[\begin{array}{cc} \cos α & \sin α \\[5pt] -\sin α & \cos α \end{array}\right]}
{= \left[\begin{array}{cc} \cos α + \cos α & -\sin α + \sin α \\[5pt] \sin α + (-\sin α) & \cos α + \cos α \end{array}\right]}
{= \left[\begin{array}{cc} 2\cos α & 0 \\[5pt] 0 & 2\cos α \end{array}\right]}
As it is given that A + A = I, we have
{\left[\begin{array}{cc} 2\cos α & 0 \\[5pt] 0 & 2\cos α \end{array}\right] = \left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]}
As the given matrices are equal, therefore, their corresponding elements must be equal. Comparing the corresponding elements, we get
{2\cos α}
= 1
{⇒ \cos α}
{= \dfrac12}
{= \cos \dfracπ3}
{α = \dfracπ3}
So, option B is the correct answer.