NCERT Solutions Class 12 Maths Chapter 3 Ex 3.3 Complete | LearnCBSEHub

Matrix Masterclass

NCERT EXERCISE 3.3 • Q1 — Q12 COMPLETE

💡 Mastery Essentials

Transpose ($A’$): Rows become columns and columns become rows.
Symmetric: Matrix is its own transpose ($A’ = A$).
Skew-Symmetric: Transpose is the negative of the matrix ($A’ = -A$).
Sum Property: $A = \frac{1}{2}(A + A’) + \frac{1}{2}(A – A’)$.

Question 01

Find the transpose of the given matrices.

(i) $\begin{bmatrix} 5 \\ 1/2 \\ -1 \end{bmatrix} \implies \mathbf{\begin{bmatrix} 5 & 1/2 & -1 \end{bmatrix}}$

(ii) $\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix} \implies \mathbf{\begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix}}$

(iii) $\begin{bmatrix} -1 & 5 & 6 \\ \sqrt{3} & 5 & 6 \\ 2 & 3 & -1 \end{bmatrix} \implies \mathbf{\begin{bmatrix} -1 & \sqrt{3} & 2 \\ 5 & 5 & 3 \\ 6 & 6 & -1 \end{bmatrix}}$

Question 02 & 03

Verify $(A \pm B)’ = A’ \pm B’$.

Verification follows by calculating the operation first then transposing, vs transposing first then operating.

Proof for Q2(i):
$A+B = \begin{bmatrix} -5 & 3 & -2 \\ 6 & 9 & 9 \\ -1 & 4 & 2 \end{bmatrix} \implies (A+B)’ = \mathbf{\begin{bmatrix} -5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \end{bmatrix}}$

$A’+B’ = \begin{bmatrix} -1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1 \end{bmatrix} + \begin{bmatrix} -4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1 \end{bmatrix} = \mathbf{\begin{bmatrix} -5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \end{bmatrix}}$

✅ Verified: LHS = RHS

Question 05

Verify Reversal Law: $(AB)’ = B’A’$.

(i) LHS: $AB = \begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix} \begin{bmatrix} -1 & 2 & 1 \end{bmatrix} = \begin{bmatrix} -1 & 2 & 1 \\ 4 & -8 & -4 \\ -3 & 6 & 3 \end{bmatrix} \implies (AB)’ = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}$

(i) RHS: $B’A’ = \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix} \begin{bmatrix} 1 & -4 & 3 \end{bmatrix} = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}$

✅ Verified: LHS = RHS

Question 07

Show (i) Symmetric and (ii) Skew-Symmetric matrices.

(i) Symmetric: Since $A’ = \begin{bmatrix} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \end{bmatrix} = A$, it is Symmetric.

(ii) Skew-Symmetric: Since $A’ = \begin{bmatrix} 0 & -1 & 1 \\ 1 & 0 & -1 \\ -1 & 1 & 0 \end{bmatrix} = -A$, it is Skew-Symmetric.

Question 09

Find $\frac{1}{2}(A + A’)$ and $\frac{1}{2}(A – A’)$ for $A = \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix}$.

$A’ = \begin{bmatrix} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{bmatrix}$

$\frac{1}{2}(A+A’) = \frac{1}{2} \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = \mathbf{\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}}$

$\frac{1}{2}(A-A’) = \frac{1}{2} \begin{bmatrix} 0 & 2a & 2b \\ -2a & 0 & 2c \\ -2b & -2c & 0 \end{bmatrix} = \mathbf{\begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix}}$

Question 10

Express $A = \begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix}$ as sum of symmetric and skew-symmetric parts.

Every square matrix $A = P + Q$ where $P = \frac{1}{2}(A+A’)$ and $Q = \frac{1}{2}(A-A’)$.

P (Symmetric): $\frac{1}{2} \begin{bmatrix} 6 & 6 \\ 6 & -2 \end{bmatrix} = \mathbf{\begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix}}$

Q (Skew-Symmetric): $\frac{1}{2} \begin{bmatrix} 0 & 4 \\ -4 & 0 \end{bmatrix} = \mathbf{\begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}}$

Result: $A = \begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix} + \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}$

Question 11 & 12 • MCQs

Critical conceptual MCQs.

11. $(AB – BA)’ = (AB)’ – (BA)’ = B’A’ – A’B’ = BA – AB = -(AB – BA)$.
Answer: (A) Skew symmetric matrix

12. $A+A’=I \implies \begin{bmatrix} 2\cos\alpha & 0 \\ 0 & 2\cos\alpha \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \implies \cos\alpha = 1/2$.
Answer: (B) $\pi/3$