NCERT Solutions Class 12 Maths Chapter 3 Ex 3.2 Full | LearnCBSEHub

Matrix Masterclass

EXERCISE 3.2 • Q1 — Q22 COMPLETE

💡 Key Concepts for Exercise 3.2

Addition/Subtraction: Possible ONLY for same order matrices. Combine $a_{ij}$ and $b_{ij}$.
Matrix Product ($AB$): Defined if Columns of A = Rows of B. Use the Row-by-Column dot product.
Multiplication Property: $AB \neq BA$ (Generally non-commutative).
Identity ($I$): $AI = IA = A$.

Question 01

Find $A+B, A-B, 3A-C, AB, BA$ for $A = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix}, B = \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix}, C = \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix}$.

(i) $A+B = \begin{bmatrix} 2+1 & 4+3 \\ 3-2 & 2+5 \end{bmatrix} = \mathbf{\begin{bmatrix} 3 & 7 \\ 1 & 7 \end{bmatrix}}$

(ii) $A-B = \begin{bmatrix} 2-1 & 4-3 \\ 3-(-2) & 2-5 \end{bmatrix} = \mathbf{\begin{bmatrix} 1 & 1 \\ 5 & -3 \end{bmatrix}}$

(iii) $3A-C = \begin{bmatrix} 6 & 12 \\ 9 & 6 \end{bmatrix} – \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix} = \mathbf{\begin{bmatrix} 8 & 7 \\ 6 & 2 \end{bmatrix}}$

(iv) $AB = \begin{bmatrix} (2)(1)+(4)(-2) & (2)(3)+(4)(5) \\ (3)(1)+(2)(-2) & (3)(3)+(2)(5) \end{bmatrix} = \mathbf{\begin{bmatrix} -6 & 26 \\ -1 & 19 \end{bmatrix}}$

Question 03 • All Products

Compute the products (i) to (vi).

(ii) Column $\times$ Row

$\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \begin{bmatrix} 2 & 3 & 4 \end{bmatrix} = \begin{bmatrix} 1(2) & 1(3) & 1(4) \\ 2(2) & 2(3) & 2(4) \\ 3(2) & 3(3) & 3(4) \end{bmatrix} = \mathbf{\begin{bmatrix} 2 & 3 & 4 \\ 4 & 6 & 8 \\ 6 & 9 & 12 \end{bmatrix}}$

(iv) 3×3 Multiplication

$\begin{bmatrix} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6 \end{bmatrix} \begin{bmatrix} 1 & -3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5 \end{bmatrix} = \mathbf{\begin{bmatrix} 14 & 0 & 42 \\ 18 & -1 & 56 \\ 22 & -2 & 70 \end{bmatrix}}$

Question 04 • Verification

Verify $A+(B-C) = (A+B)-C$.

Step 1: Compute $(B-C)$

$B-C = \begin{bmatrix} 3-4 & -1-1 & 2-2 \\ 4-0 & 2-3 & 5-2 \\ 2-1 & 0-(-2) & 3-3 \end{bmatrix} = \mathbf{\begin{bmatrix} -1 & -2 & 0 \\ 4 & -1 & 3 \\ 1 & 2 & 0 \end{bmatrix}}$

Verification is confirmed by calculating both sides: Result = $\mathbf{\begin{bmatrix} 0 & 0 & -3 \\ 9 & -1 & 5 \\ 2 & 1 & 1 \end{bmatrix}}$

Question 07 • Finding X and Y

Solve (i) $X+Y$ and $X-Y$ (ii) $2X+3Y$ and $3X+2Y$.

(ii) Simultaneous Equations

Multiply Eq(1) by 3 and Eq(2) by 2 to eliminate X:

$6X+9Y = \begin{bmatrix} 6 & 9 \\ 12 & 0 \end{bmatrix} \dots (3)$
$6X+4Y = \begin{bmatrix} 4 & -4 \\ -2 & 10 \end{bmatrix} \dots (4)$
Subtracting (4) from (3): $5Y = \begin{bmatrix} 2 & 13 \\ 14 & -10 \end{bmatrix} \implies \mathbf{Y = \begin{bmatrix} 2/5 & 13/5 \\ 14/5 & -2 \end{bmatrix}}$

Question 13 • Trigonometric Proof

Show $F(x)F(y) = F(x+y)$.

Perform Row-by-Column multiplication:

$R_1C_1: \cos x \cos y – \sin x \sin y = \mathbf{\cos(x+y)}$
$R_1C_2: -(\cos x \sin y + \sin x \cos y) = \mathbf{-\sin(x+y)}$
$R_2C_1: \sin x \cos y + \cos x \sin y = \mathbf{\sin(x+y)}$
$R_2C_2: -\sin x \sin y + \cos x \cos y = \mathbf{\cos(x+y)}$

✅ LHS = RHS. Proof complete.

Question 15 • Polynomial

Find $A^2 – 5A + 6I$.

$A^2 = \begin{bmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{bmatrix}, -5A = \begin{bmatrix} -10 & 0 & -5 \\ -10 & -5 & -15 \\ -5 & 5 & 0 \end{bmatrix}, 6I = \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix}$

Result = $\mathbf{\begin{bmatrix} 1 & -1 & -3 \\ -1 & -1 & -10 \\ -5 & 4 & 4 \end{bmatrix}}$

Question 18 • Complex Proof

Prove $I + A = (I – A) \times$ Trig Matrix.

Use substitution $t = \tan(\alpha/2)$. Use $\cos \alpha = \frac{1-t^2}{1+t^2}$ and $\sin \alpha = \frac{2t}{1+t^2}$.

$\text{RHS} = \begin{bmatrix} 1 & t \\ -t & 1 \end{bmatrix} \begin{bmatrix} \frac{1-t^2}{1+t^2} & \frac{-2t}{1+t^2} \\ \frac{2t}{1+t^2} & \frac{1-t^2}{1+t^2} \end{bmatrix} = \dots = \begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix} = I+A$

✅ LHS = RHS. Proof complete.

Question 19 • Trust Fund

Investment in ₹30,000 bonds.

Equation: $[x \quad 30000-x] \begin{bmatrix} 0.05 \\ 0.07 \end{bmatrix} = [1800]$

$0.05x + 0.07(30000-x) = 1800 \implies -0.02x = -300 \implies \mathbf{x=15000}$.

₹15,000 in each bond.

MCQs 21 — 22

Order restrictions.

21. To define $PY + WY$: Columns $P = Rows Y \implies \mathbf{k=3}$. Same order for addition $\implies \mathbf{p=n}$. (Option A)

22. Given $n=p$. Order of $7X-5Z = \mathbf{2 \times n}$. (Option B)