NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.2

1. Add corresponding elements.

1. Subtract corresponding elements.

1. \[3A = \begin{bmatrix}6 & 12 \\ 9 & 6\end{bmatrix}\], then subtract C.

1. Row 1: \((2)(1) + (4)(-2) = -6\), \((2)(3) + (4)(5) = 26\).
2. Row 2: \((3)(1) + (2)(-2) = -1\), \((3)(3) + (2)(5) = 19\).

1. Row 1: \(2 + 9 = 11\), \(4 + 6 = 10\). Row 2: \(-4 + 15 = 11\), \(-8 + 10 = 2\).

Where marks slip: \(AB \ne BA\) here: matrix multiplication is not commutative.

1. Add element by element.

1. Each sum is a perfect square, e.g. \(a^2 + b^2 + 2ab = (a + b)^2\).

1. Add element by element.

1. Every element is \(\cos^2 x + \sin^2 x = 1\).

1. \((1,1)\): \(a^2 + b^2\); \((1,2)\): \(-ab + ab = 0\); \((2,1)\): \(-ab + ab = 0\); \((2,2)\): \(b^2 + a^2\).

1. A \(3 \times 1\) times a \(1 \times 3\) gives \(3 \times 3\): entry \((i, j)\) is \(a_i b_j\).

1. Row 1: \(1 - 4,\ 2 - 6,\ 3 - 2\). Row 2: \(2 + 6,\ 4 + 9,\ 6 + 3\).

1. Row 1: \(2 + 0 + 12 = 14\), \(-6 + 6 + 0 = 0\), \(10 + 12 + 20 = 42\); rows 2 and 3 the same way.

1. \((3 \times 2)(2 \times 3) = 3 \times 3\). Row 1: \(2 - 1,\ 0 + 2,\ 2 + 1\).

1. Row 1: \(6 - 1 + 9 = 14\), \(-9 + 0 + 3 = -6\). Row 2: \(-2 + 0 + 6 = 4\), \(3 + 0 + 2 = 5\).

1. \[A + B = \begin{bmatrix}4 & 1 & -1 \\ 9 & 2 & 7 \\ 3 & -1 & 4\end{bmatrix}\], \[B - C = \begin{bmatrix}-1 & -2 & 0 \\ 4 & -1 & 3 \\ 1 & 2 & 0\end{bmatrix}\]
2. \(A + (B - C)\) and \((A + B) - C\) both work out to the matrix below.

1. \[3A = \begin{bmatrix}2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2\end{bmatrix}\] and \(5B\) is the same matrix.

1. \((1,1)\): \(\cos^2\theta + \sin^2\theta = 1\); \((1,2)\): \[\cos\theta\sin\theta - \sin\theta\cos\theta = 0\]; similarly \((2,1) = 0\), \((2,2) = 1\).

1. Add: \[2X = \begin{bmatrix}10 & 0 \\ 2 & 8\end{bmatrix}\] Subtract: \[2Y = \begin{bmatrix}4 & 0 \\ 2 & 2\end{bmatrix}\]

1. Eliminate Y: \(3(3X + 2Y) - 2(2X + 3Y) = 5X\), so \[X = \tfrac15\left(3\begin{bmatrix}2 & -2 \\ -1 & 5\end{bmatrix} - 2\begin{bmatrix}2 & 3 \\ 4 & 0\end{bmatrix}\right)\]
2. Eliminate X: \(3(2X + 3Y) - 2(3X + 2Y) = 5Y\).

1. \[\begin{aligned}2X &= \begin{bmatrix}1 & 0 \\ -3 & 2\end{bmatrix} - \begin{bmatrix}3 & 2 \\ 1 & 4\end{bmatrix} \\ &= \begin{bmatrix}-2 & -2 \\ -4 & -2\end{bmatrix}\end{aligned}\]

1. Left side: \[\begin{bmatrix}y + 2 & 6 \\ 1 & 2 x + 2\end{bmatrix}\]
2. \(2 + y = 5\), \(2x + 2 = 8\).

1. \(2x + 3 = 9\), \(2z - 3 = 15\), \(2y = 12\), \(2t + 6 = 18\).

1. \(2x - y = 10\) and \(3x + y = 5\); adding, \(5x = 15\).

1. \(3x = x + 4 \Rightarrow x = 2\); \(3w = 2w + 3 \Rightarrow w = 3\).
2. \[\begin{aligned}&3y = 6 + x + y \\ \Rightarrow\ &2y = 8 \\ \Rightarrow\ &y = 4\end{aligned}\]; \[\begin{aligned}&3z = -1 + z + w \\ \Rightarrow\ &2z = 2 \\ \Rightarrow\ &z = 1\end{aligned}\]

1. The top-left \(2 \times 2\) block of the product is \[\begin{bmatrix}- \sin{\left(x \right)} \sin{\left(y \right)} + \cos{\left(x \right)} \cos{\left(y \right)} & - \sin{\left(x \right)} \cos{\left(y \right)} - \sin{\left(y \right)} \cos{\left(x \right)} \\ \sin{\left(x \right)} \cos{\left(y \right)} + \sin{\left(y \right)} \cos{\left(x \right)} & - \sin{\left(x \right)} \sin{\left(y \right)} + \cos{\left(x \right)} \cos{\left(y \right)}\end{bmatrix}\]; the third row and column stay \(0, 0, 1\).
2. By the compound-angle formulas this is \[\begin{bmatrix}\cos{\left(x + y \right)} & - \sin{\left(x + y \right)} \\ \sin{\left(x + y \right)} & \cos{\left(x + y \right)}\end{bmatrix}\] in the block, i.e. \(F(x + y)\).

1. Left: \[\begin{bmatrix}7 & 1 \\ 33 & 34\end{bmatrix}\] Right: \[\begin{bmatrix}16 & 5 \\ 39 & 25\end{bmatrix}\]

1. Left: \[\begin{bmatrix}5 & 8 & 14 \\ 0 & -1 & 1 \\ -1 & 0 & 1\end{bmatrix}\] Right: \[\begin{bmatrix}-1 & -1 & -3 \\ 1 & 0 & 0 \\ 6 & 11 & 6\end{bmatrix}\]

1. \[A^2 = \begin{bmatrix}5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2\end{bmatrix}\]
2. \[A^2 - 5A + 6I = \begin{bmatrix}5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2\end{bmatrix} - \begin{bmatrix}10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0\end{bmatrix} + \begin{bmatrix}6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6\end{bmatrix}\]

1. \[A^2 = \begin{bmatrix}5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13\end{bmatrix}\], \[\begin{aligned}A^3 &= A^2 A \\ &= \begin{bmatrix}21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55\end{bmatrix}\end{aligned}\]
2. \(A^3 - 6A^2 + 7A + 2I\): e.g. the \((1,1)\) entry is \(21 - 30 + 7 + 2 = 0\), and every entry works out to 0.

1. \[A^2 = \begin{bmatrix}1 & -2 \\ 4 & -4\end{bmatrix}\]
2. \[kA - 2I = \begin{bmatrix}3 k - 2 & - 2 k \\ 4 k & - 2 k - 2\end{bmatrix}\] Equating: \(3k - 2 = 1\), \(-2k = -2\), \(4k = 4\), \(-2k - 2 = -4\): all give \(k = 1\).

1. Let \(t = \tan\tfrac{\alpha}{2}\): \(\cos\alpha = \dfrac{1 - t^2}{1 + t^2}\), \(\sin\alpha = \dfrac{2t}{1 + t^2}\).
2. \[I - A = \begin{bmatrix}1 & t \\ - t & 1\end{bmatrix}\] Row 1 of the product: \[\begin{aligned}\cos\alpha + t\sin\alpha &= \dfrac{1 - t^2 + 2t^2}{1 + t^2} \\ &= 1\end{aligned}\] and \[\begin{aligned}-\sin\alpha + t\cos\alpha &= \dfrac{-2t + t - t^3}{1 + t^2} \\ &= -t\end{aligned}\]
3. Row 2: \(-t\cos\alpha + \sin\alpha = t\) and \(t\sin\alpha + \cos\alpha = 1\). So the product is \[\begin{bmatrix}1 & - t \\ t & 1\end{bmatrix} = I + A\]

1. Let Rs x go in the 5% bond and Rs \(30000 - x\) in the 7% bond: \[\begin{bmatrix} x & 30000 - x\end{bmatrix}\begin{bmatrix}0.05 \\ 0.07\end{bmatrix} = [1800]\]
2. \[\begin{aligned}&0.05x + 2100 - 0.07x = 1800 \\ \Rightarrow\ &0.02x = 300\end{aligned}\]

1. \[\begin{aligned}&0.05x + 0.07(30000 - x) = 2000 \\ \Rightarrow\ &0.02x = 100\end{aligned}\]

1. Quantities \[\begin{bmatrix}120 & 96 & 120\end{bmatrix}\], prices \[\begin{bmatrix}80 \\ 60 \\ 40\end{bmatrix}\]
2. \[120 \cdot 80 + 96 \cdot 60 + 120 \cdot 40 = 9600 + 5760 + 4800\]

1. PY: \((p \times k)(3 \times k)\) needs \(k = 3\); it is \(p \times 3\).
2. WY: \((n \times 3)(3 \times 3)\) is \(n \times 3\). Adding needs \(p = n\).

1. X and Z are both \(2 \times n\) when \(n = p\), and so is the difference.