Class 12th

$cos\theta \left[\begin{array}{cc}\hfill cos\theta \hfill & \hfill sin\theta \hfill \\ \hfill -sin\theta \hfill & \hfill cos\theta \hfill \end{array}\right]+sin\theta \left[\begin{array}{cc}\hfill sin\theta \hfill & \hfill -cos\theta \hfill \\ \hfill cos\theta \hfill & \hfill sin\theta \hfill \end{array}\right]$

(E) $=\left[\begin{array}{l}co{r}^2\theta cot\theta sin\theta \\ -cos\theta sin\theta co{s}^2\theta \end{array}\right]$ $+\left[\begin{array}{l}sin2\theta -sin\theta cos\theta .\\ sin\theta cos\theta si{n}^2\theta \end{array}\right]$ $$

$=\left[\begin{array}{l}co{s}^2\theta +si{n}^2\theta cos\theta sin\theta -sin\theta cos\theta \\ -cos\theta sin\theta +sin\theta cos\theta co{s}^2\theta +si{n}^2\theta \end{array}\right]$

$=\left[\begin{array}{l}10\\ 01\end{array}\right]$

3a - 5b = 3 * $\left[\begin{array}{ccc}\hfill \frac{2}{3}\hfill & \hfill 1\hfill & \hfill \frac{5}{3}\hfill \\ \hfill \frac{1}{3}\hfill & \hfill \frac{2}{3}\hfill & \hfill \frac{4}{3}\hfill \\ \hfill \frac{7}{3}\hfill & \hfill 2\hfill & \hfill \frac{2}{3}\hfill \end{array}\right]$ - $5*\left[\begin{array}{ccc}\hfill \frac{2}{5}\hfill & \hfill \frac{3}{5}\hfill & \hfill 1\hfill \\ \hfill \frac{1}{5}\hfill & \hfill \frac{2}{5}\hfill & \hfill \frac{4}{5}\hfill \\ \hfill \frac{7}{5}\hfill & \hfill \frac{6}{5}\hfill & \hfill \frac{2}{5}\hfill \end{array}\right]$

$=\left[\begin{array}{ccc}\hfill 3*\frac{2}{3}\hfill & \hfill 3*1\hfill & \hfill 3*\frac{5}{3}\hfill \\ \hfill 3*\frac{1}{3}\hfill & \hfill 3*\frac{2}{3}\hfill & \hfill 3*\frac{4}{3}\hfill \\ \hfill 3*\frac{7}{3}\hfill & \hfill 3*2\hfill & \hfill 3*\frac{2}{3}\hfill \end{array}\right]-$ $\left[\begin{array}{ccc}\hfill 5*\frac{2}{5}\hfill & \hfill 5*\frac{3}{5}\hfill & \hfill 5*1\hfill \\ \hfill 5*\frac{1}{5}\hfill & \hfill 5*\frac{2}{5}\hfill & \hfill 5*\frac{4}{5}\hfill \\ \hfill 5*\frac{7}{5}\hfill & \hfill 5*\frac{6}{5}\hfill & \hfill 5*\frac{2}{5}\hfill \end{array}\right]$

$=\left[\begin{array}{lll}2\hfill & 3\hfill & 5\hfill \\ 1\hfill & 2\hfill & 4\hfill \\ 7\hfill & 6\hfill & 2\hfill \end{array}\right]-\left[\begin{array}{lll}2\hfill & 3\hfill & 5\hfill \\ 1\hfill & 2\hfill & 4\hfill \\ 7\hfill & 6\hfill & 2\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 2-2\hfill & \hfill 3-3\hfill & \hfill 5-5\hfill \\ \hfill 1-1\hfill & \hfill 2-2\hfill & \hfill 4-4\hfill \\ \hfill 7-7\hfill & \hfill 6-6\hfill & \hfill 2-2\hfill \end{array}\right]$ = $\left[\begin{array}{l}000\\ 600\\ 000\end{array}\right]$

A + B = $\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill -3\hfill \\ \hfill 5\hfill & \hfill 0\hfill & \hfill 2\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill 1\hfill \end{array}\right]+\left[\begin{array}{ccc}\hfill 3\hfill & \hfill -1\hfill & \hfill 2\hfill \\ \hfill 4\hfill & \hfill 2\hfill & \hfill 5\hfill \\ \hfill 2\hfill & \hfill 0\hfill & \hfill 3\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1+3\hfill & \hfill 2-1\hfill & \hfill -3+2\hfill \\ \hfill 5+4\hfill & \hfill 0+2\hfill & \hfill 2+5\hfill \\ \hfill 1+2\hfill & \hfill -1+0\hfill & \hfill 1+3\hfill \end{array}\right]$

$A+B=\left[\begin{array}{ccc}\hfill 4\hfill & \hfill 1\hfill & \hfill -1\hfill \\ \hfill 9\hfill & \hfill 2\hfill & \hfill 7\hfill \\ \hfill 3\hfill & \hfill -1\hfill & \hfill 4\hfill \end{array}\right]$

$B-C=\left[\begin{array}{lll}3\hfill & -1\hfill & 2\hfill \\ 4\hfill & 2\hfill & 5\hfill \\ 2\hfill & 0\hfill & 3\hfill \end{array}\right]-\left[\begin{array}{ccc}\hfill 4\hfill & \hfill 1\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 3\hfill & \hfill 2\hfill \\ \hfill 1\hfill & \hfill -2\hfill & \hfill 3\hfill \end{array}\right]$ $=\left[\begin{array}{l}3-4-1-12-2\\ 4-02-35-2\\ 2-10-(2)33\end{array}\right]$

$B-C=\left[\begin{array}{ccc}\hfill -1\hfill & \hfill -2\hfill & \hfill 0\hfill \\ \hfill 4\hfill & \hfill -1\hfill & \hfill 2\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 0\hfill \end{array}\right]$

Now

L.H.S. = a+ (b - c) = $\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill -3\hfill \\ \hfill 5\hfill & \hfill 0\hfill & \hfill 2\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill 1\hfill \end{array}\right]+\left[\begin{array}{ccc}\hfill -1\hfill & \hfill -2\hfill & \hfill 0\hfill \\ \hfill 4\hfill & \hfill -1\hfill & \hfill 3\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 0\hfill \end{array}\right]$ $=\left[\begin{array}{l}1-12-2-3+0\\ 5+40-12+3\\ 1+1-1+21+0\end{array}\right]$

$=\left[\begin{array}{l}00-3\\ 9-15\\ 211\end{array}\right]$

R.H.S. = (a+ b) - c = $\left[\begin{array}{ccc}\hfill 4\hfill & \hfill 1\hfill & \hfill -1\hfill \\ \hfill 9\hfill & \hfill 2\hfill & \hfill 7\hfill \\ \hfill 3\hfill & \hfill -1\hfill & \hfill 4\hfill \end{array}\right]-\left[\begin{array}{ccc}\hfill 4\hfill & \hfill 1\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 3\hfill & \hfill 2\hfill \\ \hfill 1\hfill & \hfill -2\hfill & \hfill 3\hfill \end{array}\right]$ $=\left[\begin{array}{l}4-41-1-1-3\\ 9-02-37-2\\ 3-1-1-(2)4-3\end{array}\right]$

$=\left[\begin{array}{l}00-3\\ 9-15\\ 211\end{array}\right]$ = L.H.S.

∴A + (B - C) = (A + B) - C

Kindly go through the solution

(I) $\left[\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill -b\hfill & \hfill a\hfill \end{array}\right]+\left[\begin{array}{ll}a\hfill & b\hfill \\ b\hfill & a\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill a+a\hfill & \hfill b+b\hfill \\ \hfill -b+b\hfill & \hfill a+a\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 2a\hfill & \hfill 2b\hfill \\ \hfill 0\hfill & \hfill 2a\hfill \end{array}\right]$

A $=\left[\begin{array}{ll}2\hfill & 4\hfill \\ 3\hfill & 2\hfill \end{array}\right],$ B $=\left[\begin{array}{rr}\hfill 1& \hfill 3\\ \hfill -2& \hfill 5\end{array}\right],$ C $=\left[\begin{array}{rr}\hfill -2& \hfill 5\\ \hfill 3& \hfill 4\end{array}\right]$

(i) A + B = $\left[\begin{array}{ll}2\hfill & 4\hfill \\ 3\hfill & 2\hfill \end{array}\right]+\left[\begin{array}{cc}\hfill 1\hfill & \hfill 3\hfill \\ \hfill -2\hfill & \hfill 5\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 2+1\hfill & \hfill 4+3\hfill \\ \hfill 3-2\hfill & \hfill 2+5\hfill \end{array}\right]=\left[\begin{array}{ll}3\hfill & 7\hfill \\ 1\hfill & 7\hfill \end{array}\right]$

(ii) A - B = $\left[\begin{array}{ll}2\hfill & 4\hfill \\ 3\hfill & 2\hfill \end{array}\right]-\left[\begin{array}{cc}\hfill 1\hfill & \hfill 3\hfill \\ \hfill -2\hfill & \hfill 5\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 2-1\hfill & \hfill 4-3\hfill \\ \hfill 3-(-2)\hfill & \hfill 2-5\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 3\hfill & \hfill 1\hfill \\ \hfill 5\hfill & \hfill -3\hfill \end{array}\right].$

(iii) 3A - C = 3 $\left[\begin{array}{ll}2\hfill & 4\hfill \\ 3\hfill & 2\hfill \end{array}\right]-\left[\begin{array}{cc}\hfill -2\hfill & \hfill 5\hfill \\ \hfill 3\hfill & \hfill 4\hfill \end{array}\right]=\left[\begin{array}{ll}3*2\hfill & 3*4\hfill \\ 3*3\hfill & 3*2\hfill \end{array}\right]-\left[\begin{array}{cc}\hfill -2\hfill & \hfill 5\hfill \\ \hfill 3\hfill & \hfill 4\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 6\hfill & \hfill 12\hfill \\ \hfill 9\hfill & \hfill 6\hfill \end{array}\right]\left[\begin{array}{cc}\hfill -2\hfill & \hfill 5\hfill \\ \hfill 3\hfill & \hfill 4\hfill \end{array}\right]$

$$ $=\left[\begin{array}{cc}\hfill 6-(-2)\hfill & \hfill 12-5\hfill \\ \hfill 9-3\hfill & \hfill 6-4\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 8\hfill & \hfill 7\hfill \\ \hfill 6\hfill & \hfill 2\hfill \end{array}\right]$

(iv) AB = $\left[\begin{array}{ll}2\hfill & 4\hfill \\ 3\hfill & 2\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 1\hfill & \hfill 3\hfill \\ \hfill -2\hfill & \hfill 5\hfill \end{array}\right]=\left[\begin{array}{ll}2*1+4*(-2)\hfill & 2*3+4*5\hfill \\ 3*1+2*(-2)\hfill & 3*3+2*5\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 2-8\hfill & \hfill 6+20\hfill \\ \hfill 3-4\hfill & \hfill 9+10\hfill \end{array}\right]$

= $\left[\begin{array}{rr}\hfill -6& \hfill 26\\ \hfill -1& \hfill 19\end{array}\right]$

A 3 * 3 order matrix will have 9 elements

(M) Since, the elements can be either 1 or $\mathrm{0il},$ number of choices for each element is 2 .

The required no. of arrangement = 29 (4,2 * 2 * 2 9 times ) $\Rightarrow$ 512

So, option D is correct.

For the matrices to be equal, the corresponding elements

(B) heed to be equal so,

Fora₁₁, 3x+ 7 = 0

$\Rightarrow$ 3x = -7.

$\Rightarrow$ x = $-\frac{7}{3}$

fora₁₂, 5 = y - 2 

$\Rightarrow$ y = + 5 + 2 = 7.

For a₂₁, y + 1 = 8

$$ $\Rightarrow$ y = 8 - 1 = 7.

fora₂₂, 2 - 3x = 4.

$\Rightarrow$ 3x = 2 - 4

$\Rightarrow x=-\frac{2}{3}$

As the variable x and y has more than one value which is not peacetable.

Option B is correct.

$={\left[a_{ij}\right]}_{m*n}$ is a square matrix if m = n

(E) Option C is correct.

$\left[\begin{array}{cc}\hfill a-b\hfill & \hfill 2a+c\hfill \\ \hfill 2a-b\hfill & \hfill 3c+d\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill -1\hfill & \hfill 5\hfill \\ \hfill 0\hfill & \hfill 13\hfill \end{array}\right]$

(5) Equating the corresponding elements of the matrices we get,

a - b = -1 - (i)

2a + c = 5 - (2)

2a - b = 0 - (3)

3c + d = 13 - (4)

Subtracting eqⁿ (1) from (3) we get,

2a - b (a - b) = 0- (-1)

2a - b - a+ b = 0 + 1 = 1

$\Rightarrow$  [a= 1]

Put $a=\mathrm{1\; eq}{}^n(2)$ we get,

2 * 1 + c = 5 $\Rightarrow$ c = 5 - 2 $\Rightarrow$  [c = 3]

Put $c=3q^n$  (4) we get,

3 * 3 + d = 13  => d = 13 - 9 $\Rightarrow$  [d = 4.]

put $a=1q^2(1)$ we get, 1 - b = -1 $\Rightarrow$ b = 1 + 1 $\Rightarrow$  [b= 2]