Class 12th

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]

(E) (i) a^ij = $\frac{1}{2}$ $|-3i+j|$ such that i = 1, 2, 3 and j = 1, 2, 3, 4 for 3 * 4 matrix

So, a₁₁= $\frac{1}{2}$ . $|-3.1+1|=\frac{1}{2}|-3+1|=\frac{1}{2}|-3+1|=\frac{1}{2}|-2|=\frac{2}{2}=1.$

a₁₂ = $\frac{1}{2}|-3.1+2|=\frac{1}{2}|-1|=\frac{1}{2}$

$a_{13}=\frac{1}{2}|-3.1+3|=\frac{1}{2}*0=0$

a₁₄ = $\frac{1}{2}|-3.1+4|=\frac{1}{2}|1|=\frac{1}{2}$

a₂₁ = $\frac{1}{2}|-3.2+1|=\frac{1}{2}|-6+1|=\frac{1}{2}|5|=\frac{5}{2}$

a₂₂ = $\frac{1}{2}|-3.2+2|=\frac{1}{2}|-6+2|=\frac{1}{2}|-4|=\frac{4}{2}=2$

a₂₃ = $\frac{1}{2}|-3.2+3|=\frac{1}{2}|-6+3|=\frac{1}{2}|-3|=\frac{3}{2}$

$a_{24}=\frac{1}{2}|-3\cdot 2+4|=\frac{1}{2}|-6+4|=\frac{1}{2}+4|-2|=\frac{2}{2}=1$

$a_{31}=\frac{1}{2}|-3\cdot 3+1|=\frac{1}{2}|-0+1|=\frac{1}{2}|-8|=\frac{8}{2}=4.$

$a_{32}=\frac{1}{2}|-3\cdot 2+2|=\frac{1}{2}|-9+2|=?-\frac{7}{2}=\frac{7}{2}$

$a_{33}=\frac{1}{2}|-3\cdot 3+3|=\frac{1}{2}|-9+3|=\frac{+6?}{2}=\frac{6}{2}=3$

$a_{34}=\frac{1}{2}|-3\cdot 3+4|=\frac{1}{2}|-9+4|=?\frac{\left|5\right|}{2}=\frac{5}{2}.$