Ncert Solutions Maths class 12th

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

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

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

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

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

$={\left[a_{ij}\right]}_{m\times n}$ 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&

(i) $\left[\begin{array}{ll}4\hfill & 3\hfill \\ x\hfill & 5\hfill \end{array}\right]=\left[\begin{array}{ll}y\hfill & z\hfill \\ 1\hfill & 5\hfill \end{array}\right]$

corresponding

By equating the elements of the matrices, we get,

x= 1

y= 4

z= 3.

(ii) $\left[\begin{array}{cc}\hfill x+y\hfill & \hfill 2\hfill \\ \hfill 5+z\hfill & \hfill xy\hfill \end{array}\right]=\left[\begin{array}{ll}6\hfill & 2\hfill \\ 5\hfill & 8\hfill \end{array}\right]$

By equating the corresponding elements of the matrices we get,

x+ y = 6 (I)

5 + Z = 5 $\Rightarrow z=5-5\Rightarrow z=0$

xy = 8

$\Rightarrow$ x $=\frac{8}{y}$ $\to (2)$

putting eq${}^n(2)$ in (1) we get

$\frac{8}{y}$ + y = 6.

$\Rightarrow$ 8 + y² = 6y

$\Rightarrow$ y² 6y + 8 = 0.

$\Rightarrow$ y² - 4y - 2y + 8 = 0

$\Rightarrow$

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

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

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

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

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

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

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

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

(E) (i) a^ij${\frac{\left(i+j\right)}{2}}^2$ such that i = 1, 2 and j = 1 × 2 for 2 × 2 matrix

Therefore a₁₁ = $\frac{{\left(1+1\right)}^2}{2}=\frac{2^2}{2}=2$ A _2×2 = $\left[\begin{array}{cc}\hfill a_{11}\hfill & \hfill a_{12}\hfill \\ \hfill a_{21}\hfill & \hfill a_{22}\hfill \end{array}\right]$

a₁₂ = $\frac{{\left(1+2\right)}^2}{2}=\frac{3^2}{2}=\frac{9}{2}$

a₂₁ $\frac{{\left(2+1\right)}^2}{2}=\frac{3^2}{2}=\frac{9}{2}$ $\left[\begin{array}{cc}\hfill 2\hfill & \hfill \raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill \\ \hfill \raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill & \hfill 8\hfill \end{array}\right]$

a₂₂ = ${\frac{\left(2+2\right)}{2}}^2=\frac{4^2}{2}=\frac{16}{2}=8$

As number of elements of matrix with order m * n

(E) Possible order of matrix with 18 elements are (1 * 18), (2 * 9), (3 * 6), (6 * 3), (9 * 2) and (18 * 1)

Similarly, possible order of matrix with 5 elements are (1 * 5) and (5 * 1)