31. The restriction on n, k and p so that PY + WY will be defined are:

(A) k = 3, p = n  (B) k is arbitrary, p = 2 (C) p is arbitrary, k = 3  (D) k = 2, p = 3

Let   $A=\left[\begin{array}{ccc}\hfill x_1\hfill & \hfill y_1\hfill & \hfill z_1\hfill \\ \hfill x_2\hfill & \hfill y_2\hfill & \hfill z_2\hfill \\ \hfill x_3\hfill & \hfill y_3\hfill & \hfill z_3\hfill \end{array}\right]$

Given  $A=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 2\hfill & \hfill 0\hfill & \hfill 2\hfill \end{array}\right]$   ...(1)

$\left[\begin{array}{ccc}\hfill x_1+z_1\hfill & \hfill x_2+z_2\hfill & \hfill x_3+z_3\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 2\hfill & \hfill 0\hfill & \hfill 2\hfill \end{array}\right]$

∴    x₁ + z₁ = 2                … (2)

x₂ + z₂ = 0               … (3)

x₃ + z₃ = 0    &nb

$f\left(x\right)=a{x}^2+bx+c$  

g (x) = px + q

  $f\left(g\left(x\right)\right)=a{\left(px+q\right)}^2+b\left(p\right)\left(+q\right)+c$

$\Rightarrow 8x^2-2x=a{\left(px+q\right)}^2+b\left(px+q\right)+c$

Compare 8 = ap² …………… (i)

-2 = a (2pq) + bp

0 = aq² + bq + c

  $9\left(f\left(x\right)\right)=p\left(a{x}^2+bx+c\right)+q$

=>4x² + 6x + 1 = apx² + bpx + cp + q

=> Andhra Pradesh = 4 ……………. (ii)

6 = bp

1 = cp + q

From (i) & (ii), p = 2, q = -1

=> b = 3, c = 1, a = 2

f (x) = 2x² + 3x + 1

f (2) = 8 + 6 + 1 = 15

g (x) = 2x – 1

g (2) = 3

Kindly consider the following figure

B = (I – adjA)⁵

Kindly consider the following figure

B = (I – adjA)⁵

System of equation is

$\left(\begin{array}{ccc}\hfill 2\hfill & \hfill 3\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill \left|\lambda \right|\hfill \end{array}\right)\left(\begin{array}{l}x\\ y\\ z\end{array}\right)=\left[\begin{array}{l}-2\\ 4\\ 4\lambda -4\end{array}\right]$

R₁ – 2 R₂, R₃ – R₂

$\left(\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill -3\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill -2\hfill & \hfill \left|\lambda \right|-1\hfill \end{array}\right)\left(\begin{array}{l}x\\ y\\ z\end{array}\right)=\left(\begin{array}{l}-10\\ 4\\ 4\lambda -8\end{array}\right)$

System of equation will have no solution for = -7.