Choose the correct answer from the given alternatives in the following question:

If $A = ;\begin{bmatrix} 1 & 2 \\ 2 & 1& ; \end{bmatrix}$ and $A ({\text{adj}} A) = KI$ , then the value of $k$ is 

Given,

$A=\left[\begin{array}{ll} 1 & 2 \\ 2 & 1 \end{array}\right] .$

and, $A(\operatorname{adj} A)=K I$

$\begin{array}{l} \Rightarrow\left[\begin{array}{ll} 1 & 2 \\ 2 & 1 \end{array}\right] \times\left[\begin{array}{cc} 1 & -2 \\ -2 & 1 \end{array}\right]=k\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\ \Rightarrow\left[\begin{array}{cc} 1-4 & -2+2 \\ 2-2 & -4+1 \end{array}\right]=k\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\ \Rightarrow\left[\begin{array}{cc} -3 & 0 \\ 0 & -3 \end{array}\right]=k\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\ \Rightarrow-3\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right]=k\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\ \Rightarrow k=-3 . \end{array}$