If $A = \begin{bmatrix} x & 0 \\ 1 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 4 & 0 \\ -1 & 1 \end{bmatrix}$, then value of $x$ for which $A^2 = B$ is:

$-2$

The correct answer is Option (1) → $-2$ ##

$A = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} x & 0 \\ 1 & 1 \end{bmatrix}$

Given that,  $A^2 = B$

$\begin{bmatrix} x & 0 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} x & 0 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ -1 & 1 \end{bmatrix}$

$\begin{bmatrix} x^2 & 0 \\ x+1 & 1 \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ -1 & 1 \end{bmatrix}$

$x + 1 = -1$

$⇒x = -2$ ...(i)

And $x^2 = 4$

$⇒x = \pm 2$ ...(ii)

From Eq. (i) and (ii), we conclude $x = -2$.