Question
If $A=\left[\begin{array}{ll}x & 0 \\ 1 & 1\end{array}\right]$ and $B=\left[\begin{array}{cc}4 & 0 \\ -1 & 1\end{array}\right]$, then value of $x$ for which $A^2=B$ is

Answer

Given, $A=\left[\begin{array}{ll}x & 0 \\ 1 & 1\end{array}\right]$ and $B=\left[\begin{array}{cc}4 & 0 \\ -1 & 1\end{array}\right]$
Now, $A^2=\left[\begin{array}{ll}x & 0 \\ 1 & 1\end{array}\right]\left[\begin{array}{ll}x & 0 \\ 1 & 1\end{array}\right]=\left[\begin{array}{cc}x^2 & 0 \\ x+1 & 1\end{array}\right]$
$\begin{array}{c}
A^2=B \\
\Rightarrow\left[\begin{array}{cc}
x^2 & 0 \\
x+1 & 1
\end{array}\right]=\left[\begin{array}{cc}
4 & 0 \\
-1 & 1
\end{array}\right]
\end{array}$
On comparing, we get $x^2=4$ and $x+1=-1$
$\Rightarrow x= \pm 2$ and $x=-2$
$\therefore \quad$ Required value of $x$ is -2 .

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