Question

If $A=\left[\begin{array}{cc}3& x\\ 0& 1\end{array}\right]$ and $B=\left[\begin{array}{cc}9& 16\\ 0& -y\end{array}\right]$, find $x$ and $y$ when $A^2=B$.

Answer 1

Given : $A=\left[\begin{array}{cc}3& x\\ 0& 1\end{array}\right]$ and $B=\left[\begin{array}{cc}9& 16\\ 0& -y\end{array}\right]$
$A^2=B$
$\left[\begin{array}{cc}3& x\\ 0& 1\end{array}\right]\left[\begin{array}{cc}3& x\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}9& 16\\ 0& -y\end{array}\right]$

$\left[\begin{array}{cc}3\times 3+x\times 0& 3\times x+x\times 1\\ 0\times 3+1\times 0& 0\times x+1\times 1\end{array}\right]=\left[\begin{array}{cc}9& 16\\ 0& -y\end{array}\right]$

$\left[\begin{array}{cc}9& 4x\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}9& 16\\ 0& -y\end{array}\right]$
By equality of matrices,
$4x=16$ and $1=-y$
$x=4$ and $y=-1$