How would I find the values of $x$ and $y$ for which $A^2=A$ [closed]
By Sarah Scott •
$$A=\begin{pmatrix}x&-3\\ y&3\end{pmatrix}$$
$A $ is the matrix above.
$\endgroup$ 11 Answer
$\begingroup$Calculate $A^2$
$$\begin{bmatrix}x&-3\\ y&3\end{bmatrix}\begin{bmatrix}x&-3\\ y&3\end{bmatrix}$$
$$=\begin{bmatrix}x^2-3y&-3x-9\\ xy+3y&-3y+9\end{bmatrix}$$
Then solve
$$x^2-3y=x$$
$$-3x-9=-3$$
$$xy+3y=y$$
$$-3y+9=3$$
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