A is a 2 x 2 matrix such that $\mathrm{A}\left[\begin{array}{c}1\\ -1\end{array}\right]=\left[\begin{array}{c}-1\\ 2\end{array}\right]\text{ and }{\mathrm{A}}^2\left[\begin{array}{c}1\\ -1\end{array}\right]=\left[\begin{array}{l}1\\ 0\end{array}\right]$. The sum of the elements of A is

$\begin{array}{l}\mathrm{A}\left[\begin{array}{c}1\\ -1\end{array}\right]=\left[\begin{array}{c}-1\\ 2\end{array}\right] \left(1\right)\\ {\mathrm{A}}^2\left[\begin{array}{c}1\\ -1\end{array}\right]=\left[\begin{array}{l}1\\ 0\end{array}\right] \left(2\right)\end{array}$

Let A be given by $\mathrm{A}=\left[\begin{array}{l}\mathrm{a}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{b}\\ \mathrm{c}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{d}\end{array}\right]$. The first equation gives

$\begin{array}{l}\mathrm{a}-\mathrm{b}=-1 \left(3\right)\\ \mathrm{c}-\mathrm{d}=2 \left(4\right)\end{array}$

For the second equation,

${\mathrm{A}}^2\left[\begin{array}{c}1\\ -1\end{array}\right]=\mathrm{A}\left(\mathrm{A}\left[\begin{array}{c}1\\ -1\end{array}\right]\right)=\mathrm{A}\left(\left[\begin{array}{c}-1\\ 2\end{array}\right]\right)=\left[\begin{array}{l}1\\ 0\end{array}\right]$

This gives

$\begin{array}{l}-\mathrm{a}+2\mathrm{b}=1 \left(5\right)\\ -\mathrm{c}+2\mathrm{d}=0 \left(6\right)\end{array}$

Eqs. (3) + (5) gives b = 0 and a = -1

Eqs. (4) + (6) gives d = 2 and c = 4

So the sum a + b + c + d = 5.