Evaluating matrix equation A $2x2$ matrix $M$ satisfies the conditions $$M\begin{bmatrix}
       -8 \\
       1
       \end{bmatrix} = \begin{bmatrix}
       3 \\
       8
       \end{bmatrix}$$
and
$$M\begin{bmatrix}
       1 \\
       5
       \end{bmatrix} = \begin{bmatrix}
       -8 \\
       7
       \end{bmatrix}.$$
Evaluate $$M\begin{bmatrix}
       1 \\
       1
       \end{bmatrix}.$$
What is the question essentially asking? Isn't this just a matrix equation?

If you find two numbers $x,y\mathbb{R}$ such that $x\begin{bmatrix}-8\\1\end{bmatrix}+y\begin{bmatrix}1\\5\end{bmatrix}=\begin{bmatrix}1\\1\end{bmatrix}$, then, make a left  multiply by $M$:
$$xM\begin{bmatrix}-8\\1\end{bmatrix}+yM\begin{bmatrix}1\\5\end{bmatrix}=M\begin{bmatrix}1\\1\end{bmatrix}$$
finally,
$$ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~x\begin{bmatrix}3\\8\end{bmatrix}+y \begin{bmatrix}-8\\7\end{bmatrix} = M\begin{bmatrix}1\\1\end{bmatrix}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\cdots(I)$$
So, all your problem is reduced to $x\begin{bmatrix}-8\\1\end{bmatrix}+y\begin{bmatrix}1\\5\end{bmatrix}=\begin{bmatrix}1\\1\end{bmatrix}$
which is equals to
$$\begin{cases} -8x + y & =1\\ x+5y  & = 1\end{cases}
$$
with solution $x=-\frac{4}{41},~y=\frac{9}{41}$. Just reemplace $x,y$ in equation $(I)$.
