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?
 A: We know that
$$M\begin{pmatrix}
       -8 & 1 \\
       1 & 5
       \end{pmatrix} = \begin{pmatrix}
       3 &-8\\
       8 & 7
       \end{pmatrix}.$$
Multiply that equation from right by
$$
\begin{pmatrix}
       -8 & 1 \\
       1 & 5
       \end{pmatrix}^{-1}=\frac{1}{-41}\begin{pmatrix}
       5 & -1 \\
       -1 & -8
       \end{pmatrix}
$$
to get $M$.
A: Note that the given input vectors form a basis of $\Bbb{R}^2$. So we can express any vector in $\Bbb{R}^2$ in terms of the given vectors such as:
$$\begin{bmatrix}a\\b\end{bmatrix}={\small\left(\frac{b-5a}{41}\right)}\begin{bmatrix}-8\\1\end{bmatrix}+{\small \left(\frac{a+8b}{41}\right)}\begin{bmatrix}1\\5\end{bmatrix}. \tag{1}$$
In particular,
$$\begin{bmatrix}1\\1\end{bmatrix}={\small\frac{-4}{41}}\begin{bmatrix}-8\\1\end{bmatrix}+{\small\frac{9}{41}}\begin{bmatrix}1\\5\end{bmatrix}.$$
Thus
\begin{align*}
M\begin{bmatrix}1\\1\end{bmatrix}& ={\small \frac{-4}{41}}\color{red}{M\begin{bmatrix}-8\\1\end{bmatrix}}+{\small \frac{9}{41}}\color{blue}{M\begin{bmatrix}1\\5\end{bmatrix}}\\
&={\small \frac{-4}{41}}\color{red}{\begin{bmatrix}3\\8\end{bmatrix}}+{\small\frac{9}{41}}\color{blue}{\begin{bmatrix}-8\\7\end{bmatrix}}\\
&=\begin{bmatrix}\frac{-84}{41}\\\frac{31}{41}\end{bmatrix}
\end{align*}
Generalization:
In fact, we can answer more generally as to what will $M$ do to any vector in $\Bbb{R}^2$. From equation (1)
\begin{align*}
M\begin{bmatrix}a\\b\end{bmatrix}&={\small\left(\frac{b-5a}{41}\right)}\color{red}{M\begin{bmatrix}-8\\1\end{bmatrix}}+{\small \left(\frac{a+8b}{41}\right)}\color{blue}{M\begin{bmatrix}1\\5\end{bmatrix}}\\
&={\small\left(\frac{b-5a}{41}\right)}\color{red}{\begin{bmatrix}3\\8\end{bmatrix}}+{\small \left(\frac{a+8b}{41}\right)}\color{blue}{\begin{bmatrix}-8\\7\end{bmatrix}}\\
&={\small \frac{1}{41}}\begin{bmatrix}-23a-61b\\-33a+64b\end{bmatrix}
\end{align*}
This also helps us find $M$ as
$$M\begin{bmatrix}a\\b\end{bmatrix}={\small \frac{1}{41}}\begin{bmatrix}-23a-61b\\-33a+64b\end{bmatrix}=\begin{bmatrix}\frac{-23}{41}&\frac{-61}{41}\\ \frac{-33}{41}&\frac{64}{41}\end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix}.$$
A: 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 $(I)$.
