<s>[INST] How would I solve the following? $$M = \begin{pmatrix}1&-4\\ -6&-6\end{pmatrix}$$
Find $c_1$ and $c_2$ such that $M^2 + c_1M +c_2I_2 = 0$, where $I_2$ is the identity $2 \times 2$ matrix.
The matrix above is $M$.
I got stuck after I merged the $c_1$ and $c_2$ into their respective matrices and calculated the square of $M$.
 [/INST] $M^2=\left(
\begin{array}{cc}
 25 & 20 \\
 30 & 60 \\
\end{array}
\right)$
$M^2 + c_1M +c_2I_2 = 0$ becomes
$\left(
\begin{array}{cc}
 25 & 20 \\
 30 & 60 \\
\end{array}
\right)+\left(
\begin{array}{cc}
 c_1 & -4 c_1 \\
 -6 c_1 & -6 c_1 \\
\end{array}
\right)+\left(
\begin{array}{cc}
 c_2 & 0 \\
 0 & c_2 \\
\end{array}
\right)=\left(
\begin{array}{cc}
 0 & 0 \\
 0 & 0 \\
\end{array}
\right)$
that is
$\left(
\begin{array}{rr}
 c_1+c_2+25=0 & 20-4 c_1=0 \\
 30-6 c_1=0 & -6 c_1+c_2+60=0 \\
\end{array}
\right)$
from $20-4c_1=0$ we get $c_1=5$ and substituting both in $c_1+c_2+25=0$ and in $-6c_1+c_2+60=0$ we get $c_2=-30$
Solution is $c_1=5;\;c_2=-30$
</s>