Answer
$c_1 = -3$
$c_2 = -1$
$c_3 = 2$
Work Step by Step
Given that
\begin{equation}
\begin{pmatrix} 4 & -3 & 1 \\ 5 & -2 & 5 \\ -6 & 2 & -3 \end{pmatrix}
\begin{pmatrix} -3 \\ -1 \\ 2 \end{pmatrix}
= \begin{pmatrix} -7 \\ -3 \\ 10 \end{pmatrix}
\end{equation}
By matrix-vector multiplication, we rearrange the equation as:
\begin{equation}
-3*\begin{pmatrix} 4 \\ 5 \\ -6 \end{pmatrix} +
-1*\begin{pmatrix} -3 \\ -2 \\ 2 \end{pmatrix} +
2*\begin{pmatrix} 1 \\ 5 \\ -3 \end{pmatrix} +
= \begin{pmatrix} -7 \\ -3 \\ 10 \end{pmatrix}
\end{equation}
Therefore,
$c_1 = -3$
$c_2 = -1$
$c_3 = 2$
