Solution:
$D = \quad
\begin{vmatrix}
1 & 1 & 1\\
1 & -1 & 1\\
1 & 1 & -1
\end{vmatrix}$
=1(1-1)-1(-1-1)+1(1+1)= 4
$D_{V1} = \quad
\begin{vmatrix}
9 & 1 & 1\\
3 & -1 & 1\\
1 & 1 & -1
\end{vmatrix}$
= 9(1-1)-1(-3-1)+1(3+1) = 8
$V_{1} = \frac{D_{V1}}{D} = \frac{8}{4} = 2$
$D_{V2} = \quad
\begin{vmatrix}
1 & 9 & 1\\
1 & 3 & 1\\
1 & 1 & -1
\end{vmatrix}$
= 1(-3-1)-9(-1-1)+1(1-3) = 12
$V_{2} = \frac{D_{V2}}{D} = \frac{12}{4} = 3$
$D_{V3} = \quad
\begin{vmatrix}
1 & 1 & 9\\
1 & -1 & 3\\
1 & 1 & -1
\end{vmatrix}$
$V_{3} = \frac{D_{V3}}{D} = \frac{16}{4} = 4$