In matrix form, the system is
[tex]\begin{bmatrix}1&2&-1\\1&2&1\\-1&-1&2\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=\begin{bmatrix}-4\\2\\6\end{bmatrix}[/tex]
Solving this "using matrices" is a bit ambiguous but brings to mind two standard methods.
Compute the inverse of the coefficient matrix using the formula
[tex]\mathbf A^{-1}=\dfrac1{\det\mathbf A}\mathbf C^\top[/tex]
where [tex]\mathbf A[/tex] is the coefficient matrix, [tex]\det\mathbf A[/tex] is its determinant, [tex]\mathbf C[/tex] is the cofactor matrix, and [tex]\top[/tex] denotes the matrix transpose.
We compute the determinant by a Laplace expansion along the first column:
[tex]\det\mathbf A=\begin{vmatrix}1&2&-1\\1&2&1\\-1&-1&2\end{vmatrix}[/tex]
[tex]\det\mathbf A=\begin{vmatrix}2&1\\-1&2\end{vmatrix}-\begin{vmatrix}2&-1\\-1&2\end{vmatrix}-\begin{vmatrix}2&-1\\2&1\end{vmatrix}[/tex]
[tex]\det\mathbf A=5-3-4=-2[/tex]
The cofactor matrix is
[tex]\mathbf C=\begin{bmatrix}5&-3&1\\-3&1&-1\\4&-2&0\end{bmatrix}\implies\mathbf C^\top=\begin{bmatrix}5&-3&4\\-3&1&-2\\1&-1&0\end{bmatrix}[/tex]
which makes the inverse
[tex]\mathbf A^{-1}=\begin{bmatrix}-5/2&3/2&-2\\3/2&-1/2&1\\-1/2&1/2&0\end{bmatrix}[/tex]
Finally,
[tex]\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=\mathbf A^{-1}\begin{bmatrix}-4\\2\\6\end{bmatrix}\implies\boxed{x_1=1,x_2=-1,x_3=3}[/tex]
Take the augmented matrix
[tex]\begin{bmatrix}1&2&-1&-4\\1&2&1&2\\-1&-1&2&6\end{bmatrix}[/tex]
Subtract row 1 from row 2, and -(row 1) from row 3:
[tex]\begin{bmatrix}1&2&-1&-4\\0&0&2&6\\0&1&1&2\end{bmatrix}[/tex]
Multiply row 2 by 1/2:
[tex]\begin{bmatrix}1&2&-1&-4\\0&0&1&3\\0&1&1&2\end{bmatrix}[/tex]
The second row tells us that
[tex]x_3=3[/tex]
Then in the third row,
[tex]x_2+x_3=2\implies x_2=-1[/tex]
Then in the first row,
[tex]x_1+2x_2-x_3=-4\implies x_1=1[/tex]
