In the following exercise, solve each system of equations using a matrix.

$\left\{\begin{array}{l}x+2y-3z=-1\\ x-3y+z=1\\ 2x-y-2z=2\end{array}\right.$

The given system of equations is:

$\left\{\begin{array}{l}x+2y-3z=-1\\ x-3y+z=1\\ 2x-y-2z=2\end{array}\right.$

The augmented matrix of the given system of equations is given as:

$\left[\begin{array}{rrrr}1& 2& -3& -1\\ 1& -3& 1& 1\\ 2& -1& -2& 2\end{array}\right]$

In the augmented matrix, the first equation gives the first row and the second equation gives the second row. The vertical line replaces the equal signs.

Perform row operation $R_2\to R_1-R_2$

role="math" localid="1644579738873" $\left[\begin{array}{rrrr}1& 2& -3& -1\\ 1& -3& 1& 1\\ 2& -1& -2& 2\end{array}\right]\stackrel{R_2\to R_1-R_2}{⟶}\left[\begin{array}{rrrr}1& 2& -3& -1\\ 0& 5& -4& -2\\ 2& -1& -2& 2\end{array}\right]$

Performing row operation $R_3\to 2R_1-R_3$:

$\left[\begin{array}{rrrr}1& 2& -3& -1\\ 0& 5& -4& -2\\ 2& -1& -2& 2\end{array}\right]\stackrel{R_3\to 2R_1-R_3}{\to }\left[\begin{array}{lllr}1& 2& -3& -1\\ 0& 5& -4& -2\\ 0& 5& -4& -4\end{array}\right]$

Perform row operation $R_3\to R_2-R_3$

$\left[\begin{array}{lllr}1& 2& -3& -1\\ 0& 5& -4& -2\\ 0& 5& -4& -4\end{array}\right]\stackrel{R_1\to R_2-R_3}{\to }\left[\begin{array}{rrrr}1& 2& -3& -1\\ 0& 5& -4& -2\\ 0& 0& 0& 2\end{array}\right]$

The echelon form of the matrix is:

$\left[\begin{array}{rrrr}1& 2& -3& -1\\ 0& 5& -4& -2\\ 0& 0& 0& 2\end{array}\right]$

The system of equation by the above matrix is:

$$\begin{gathered} x+2y-3z=-1 \\ 5y-4z=-2 \\ 0=2 \end{gathered}$$

But we know that $0\ne 2$

So,

This system of equation has no solution.