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

$\left\{\begin{array}{l}x+y-3z=-1\\ y-z=0\\ -x+2y=1\end{array}\right.$

The given system of equations is:

$\left\{\begin{array}{l}x+y-3z=-1\\ y-z=0\\ -x+2y=1\end{array}\right.$

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

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

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 $R_3\to R_1+R_3$

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

Perform row operation $R_3\to \frac{1}{3}{R}_3$

$\left[\begin{array}{rrrr}1& 1& -3& -1\\ 0& 1& -1& 0\\ 0& 3& -3& 0\end{array}\right]\stackrel{R_3\to \frac{1}{3}{R}_3}{\to }\left[\begin{array}{llrr}1& 1& -3& -1\\ 0& 1& -1& 0\\ 0& 1& -1& 0\end{array}\right]$

Perform a row operation $R_3\to R_2-R_3$

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

The echelon form of the matrix is:

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

The system of equation by the above matrix is:

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

So, this system has infinitely many solutions. So we this system of equation in terms of $z$.

Expressing $y$in terms of $z$, we get:

localid="1644581444721" $$\begin{gathered} -y-z=0 \\ y=-z \end{gathered}$$

Now, substituting $y=-z$in the equation$x+y-3z=-1$, we get

localid="1644581617873" $$\begin{gathered} x+y-3z=-1 \\ x-z-3z=-1 \\ x-4z=-1 \\ x=-1+4z \end{gathered}$$

So the solution of the given system of equation is:

$$\begin{gathered} x=-1+4z \\ y=-z \end{gathered}$$

Where $z$is any real number.