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

$\left\{\begin{array}{l}2y+3z=-1\\ 5x+3y=-6\\ 7x+z=1\end{array}\right.$

The given system of equations is:

$\left\{\begin{array}{l}2y+3z=-1\\ 5x+3y=-6\\ 7x+z=1\end{array}\right.$

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

$\left[\begin{array}{rrrr}0& 2& 3& -1\\ 5& 3& 0& -6\\ 7& 0& 1& 1\end{array}\right]$

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 $R_3\to 7R_2-5R_3$:

$\left[\begin{array}{rrrr}0& 2& 3& -1\\ 5& 3& 0& -6\\ 7& 0& 1& 1\end{array}\right]\stackrel{R_3\to 7R_2-5R_3}{⟶}\left[\begin{array}{cccc}0& 2& 3& -1\\ 5& 3& 0& -6\\ 0& 21& -5& -47\end{array}\right]$

Performing row operation$R_3\to 5R_1+3R_3$:

$\left[\begin{array}{cccc}0& 2& 3& -1\\ 5& 3& 0& -6\\ 0& 21& -5& -47\end{array}\right]\stackrel{R_3\to 5R_1+3R_3}{⟶}\left[\begin{array}{rrrl}0& 2& 3& -1\\ 5& 3& 0& -6\\ 0& 73& 0& -146\end{array}\right]$

Dividing row 3 by $73$, we get:

$\left[\begin{array}{rrrl}0& 2& 3& -1\\ 5& 3& 0& -6\\ 0& 73& 0& -146\end{array}\right]\stackrel{R_3\to \frac{1}{73}{R}_3}{\to }\left[\begin{array}{llll}0& 2& 3& -1\\ 5& 3& 0& -6\\ 0& 1& 0& -2\end{array}\right]$

The echelon form of the matrix is:

$\left[\begin{array}{llll}0& 2& 3& -1\\ 5& 3& 0& -6\\ 0& 1& 0& -2\end{array}\right]$

The system of equation by the above matrix is:

$$\begin{gathered} 2y+3z=-1 \\ 5x+3y=-6 \\ y=-2 \end{gathered}$$

Substituting $y=-2$in the equation

$5x+3y=-6$:

$$\begin{gathered} 5x+3(-2)=-6 \\ 5x-6=-6 \\ 5x=0 \\ x=0 \end{gathered}$$

Again substituting $y=-2$in the equation $2y+3z=-1$:

$$\begin{gathered} 2(-2)+3z=-1 \\ -4+3z=-1 \\ 3z=-1+4 \\ 3z=3 \\ z=1 \end{gathered}$$

Thus, the solution of the system of equation is given as:

$$\begin{gathered} x=0 \\ y=-2 \\ z=1 \end{gathered}$$

The solution of the system in the ordered triad is$\left(0,-2,1\right)$

Substitute the ordered triad $\left(0,-2,1\right)$in the equation $2y+3z=-1$:

$$\begin{gathered} 2y+3z=-1 \\ 2(-2)+3\left(1\right)=-1 \\ -4+3=-1 \\ -1=-1 \end{gathered}$$

This is true

Substituting the ordered triad $\left(-2,5,2\right)$in the equation $5x+3y=-6$:

$$\begin{gathered} 5x+3y=-6 \\ 5\left(0\right)+3(-2)=-6 \\ -6=-6 \end{gathered}$$

This is true.

Substituting the ordered triad $\left(0,-2,1\right)$in the equation $7x+z=1$:

$$\begin{gathered} 7x+z=1 \\ 7\left(0\right)+1=1 \\ 1=1 \end{gathered}$$

This is true.

Thus the ordered triad $\left(0,-2,1\right)$is the solution of the system of equations.