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

$\left\{\begin{array}{l}2x-y+3z=-3\\ -x+2y-z=10\\ x+y+z=5\end{array}\right.$

The given system of equations is:

$\left\{\begin{array}{l}2x-y+3z=-3\\ -x+2y-z=10\\ x+y+z=5\end{array}\right.$

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

$\left[\begin{array}{rrrr}2& -1& 3& -3\\ -1& 2& -1& 10\\ 1& 1& 1& 5\end{array}\right]$

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.

Interchanging the row $1$and $3$, we get:

$\left[\begin{array}{rrrr}2& -1& 3& -3\\ -1& 2& -1& 10\\ 1& 1& 1& 5\end{array}\right]\stackrel{R_1\leftrightarrow R_3}{⟶}\left[\begin{array}{rrrr}1& 1& 1& 5\\ -1& 2& -1& 10\\ 2& -1& 3& -3\end{array}\right]$Add row $1$and row $2$, we get:

$\left[\begin{array}{rrrr}1& 1& 1& 5\\ -1& 2& -1& 10\\ 2& -1& 3& -3\end{array}\right]\stackrel{R_1+R_2}{⟶}\left[\begin{array}{rrrr}1& 1& 1& 5\\ 0& 3& 0& 15\\ 2& -1& 3& -3\end{array}\right]$

Multiply row $1$by $-2$and then add to row $3$, we get:

$\left[\begin{array}{rrrr}1& 1& 1& 5\\ 0& 3& 0& 15\\ 2& -1& 3& -3\end{array}\right]\stackrel{-2R_1+R_3}{\to }\left[\begin{array}{rrrr}1& 1& 1& 5\\ 0& 3& 0& 15\\ 2& -3& 1& -13\end{array}\right]$

Dividing row $2$by $3$, we get:

$\left[\begin{array}{rrrr}1& 1& 1& 5\\ 0& 3& 0& 15\\ 2& -3& 1& -13\end{array}\right]\stackrel{\frac{R_2}{3}}{\to }\left[\begin{array}{rrrr}1& 1& 1& 5\\ 0& 1& 0& 5\\ 0& -3& \text{'}1& -13\end{array}\right]$

The echelon form of the matrix is:

$\left[\begin{array}{rrrr}1& 1& 1& 5\\ 0& 1& 0& 5\\ 0& -3& 1& -13\end{array}\right]$

Equation from the matrix is given as:

$$\begin{gathered} x+y+z=5 \\ y=5 \\ -3y+z=-13 \end{gathered}$$

Substituting the value $y=5$in the equation $-3y+z=-13$:

$$\begin{gathered} -3\left(5\right)+z=-13 \\ -15+z=-13 \\ z=-13+15 \\ z=2 \end{gathered}$$

Put $$\begin{gathered} y=5 \\ z=2 \end{gathered}$$in the equation

$x+y+z=5$:

$$\begin{gathered} x+5+2=5 \\ x+7=5 \\ x=5-7 \\ x=-2 \end{gathered}$$

Thus, the solution to the given system of equations is :

$$\begin{gathered} x=-2 \\ y=5 \\ z=2 \end{gathered}$$

The solution of the system in the ordered triad is$(-2,5,2)$

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

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

This is true

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

$$\begin{gathered} -x+2y-z=10 \\ -(-2)+2\left(5\right)-2=10 \\ 10=10 \end{gathered}$$

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

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

This is also true.

Thus the ordered triad$\left(-2,5,2\right)$is the solution of the system of equations.