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

$\left\{\begin{array}{l}4x+y=6\\ x-y=4\end{array}\right.$

The given system of equations is:

$\left\{\begin{array}{l}4x+y=6\\ x-y=4\end{array}\right.$

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

$\left[\begin{array}{rrr}4& 1& 6\\ 1& -1& 4\end{array}\right]$

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 $2$, we get:

$\left[\begin{array}{rrr}4& 1& 6\\ 1& -1& 4\end{array}\right]\stackrel{R_1\leftrightarrow R_2}{\to }\left[\begin{array}{rrr}1& -1& 4\\ 4& 1& 6\end{array}\right]$

Then we multiply the row $1$by $-4$and add it to row $2$, we get:

role="math" localid="1644565421856" $$\begin{gathered} \left[\begin{array}{rrr}1& -1& 4\\ 4& 1& 6\end{array}\right]\stackrel{-4R_1+R_2}{⟶}\left[\begin{array}{lll}1& -1& 4\\ (-4·1)+4& (-4(-1\left)\right)+1& (-4·4)+6\end{array}\right] \\ =\left[\begin{array}{rrr}1& -1& 4\\ 0& 5& -10\end{array}\right] \end{gathered}$$

Dividing row $2$by $5$, we get:

$\left[\begin{array}{rrr}1& -1& 4\\ 0& 5& -10\end{array}\right]\stackrel{\frac{R_2}{5}}{\to }\left[\begin{array}{rrr}1& -1& 4\\ 0& 1& -2\end{array}\right]$

The echelon form of the matrix is:

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

The system of equation by the above matrix is:

$$\begin{gathered} x-y=4 \\ y=-2 \end{gathered}$$

Substituting the value $y=-2$in the equation

$$\begin{gathered} x-y=4 \\ x-(-2)=4 \\ x+2=4 \\ x=4-2 \\ x=2 \end{gathered}$$

So, the solution of the given system of equation is :

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

Substituting $$\begin{gathered} x=2 \\ y=-2 \end{gathered}$$in the equation, we get:

$$\begin{gathered} 4x+y=6 \\ 4\left(2\right)-2=6 \\ 6=6 \end{gathered}$$

This is true.

Substituting $$\begin{gathered} x=2 \\ y=-2 \end{gathered}$$in the equation, we get:

$$\begin{gathered} x-y=4 \\ 2-(-2)=4 \\ 4=4 \end{gathered}$$

This is true.

So the ordered pair$\left(2,2\right)$is the solution.