Question

Solve the system of equations using a matrix.

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

Short answer

The solution of system of equation is (4,-1).

Step-by-step solution

Step 1. Given information.

Given linear equations:

$$\begin{gathered} 2x+y=7 \\ x-2y=6 \end{gathered}$$

Step 2. Write the concept.

Let's write the equation in augmented form.

The augmented matrix is as follows:

$\left[\begin{array}{ccc}2& 1& 7\\ 1& -2& 6\end{array}\right]$

Calculation

Solve the augmented matrix

$\left[\begin{array}{ccc}2& 1& 7\\ 1& -2& 6\end{array}\right]$

Let us interchange row 1 and row 2.

$\left[\begin{array}{ccc}2& 1& 7\\ 1& -2& 6\end{array}\right]\stackrel{R_1\leftrightarrow R_2}{\to }\left[\begin{array}{ccc}1& -2& 6\\ 2& 1& 7\end{array}\right]$

Multiply row 1 by $-2$and then add it to row 2 to make it zero.

$$\begin{gathered} \stackrel{-2R_1+R_2}{\to }\left[\begin{array}{ccc}1& -2& 6\\ -2\left(1\right)+2& (-2)(-2)+1& -2\left(6\right)+7\end{array}\right] \\ \stackrel{-2R_1+R_2}{\to }\left[\begin{array}{ccc}1& -2& 6\\ 0& 5& -5\end{array}\right] \\ \stackrel{\frac{1}{5}{R}_2}{\to }\left[\begin{array}{ccc}1& -2& 6\\ 0& 1& -1\end{array}\right] \end{gathered}$$

Step 4. Echelon form

Matrix in echelon form is

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

Therefore the corresponding system of equation is

$$\begin{gathered} x-2y=6 \\ y=-1 \end{gathered}$$

Step 5. Substitution method for remaining values.

Put $y=-1$in the equation$$\begin{gathered} x-2y=6 \\ x+2=6 \\ x=6-2 \\ x=4 \end{gathered}$$