Q 4.81 - Intermediate Algebra

Question

Solve the system of equations using a matrix: $\left\{\begin{array}{l}2x+y=7\\ x-2y=6\end{array}\right.$

Step-by-Step Solution

Answer

The solution for the system of linear equation is $\left(4,-1\right)$.

Step 1. Given information.

Consider the given system of equations,

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

Step 2. Write in augmented form.

The augmented matrix for the given system of equations is

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

Step 3. Apply row operations.

Apply $\frac{R_1}{2}\to R_1$,

$\left[\begin{array}{rrr}1& \frac{1}{2}& \frac{7}{2}\\ 0& -2& 6\end{array}\right]$

Apply $R_2-R_1\to R_2$,

$\left[\begin{array}{rrr}1& \frac{1}{2}& \frac{7}{2}\\ 0& -\frac{5}{2}& \frac{5}{2}\end{array}\right]$

Apply $\left(-\frac{2}{5}\right)\times R_2\to R_2$,

$\left[\begin{array}{rrr}1& \frac{1}{2}& \frac{7}{2}\\ 0& 1& -1\end{array}\right]$

Now, the matrix is in row-echelon form.

Step 4. Write in system of equations.

Writing the corresponding system of equations,

$$\begin{gathered} x+\frac{1}{2}y=\frac{7}{2} ...... \left(i\right) \\ y=-1 ...... \left(ii\right) \end{gathered}$$

Substitute $y$ in equation (i),

$$\begin{gathered} x+\frac{1}{2}\left(-1\right)=\frac{7}{2} \\ x=\frac{7}{2}+\frac{1}{2} \\ x=\frac{8}{2} \\ x=4 \end{gathered}$$

Step 5. Check the answers.

Substitute the values $x,y$ in equation (i),

$$\begin{gathered} 4+\frac{1}{2}\left(-1\right)=\frac{7}{2} \\ 4-\frac{1}{2}=\frac{7}{2} \\ \frac{7}{2}=\frac{7}{2} \end{gathered}$$

This is true.