Question

In Problems 20–25, solve each system of equations using matrices. If the system has no solution, say that it is inconsistent.

$\left\{\begin{array}{r}3x-2y=1\\ 10x+10y=5\end{array}\right.$

Short answer

The solution in ordered pair form is$\left(\frac{2}{5},\frac{1}{10}\right)$

Step-by-step solution

Step 1. Given information

The given system of equations is$\left\{\begin{array}{r}3x-2y=1\\ 10x+10y=5\end{array}\right.$

Step 2. Solve system of equations using matrix

Write the augmented matrix of the system,

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

Perform the operation $R_1=\frac{R_1}{3}$,

$\left[\begin{array}{lll}3& -2& 1\\ 2& 2& 1\end{array}\right]\to \left[\begin{array}{ccc}1& -\frac{{\displaystyle 2}}{{\displaystyle 3}}& \frac{{\displaystyle 1}}{{\displaystyle 3}}\\ 2& 2& 1\end{array}\right]$

Now, perform the operation $R_2=R_2-2R_1$,

$\left[\begin{array}{ccc}1& -\frac{{\displaystyle 2}}{{\displaystyle 3}}& \frac{{\displaystyle 1}}{{\displaystyle 3}}\\ 2& 2& 1\end{array}\right]\to \left[\begin{array}{ccc}1& -\frac{{\displaystyle 2}}{{\displaystyle 3}}& \frac{{\displaystyle 1}}{{\displaystyle 3}}\\ 0& \frac{{\displaystyle 10}}{{\displaystyle 3}}& \frac{{\displaystyle 1}}{{\displaystyle 3}}\end{array}\right]$

Now, perform the operation $R_2=R_2·\frac{3}{10}$,

$\left[\begin{array}{ccc}1& -\frac{{\displaystyle 2}}{{\displaystyle 3}}& \frac{{\displaystyle 1}}{{\displaystyle 3}}\\ 0& \frac{{\displaystyle 10}}{{\displaystyle 3}}& \frac{{\displaystyle 1}}{{\displaystyle 3}}\end{array}\right]\to \left[\begin{array}{ccc}1& -\frac{{\displaystyle 2}}{{\displaystyle 3}}& \frac{{\displaystyle 1}}{{\displaystyle 3}}\\ 0& 1& \frac{{\displaystyle 1}}{{\displaystyle 10}}\end{array}\right]$

Now, write the system of equations for this obtained augmented matrix,

$\left\{\begin{array}{l}x-\frac{2}{3}y=\frac{1}{3}\\ y=\frac{1}{10}\end{array}\right.$

Now,

role="math" localid="1646998275861" $$\begin{gathered} x-\frac{2}{3}·\frac{1}{10}=\frac{1}{3} \\ x=\frac{1}{3}+\frac{1}{15} \\ x=\frac{2}{5} \end{gathered}$$

So, The solution in ordered pair form is$\left(\frac{2}{5},\frac{1}{10}\right)$