Chapter C, Problem 1DE

Solve using matrices:

$\begin{array}{l}5x-2y=-44,\\ 2x+5y=-6.\end{array}$

To determine

To calculate: The solution of the linear equation by using matrices:

$\begin{array}{c}5x-2y=-44\\ 2x+5y=-6\end{array}$

Answer to Problem 1DE

Solution:

The solution of the linear equation $\begin{array}{c}5x-2y=-44\\ 2x+5y=-6\end{array}$ is $\underset{\_}{\left(-8,2\right)}$.

Explanation of Solution

Given information:

The set of linear equation:

$\begin{array}{c}5x-2y=-44\\ 2x+5y=-6\end{array}$

Formula Used:

Steps to convert the linear equation of two variable in matrix form:

Step 1: A set of linear equation should be written in the form of matrix where x coefficients should be written in first column, y coefficients should be written in second column and constants must be written in third column. Separate constants from x and y coefficients by dash line.

Step 2: The main goal is to transform the above matrix into below form:

$\left[\begin{array}{ccc}a& b& c\\ 0& d& e\end{array}\right]$

Step 3: verified the results.

Calculation:

Consider the linear equation is,

$\begin{array}{c}5x-2y=-44\\ 2x+5y=-6\end{array}$

The above linear equation in matrix form can be written as,

$\left[\begin{array}{ccc}5& -2& -44\\ 2& 5& -6\end{array}\right]$

Multiply row 1 by $\frac{1}{5}$,

$\begin{array}{c}\text{New row 1}=\frac{1}{5}\left(\text{Row 1 from step above}\right)\\ =\frac{1}{5}\left(\begin{array}{ccc}5& -2& -44\end{array}\right)\\ =\left(\begin{array}{ccc}1& -\frac{2}{5}& -\frac{44}{5}\end{array}\right)\end{array}$

The set of linear equations become,

$\left[\begin{array}{ccc}1& -\frac{2}{5}& -\frac{44}{5}\\ 2& 5& -6\end{array}\right]$

Add $-2$ times the row 1 to the row 2,

$\begin{array}{c}\text{New row 2}=-2\left(\begin{array}{ccc}1& -\frac{2}{5}& -\frac{44}{5}\end{array}\right)+\left(\begin{array}{ccc}2& 5& -6\end{array}\right)\\ =\left(\begin{array}{ccc}-2& \frac{4}{5}& \frac{88}{5}\end{array}\right)+\left(\begin{array}{ccc}2& 5& -6\end{array}\right)\\ =\left(\begin{array}{ccc}0& \frac{29}{5}& \frac{58}{5}\end{array}\right)\end{array}$

The set of linear equations become,

$\left[\begin{array}{ccc}1& -\frac{2}{5}& -\frac{44}{5}\\ 0& \frac{29}{5}& \frac{58}{5}\end{array}\right]$

Multiply row 2 by $\frac{5}{29}$,

$\begin{array}{c}\text{New row 2}=\frac{5}{29}\left(\text{Row 2 from step above}\right)\\ =\frac{5}{29}\left(\begin{array}{ccc}0& \frac{29}{5}& \frac{58}{5}\end{array}\right)\\ =\left(\begin{array}{ccc}0& 1& 2\end{array}\right)\end{array}$

The set of linear equations become,

$\left[\begin{array}{ccc}1& -\frac{2}{5}& -\frac{44}{5}\\ 0& 1& 2\end{array}\right]$

Add $\frac{2}{5}$ times the row 2 to the row 1,

$\begin{array}{c}\text{New row 1}=\frac{2}{5}\left(\begin{array}{ccc}0& 1& 2\end{array}\right)+\left(\begin{array}{ccc}1& -\frac{2}{5}& -\frac{44}{5}\end{array}\right)\\ =\left(\begin{array}{ccc}0& \frac{2}{5}& \frac{4}{5}\end{array}\right)+\left(\begin{array}{ccc}1& -\frac{2}{5}& -\frac{44}{5}\end{array}\right)\\ =\left(\begin{array}{ccc}1& 0& -8\end{array}\right)\end{array}$

The set of linear equations become,

$\left[\begin{array}{ccc}1& 0& -8\\ 0& 1& 2\end{array}\right]$

Reinsert the variables to have a set of linear equations,

$\begin{array}{c}x=-8\\ y=2\end{array}$

The solution is $\left(-8,2\right)$.

To check the answer:

Now, substitute the value $x=-8,y=2$ in the equation $5x-2y=-44$

Therefore,

$\begin{array}{r}5x-2y=-44\\ 5\left(-8\right)-2\left(2\right)\stackrel{?}{=}-44\\ -40-4\stackrel{?}{=}-44\\ -44\stackrel{?}{=}-44\end{array}$

Solution verified.

Hence, the solution of the linear equation $\begin{array}{c}5x-2y=-44\\ 2x+5y=-6\end{array}$ is $\left(-8,2\right)$.