Chapter 8, Problem 17RE

To determine

To solve: the system of linear equations $\{\begin{array}{c}0.3x-0.1y=-0.13\\ 0.2x-0.3y=-0.25\end{array}$

Answer to Problem 17RE

  $x=-0.2,y=0.7$

Explanation of Solution

Calculation:

First write the augmented matrix of the given system of equations.

Consider the system of equations,

  $\{\begin{array}{c}0.3x-0.1y=-0.13\\ 0.2x-0.3y=-0.25\end{array}$

Now, use the coefficients and constant terms as the matrix entries.

So, the augmented matrix for the above system is:

  $\left[\begin{array}{ccc}0.3& -0.1& -0.13\\ 0.2& -0.3& -0.25\end{array}\right]$

Now perform the row operation $R_1\to 100R_1$ and $R_2\to 100R_2$

  $\left[\begin{array}{ccc}30& -10& -13\\ 20& -30& -25\end{array}\right]$

Now perform the row operation $R_1\to \frac{R_1}{30}$

  $\left[\begin{array}{ccc}1& -\frac{1}{3}& -\frac{13}{30}\\ 20& -30& -25\end{array}\right]$

Now perform the row operation $R_2\to R_2-20R_1$

  $\left[\begin{array}{ccc}1& -\frac{1}{3}& -\frac{13}{30}\\ 0& -\frac{70}{3}& -\frac{49}{3}\end{array}\right]$

The matrix is now in row-echelon form, and the corresponding system is:

  $\{\begin{array}{c}x-\frac{1}{3}y=-\frac{13}{30}\\ -\frac{70}{3}y=-\frac{49}{3}\end{array}$

Now use back substitution to find x and y ,

  $\begin{array}{c}-\frac{70}{3}y=-\frac{49}{3}\\ y=\left(-\frac{49}{3}\right)\times \left(-\frac{3}{70}\right)\\ y=\frac{7}{10}\\ y=0.7\end{array}$

Substitute

  $y=\frac{7}{10}$ in the first equation,

  $\begin{array}{c}x-\frac{1}{3}\left(\frac{7}{10}\right)=-\frac{13}{30}\\ x-\frac{7}{30}=-\frac{13}{30}\\ x=-\frac{13}{30}+\frac{7}{30}\end{array}$

  $\begin{array}{c}x=-\frac{6}{30}\\ x=-\frac{1}{5}\\ x=-0.2\end{array}$