Chapter 6, Problem 63SGR

To determine

To solve: each system of equation using an augmented matrix.

Answer to Problem 63SGR

The solution is: $x=3,y=-7$ .

Explanation of Solution

Given:

  $\begin{array}{l}x-3y=24\\ 5x-2y=29\end{array}$

Calculation:

The strategy is to perform the indicated operation.

Here, the augmented matrices are:

  $\left[\begin{array}{cc}1& -3\text{ |}\\ 5& \text{ -2 |}\end{array}\begin{array}{c}\text{ 24}\\ \text{ 29}\end{array}\right]$

Now the matrix in row reduced form will be:

  $\begin{array}{l}\left[\begin{array}{cc}1& -3\text{ |}\\ 5& \text{ -2 |}\end{array}\begin{array}{c}\text{ 24}\\ \text{ 29}\end{array}\right]\\ \stackrel{R_2-5R_1}{\to }\left[\begin{array}{cc}1& -3\text{ |}\\ 0& \text{ 13 |}\end{array}\begin{array}{c}24\\ \text{ -91}\end{array}\right]\\ \stackrel{\frac{1}{13}{R}_2}{\to }\left[\begin{array}{cc}1& -3\text{ |}\\ 0& \text{ 1 |}\end{array}\begin{array}{c}24\\ \text{ -7}\end{array}\right]\\ \stackrel{R_1+3R_2}{\to }\left[\begin{array}{cc}1& \text{0 |}\\ 0& \text{ 1 |}\end{array}\begin{array}{c}\text{ 3}\\ \text{ -7}\end{array}\right]\end{array}$

Conclusion:

Hence, the solution is: $x=3,y=-7$ .