Chapter 11, Problem 14CT

To determine

The solution of the system of linear equations using matrices, $\{\begin{array}{l}x+\frac{1}{4}y=7\\ 8x+2y=56\end{array}$.

Answer to Problem 14CT

Solution:

The solution of the system $\{\begin{array}{l}x+\frac{1}{4}y=7\\ 8x+2y=56\end{array}$ is $\left\{\left(x,y\right)|x\text{is}\text{a}\text{real}\text{number},y=28-4x\right\}$.

Explanation of Solution

Given information:

The system of linear equations, $\{\begin{array}{l}x+\frac{1}{4}y=7\\ 8x+2y=56\end{array}$

Explanation:

The augmented matrix of the system is,

$\left[\begin{array}{ccc}1& \frac{1}{4}& 7\\ 8& 2& 56\end{array}\right]$

Perform the row operation $R_2=-8R_1+r_2$,

$\Rightarrow \left[\begin{array}{ccc}1& \frac{1}{4}& 7\\ 0& 0& 0\end{array}\right]$

The augmented matrix is in row echelon form as the bottom row consists zeros.

The system consists of one equation $x+\frac{1}{4}y=7$ in two variables. The system is dependent and therefore has infinitely many solutions.

To solve $x+\frac{1}{4}y=7$ for $y$,

$\Rightarrow y=28-4x$

Any order pair satisfying $\left\{\left(x,y\right)|x\text{is}\text{a}\text{real}\text{number},y=28-4x\right\}$.

Hence, the solution of system of equation $\{\begin{array}{l}x+\frac{1}{4}y=7\\ 8x+2y=56\end{array}$ is $\left\{\left(x,y\right)|x\text{is}\text{a}\text{real}\text{number},y=28-4x\right\}$.