Chapter B, Problem 1E

Solve each system of equations by the matrix method. Round your answers to the nearest tenth when necessary.

$\begin{array}{l}2x+3y=5\\ 5x+y=19\end{array}$

To determine

To calculate: The values of x and y by the use of the matrix method for the equations, $5x+y=19$ and $2x+3y=5$.

Answer to Problem 1E

Solution:

The value of x is 4, and the value of y is $-1$.

Explanation of Solution

Given:

The system of equations is

$\begin{array}{c}2x+3y=5\\ 5x+y=19\end{array}$

Formula Used:

Reduce $2\times 2$ matrix to reduced row echelon form,

$\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

Calculation:

Consider the equations,

$2x+3y=5$ …… (1)

$5x+y=19$ …… (2)

The augmented matrix representation of the equations is,

$\left[\begin{array}{cc}2& 3\\ 5& 1\end{array}|\begin{array}{c}5\\ 19\end{array}\right]$

The above matrix is converted to therow echelon formwith the help of matrix row operations,

Apply $R_1\to R_1-3R_2$,

$\left[\begin{array}{cc}-13& 0\\ 5& 1\end{array}|\begin{array}{c}-52\\ 19\end{array}\right]$

Apply $R_2\to R_2+\frac{5R_1}{13}$,

$\left[\begin{array}{cc}-13& 0\\ 0& 1\end{array}|\begin{array}{c}-52\\ -1\end{array}\right]$

Apply $R_1\to \frac{R_1}{-13}$,

$\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}|\begin{array}{c}4\\ -1\end{array}\right]$

As the row echelon form is achieved. Thus, the variables are

$\begin{array}{c}1\left(x\right)+0\left(y\right)=4\\ x=4\end{array}$

And,

$\begin{array}{c}0\left(x\right)+1\left(y\right)=-1\\ y=-1\end{array}$

Hence, the value of x and y are 4 and $-1$, respectively.