Chapter 11.4, Problem 76AYU

In Problems 79-86, algebraically solve each system of equations using any method you wish.

$\{\begin{array}{l}2x+8y=-8\\ x+7y=-13\end{array}$

To determine

To solve: The system of equations algebraically by any method.

Answer to Problem 76AYU

Solution:

$x=8,y=-3$

Explanation of Solution

Given:

$2x+8y=-8$

$x+7y=-13$

Calculation:

The original system of equations can be written compactly as a matrix equation.

$AX=B$

Where,

$A=\left[\begin{array}{cc}2& 8\\ 1& 7\end{array}\right],X=\left[\begin{array}{c}x\\ y\end{array}\right]\text{ and }B=\left[\begin{array}{c}-8\\ -13\end{array}\right]$

To solve the for $x,\text{ and }y$ the solution of $X$ has to be found as given below:

$AX=B$

Multiply both sides by $A^{-1}$.

$A^{-1}\left(AX\right)=A^{-1}B$

By association property of matrix multiplication, the above equation can be rewritten as:

$(A^{-1}A)X=A^{-1}B$

By the definition of inverse matrix $(A^{-1}A)=I_n$.

$I_{n}X=A^{-1}B$

By the property of Identity matrix, $I_{n}X=X$.

Hence,

$X=A^{-1}B$

i.e. $\left[\begin{array}{c}x\\ y\end{array}\right]=A^{-1}B$.

Determining $A^{-1}$.

Step 1: Construction of $\left[A|I_n\right]$.

$\left[A|I_n\right]=[\begin{array}{cc}2& 8\\ 1& 7\end{array}|\begin{array}{cc}1& 0\\ 0& 1\end{array}]$

Step 2: Transform the matrix $\left[A|I_n\right]$ into reduced row echelon form.

$=[\begin{array}{cc}2& 8\\ 1& 7\end{array}|\begin{array}{cc}1& 0\\ 0& 1\end{array}]$

$\text{Interchanging Rows}$,

$\underset{R_1\leftrightarrow R_2}{\to }[\begin{array}{cc}1& 7\\ 2& 8\end{array}|\begin{array}{cc}1& 0\\ 0& 1\end{array}]$

$\underset{R_2=r_2-2r_1}{\to }[\begin{array}{cc}1& 7\\ 0& -6\end{array}|\begin{array}{cc}0& 1\\ 1& -2\end{array}]$

$\underset{R_2=\left(-\frac{1}{6}\right)r_2}{\to }[\begin{array}{cc}1& 7\\ 0& 1\end{array}|\begin{array}{cc}0& 1\\ -\frac{1}{6}& \frac{1}{3}\end{array}]$

$\underset{R_1=r_1-7r_2}{\to }[\begin{array}{cc}1& 0\\ 0& 1\end{array}|\begin{array}{cc}\frac{7}{6}& -\frac{4}{3}\\ -\frac{1}{6}& \frac{1}{3}\end{array}]$

Step 3: Separate the matrix on the right of the vertical bar which is the inverse of $A$.

$A^{-1}=\left[\begin{array}{cc}\frac{7}{6}& -\frac{4}{3}\\ -\frac{1}{6}& \frac{1}{3}\end{array}\right]$

Solving $x\text{ and }y$.

$\left[\begin{array}{c}x\\ y\end{array}\right]=A^{-1}B$

$=\left[\begin{array}{cc}\frac{7}{6}& -\frac{4}{3}\\ -\frac{1}{6}& \frac{1}{3}\end{array}\right]\left[\begin{array}{c}-8\\ -13\end{array}\right]$

$=\left[\begin{array}{c}-\frac{56}{6}+\frac{52}{3}\\ \frac{8}{6}-\frac{13}{3}\end{array}\right]$

$=\left[\begin{array}{c}8\\ -3\end{array}\right]$

The solution for the given system of equations is $x=8,y=-3$.