College Algebra, Chapter 7, 7.3, Section 7.3, Problem 28

Solve the system of equations $\left\{
\begin{equation}
\begin{aligned}

-7x + 4y =& 0
\\
8x -5y =& 100

\end{aligned}
\end{equation}
\right.
$, $\displaystyle \left[ \begin{array}{cc}
\displaystyle \frac{-5}{3} & \displaystyle \frac{-4}{3} \\
\displaystyle \frac{-8}{3} & \displaystyle \frac{-7}{3}
\end{array} \right] $ by converting to a matrix equation and using the inverse of the coefficient matrix $\left[ \begin{array}{cc}
-9 & 4 \\
7 & -3
\end{array} \right]$

We write the system as a matrix equation of the form $AX = B$







Using the rule for finding the inverse of a $2 \times 2$ matrix, we get

$\displaystyle A^{-1} = \left[ \begin{array}{cc}
-7 & 4 \\
8 & -5
\end{array} \right]^{-1} = \frac{1}{-7 (-5) - 4(8)} \left[ \begin{array}{cc}
-5 & -4 \\
-8 & -7
\end{array} \right] = \frac{1}{3} \left[ \begin{array}{cc}
-5 & -4 \\
-8 & -7
\end{array} \right] $

Multiplying each side of the matrix equation by this inverse matrix, we get


$
\begin{equation}
\begin{aligned}

\left[ \begin{array}{c}
x \\
y
\end{array} \right] =&
\frac{1}{3}
\left[ \begin{array}{cc}
-5 & -4 \\
-8 & -7
\end{array} \right]

\left[ \begin{array}{c}
0 \\
100
\end{array} \right] = \left[ \begin{array}{c}
\displaystyle \frac{-5}{3} \cdot 0 + \left( \frac{-4}{3} \right) \cdot 100 \\
\displaystyle \frac{-8}{3} \cdot 0 + \left( \frac{-7}{3} \right) \cdot 100
\end{array} \right]
=
\left[ \begin{array}{c}
\displaystyle \frac{-400}{3} \\
\displaystyle \frac{-700}{3}
\end{array} \right]

X =& A^{-1} \qquad B

\end{aligned}
\end{equation}
$


So $\displaystyle x = \frac{-400}{3}$ and $\displaystyle y = \frac{-700}{3}$