College Algebra, Chapter 7, 7.3, Section 7.3, Problem 26

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

3x + 4y =& 10
\\
7x + 9y =& 20

\end{aligned}
\end{equation}
\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}
3 & 4 \\
7 & 9
\end{array} \right]^{-1} = \frac{1}{3(9) - 4(7)} \left[ \begin{array}{cc}
9 & -4 \\
-7 & 3
\end{array} \right] = -1 \left[ \begin{array}{cc}
9 & -4 \\
-7 & 3
\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] =&

\left[ \begin{array}{cc}
-9 & 4 \\
7 & -3
\end{array} \right]

\left[ \begin{array}{c}
10 \\
20
\end{array} \right] = \left[ \begin{array}{c}
9 \cdot 10 + (-4 \cdot 20) \\
-7 \cdot 10 + 3 \cdot 20
\end{array} \right]
=
\left[ \begin{array}{c}
10 \\
-10
\end{array} \right]

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

\end{aligned}
\end{equation}
$


So $x = 10$ and $y = -10$