Express the system of linear equations $\left\{\begin{equation}
\begin{aligned}
6x-5y =& 1
\\
8x-7y =& -1
\end{aligned}
\end{equation} \right.$ as a matrix equation then solve the matrix equation by multiplying each side by the inverse of the coefficient matrix.
The equivalent matrix equation of the system is
$\left[ \begin{array}{cc}
6 & -5 \\
8 & -7
\end{array} \right] \left[ \begin{array}{c}
x \\
y
\end{array} \right] = \left[ \begin{array}{c}
1 \\
-1
\end{array} \right]$
If we let
$\displaystyle A = \left[ \begin{array}{cc}
6 & -5 \\
8 & -7
\end{array} \right] \qquad X = \left[ \begin{array}{c}
x \\
y
\end{array} \right] \qquad B = \left[ \begin{array}{c}
1 \\
-1
\end{array} \right]$
Then the matrix can be written as
$AX = B$
We solve this matrix equation by multiplying each side by the inverse of $A$
$
\begin{equation}
\begin{aligned}
AX =& B
&&
\\
A^{-1} (AX) =& A^{-1} B
&& \text{Multiply each side by } A^{-1}
\\
(A^{-1} A) X =& A^{-1} B
&& \text{Associate Property}
\\
I_3 X =& A^{-1} B
&& \text{Property of Inverses}
\\
X =& A^{-1} B
&& \text{Property of Identity Matrix}
\end{aligned}
\end{equation}
$
Solving for the inverse of $A$
$\displaystyle A^{-1} = \left[ \begin{array}{cc}
6 & -5 \\
8 & -7
\end{array} \right]^{-1} = \frac{1}{(6)(-7) - (-5)(8)} \left[ \begin{array}{cc}
-7 & 5 \\
-8 & 6
\end{array} \right] = \frac{-1}{2} \left[ \begin{array}{cc}
-7 & 5 \\
-8 & 6
\end{array} \right] = \left[ \begin{array}{cc}
\displaystyle \frac{7}{2} & \displaystyle \frac{-5}{2} \\
4 & -3
\end{array} \right]$
So
$
\begin{equation}
\begin{aligned}
X =& A^{-1} B
\\
\\
X =& \left[ \begin{array}{cc}
\displaystyle \frac{7}{2} & \displaystyle \frac{-5}{2} \\
4 & -3
\end{array} \right] \left[ \begin{array}{c}
1 \\
-1
\end{array} \right]
\\
\\
X =& \left[ \begin{array}{c}
\displaystyle \frac{7}{2} \cdot 1 + \left( \frac{-5}{2} \right) \cdot (-1) \\
4 \cdot 1 + (-3) \cdot (-1)
\end{array} \right]
=
\left[ \begin{array}{c}
6 \\
7
\end{array} \right]
\end{aligned}
\end{equation}
$
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