College Algebra, Chapter 7, 7.2, Section 7.2, Problem 38

Find $x$ and $y$ if $\displaystyle
\left[ \begin{array}{cc}
x & y \\
-y & x
\end{array} \right] -
\left[ \begin{array}{cc}
y & x \\
x & -y
\end{array} \right]
=
\left[ \begin{array}{cc}
4 & -4 \\
-6 & 6
\end{array} \right]$

Since the matrices are equal, corresponding entries must be the same. So we must have $x - y = 4, y - x = -4, -y - x = -6$ and $x + y = 6$.

We notice that the equations in first row are equal, also in the second row. Now we use

$x - y = 7$ and $x + y = 6 $ to solve for $x$ and $y$

We write it as the system of equations


$
\left\{
\begin{equation}
\begin{aligned}

x - y =& 4
&& \text{Equation 1}
\\
x + y =& 6
&& \text{Equation 2}

\end{aligned}
\end{equation}
\right.
$


We use elimination method to solve the system


$
\left\{
\begin{equation}
\begin{aligned}

x - y =& 4
&&
\\
\\
-x - y =& -6
&& \text{Equation 2 } + (-1) \times \text{ Equation 1}
\\
\\
\end{aligned}
\end{equation}
\right.
$


$
\begin{equation}
\begin{aligned}
\hline
\\
\\
-2y =& -2
&& \text{Add}
\\
\\
y =& \frac{-2}{-2}
&& \text{Divide by } -2
\\
\\
y =& 1
&&


\end{aligned}
\end{equation}
$


We back substitute $y = 1$ into the first equation and solve for $x$.


$
\begin{equation}
\begin{aligned}

x - 1 =& 4
&& \text{Back-substitute } y = 1
\\
x =& 4 + 1
&& \text{Add } 1
\\
x =& 5
&&

\end{aligned}
\end{equation}
$


So,


$
\begin{equation}
\begin{aligned}

\left[ \begin{array}{cc}
x & y \\
-y & x
\end{array} \right]

-

\left[ \begin{array}{cc}
y & x \\
x & -y
\end{array} \right]

=&

\left[ \begin{array}{cc}
4 & -4 \\
-6 & 6
\end{array} \right]

\\
\\
\\
\\

\left[ \begin{array}{cc}
5 & 1 \\
-1 & 5
\end{array} \right]

-

\left[ \begin{array}{cc}
1 & 5 \\
5 & -1
\end{array} \right]

=&

\left[ \begin{array}{cc}
4 & -4 \\
-6 & 6
\end{array} \right]

\end{aligned}
\end{equation}
$