Question
If $A=\left[\begin{array}{ll}3 & 2 \\ 0 & 5\end{array}\right]$ and $B=\left[\begin{array}{ll}1 & 0 \\ 1 & 2\end{array}\right]$ find the each of the following and state it they are equal: $A^2-B^2$
✓
Answer
Given
$
A=\left[\begin{array}{ll}
3 & 2 \\
0 & 5
\end{array}\right]
$
and
$
\begin{aligned}
& B=\left[\begin{array}{ll}
1 & 0 \\
1 & 2
\end{array}\right] \\
& A^2-B^2
\end{aligned}
$
$
=\left[\begin{array}{ll}
3 & 2 \\
0 & 5
\end{array}\right] \times\left[\begin{array}{ll}
3 & 2 \\
0 & 5
\end{array}\right]-\left[\begin{array}{ll}
1 & 0 \\
1 & 2
\end{array}\right] \times\left[\begin{array}{ll}
1 & 0 \\
1 & 2
\end{array}\right]
$
$=\left[\begin{array}{ll}9+0 & 6+10 \\ 0+0 & 0+25\end{array}\right]-\left[\begin{array}{ll}1+0 & 0+0 \\ 1+2 & 0+4\end{array}\right]$
$=\left[\begin{array}{ll}9 & 16 \\ 0 & 25\end{array}\right]-\left[\begin{array}{ll}1 & 0 \\ 3 & 4\end{array}\right]$
$=\left[\begin{array}{ll}9-1 & 16-0 \\ 0-3 & 25-4\end{array}\right]$
$=\left[\begin{array}{cc}8 & 16 \\ -3 & 21\end{array}\right]$
We see that $(A+B)(A-B) \neq A^2-B^2$.
$
A=\left[\begin{array}{ll}
3 & 2 \\
0 & 5
\end{array}\right]
$
and
$
\begin{aligned}
& B=\left[\begin{array}{ll}
1 & 0 \\
1 & 2
\end{array}\right] \\
& A^2-B^2
\end{aligned}
$
$
=\left[\begin{array}{ll}
3 & 2 \\
0 & 5
\end{array}\right] \times\left[\begin{array}{ll}
3 & 2 \\
0 & 5
\end{array}\right]-\left[\begin{array}{ll}
1 & 0 \\
1 & 2
\end{array}\right] \times\left[\begin{array}{ll}
1 & 0 \\
1 & 2
\end{array}\right]
$
$=\left[\begin{array}{ll}9+0 & 6+10 \\ 0+0 & 0+25\end{array}\right]-\left[\begin{array}{ll}1+0 & 0+0 \\ 1+2 & 0+4\end{array}\right]$
$=\left[\begin{array}{ll}9 & 16 \\ 0 & 25\end{array}\right]-\left[\begin{array}{ll}1 & 0 \\ 3 & 4\end{array}\right]$
$=\left[\begin{array}{ll}9-1 & 16-0 \\ 0-3 & 25-4\end{array}\right]$
$=\left[\begin{array}{cc}8 & 16 \\ -3 & 21\end{array}\right]$
We see that $(A+B)(A-B) \neq A^2-B^2$.
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