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\begin{document}


{\bf Question}

Let $$A = \left( \begin{array}{cc} -2 &amp; 3 \\ 4 &amp; -5
\end{array}\right) \hspace{.2in} b = \left( \begin{array}{cc} 1 &amp;
0 \\ 3 &amp; -2 \end{array}\right).$$

Evaluate $AB$, $BA, a, A^2 - B^2, \, (A-B)(A+B), \, (A+B)(A-B).$

${}$

Verify that $$\det AB = \det A \det B$$ and that $$\det(A-B)(A+B)
= \det(A_B)\det(A+B)$$

\vspace{.25in}

{\bf Answer}

$AB = \left( \begin{array}{cc} 7 &amp; -6 \\ -11 &amp; 10 \end{array}
\right)$

$BA = \left( \begin{array}{cc} -2 &amp; 3 \\ -14 &amp; 19  \end{array}
\right)$

$A^2 = \left( \begin{array}{cc} 16 &amp; -21 \\ -28 &amp; 37 \end{array}
\right)$

$ B^2 =  \left( \begin{array}{cc} 1 &amp; 0 \\ -3 &amp; 4 \end{array}
\right)$

$A^2 - B^2 = \left( \begin{array}{cc} 15 &amp; -21 \\ -25 &amp; 33
\end{array} \right)$

$A - B =\left( \begin{array}{cc} -3 &amp; 3 \\ 1 &amp; -3\end{array}
\right)$

$A + B = \left( \begin{array}{cc} -1 &amp; 3 \\ 7 &amp; -7 \end{array}
\right)$

$(A-B)(A+B) = \left( \begin{array}{cc} 24 &amp; -30 \\ -22 &amp; 24
\end{array} \right)$

$(A+B)(A-B) =\left( \begin{array}{cc} 6 &amp; -12 \\ -28 &amp; 42
\end{array} \right)$

$\det A = -2 \hspace{.1in} \det B = -2 \hspace{.1in} \det AB = 4 =
(-2)(-2) \hspace{.1in} $

$\det(A-B) = 6 \hspace{.1in} \det(A+B) = -14 \hspace{.1in}
\det(A-B)(A+B) = -84 = 6 \times (-14)$
\end{document}
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