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{\bf Question}

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

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}

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

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

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

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

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

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

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

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

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

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

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