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(A) First, calculate the value of $A$:$A = -1(1 - (-12)) - 7(2 - (-9)) + 0 = -1(13) - 7(11) = -13 - 77 = -90$.Now, let $B = \left|\begin{array}{ccc}13

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$A^{2}$

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(A) First, calculate the value of $A$:$A = -1(1 - (-12)) - 7(2 - (-9)) + 0 = -1(13) - 7(11) = -13 - 77 = -90$.Now, let $B = \left|\begin{array}{ccc}13 & -11 & 5 \ -7 & -1 & 25 \ -21 & -3 & -15\end{array}ight|$.Taking $5$ common from $C_3$ and $3$ common from $R_3$:$B = 5 	imes 3 \left|\begin{array}{ccc}13 & -11 & 1 \ -7 & -1 & 5 \ -7 & -1 & -1\end{array}ight| = 15 \left|\begin{array}{ccc}13 & -11 & 1 \ -7 & -1 & 5 \ 0 & 0 & -6\end{array}ight|$ (using $R_3 ightarrow R_3 - R_2$).Expanding along $R_3$:$B = 15 	imes (-6) \left|\begin{array}{cc}13 & -11 \ -7 & -1\end{array}ight| = -90 	imes (-13 - 77) = -90 	imes (-90) = 8100$.Since $A = -90$, then $A^2 = (-90)^2 = 8100$.Therefore, $B = A^2$.

If $\left|\begin{array}{ccc}-1 & 7 & 0 \\ 2 & 1 & -3 \\ 3 & 4 & 1\end{array}\right|=A$, then the value of $\left|\begin{array}{ccc}13 & -11 & 5 \\ -7 & -1 & 25 \\ -21 & -3 & -15\end{array}\right|$ is:

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