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Question
Evaluate the following determinant:
$$ det \left( \begin{matrix}3&2&3\\-6&2&13\\4&0&18\end{matrix} \right) $$
Answer
$$ det \left( \begin{matrix}3&2&3\\-6&2&13\\4&0&18\end{matrix} \right) = 404 $$
Explanation
To find the 3x3 determinant, we can use the Rule of Sarrus. .
$$ \begin{aligned} det \left( \begin{matrix}3&2&3\\-6&2&13\\4&0&18\end{matrix} \right) &= \left[
\begin{array}{ccc|cc} \cssId{i00}{3} & \cssId{i01}{2} & \cssId{i02}{3} & \cssId{i03}{3} & \cssId{i04}{2} \\
\cssId{i10}{-6} & \cssId{i11}{2} & \cssId{i12}{13} & \cssId{i13}{-6} & \cssId{i14}{2} \\
\cssId{i20}{4} & \cssId{i21}{0} & \cssId{i22}{18} & \cssId{i23}{4} & \cssId{i24}{0}
\end{array} \right | = \\\\
&= \cssId{u0}{3 \cdot 2 \cdot 18} \cssId{u1}{+2 \cdot 13 \cdot 4} \cssId{u2}{+3 \cdot \left(-6\right) \cdot 0} \cssId{u3}{-4 \cdot 2 \cdot 3} \cssId{u4}{-0 \cdot 13 \cdot 3} \cssId{u5}{-18 \cdot \left(-6\right) \cdot 2} = \\\\
&= 108 + 104 + 0 - 24 - 0 - \left( -216\right) = 404
\end{aligned}
$$
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