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Question
Evaluate the following determinant:
$$ det \left( \begin{matrix}6&132&-2\\-3&0&-10\\-1&0&4\end{matrix} \right) $$
Answer
$$ det \left( \begin{matrix}6&132&-2\\-3&0&-10\\-1&0&4\end{matrix} \right) = 2904 $$
Explanation
To find the 3x3 determinant, we can use the Rule of Sarrus. .
$$ \begin{aligned} det \left( \begin{matrix}6&132&-2\\-3&0&-10\\-1&0&4\end{matrix} \right) &= \left[
\begin{array}{ccc|cc} \cssId{i00}{6} & \cssId{i01}{132} & \cssId{i02}{-2} & \cssId{i03}{6} & \cssId{i04}{132} \\
\cssId{i10}{-3} & \cssId{i11}{0} & \cssId{i12}{-10} & \cssId{i13}{-3} & \cssId{i14}{0} \\
\cssId{i20}{-1} & \cssId{i21}{0} & \cssId{i22}{4} & \cssId{i23}{-1} & \cssId{i24}{0}
\end{array} \right | = \\\\
&= \cssId{u0}{6 \cdot 0 \cdot 4} \cssId{u1}{+132 \cdot \left(-10\right) \cdot \left(-1\right)} \cssId{u2}{+\left(-2\right) \cdot \left(-3\right) \cdot 0} \cssId{u3}{-\left(-1\right) \cdot 0 \cdot \left(-2\right)} \cssId{u4}{-0 \cdot \left(-10\right) \cdot 6} \cssId{u5}{-4 \cdot \left(-3\right) \cdot 132} = \\\\
&= 0 + 1320 + 0 - 0 - 0 - \left( -1584\right) = 2904
\end{aligned}
$$
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