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Question
Evaluate the following determinant:
$$ det \left( \begin{matrix}3&-4&8\\4&1&-2\\-6&-13&20\end{matrix} \right) $$
Answer
$$ det \left( \begin{matrix}3&-4&8\\4&1&-2\\-6&-13&20\end{matrix} \right) = -114 $$
Explanation
To find the 3x3 determinant, we can use the Rule of Sarrus. .
$$ \begin{aligned} det \left( \begin{matrix}3&-4&8\\4&1&-2\\-6&-13&20\end{matrix} \right) &= \left[
\begin{array}{ccc|cc} \cssId{i00}{3} & \cssId{i01}{-4} & \cssId{i02}{8} & \cssId{i03}{3} & \cssId{i04}{-4} \\
\cssId{i10}{4} & \cssId{i11}{1} & \cssId{i12}{-2} & \cssId{i13}{4} & \cssId{i14}{1} \\
\cssId{i20}{-6} & \cssId{i21}{-13} & \cssId{i22}{20} & \cssId{i23}{-6} & \cssId{i24}{-13}
\end{array} \right | = \\\\
&= \cssId{u0}{3 \cdot 1 \cdot 20} \cssId{u1}{+\left(-4\right) \cdot \left(-2\right) \cdot \left(-6\right)} \cssId{u2}{+8 \cdot 4 \cdot \left(-13\right)} \cssId{u3}{-\left(-6\right) \cdot 1 \cdot 8} \cssId{u4}{-\left(-13\right) \cdot \left(-2\right) \cdot 3} \cssId{u5}{-20 \cdot 4 \cdot \left(-4\right)} = \\\\
&= 60 + \left( -48\right) + \left( -416\right) - \left( -48\right) - 78 - \left( -320\right) = -114
\end{aligned}
$$
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