Question
Evaluate the following determinant:
$$ det \left( \begin{matrix}-1&-2&2\\-3&4&1\\5&-1&4\end{matrix} \right) $$Answer
$$ det \left( \begin{matrix}-1&-2&2\\-3&4&1\\5&-1&4\end{matrix} \right) = -85 $$
Explanation
To find the 3x3 determinant, we can use the Rule of Sarrus. .
$$ \begin{aligned} det \left( \begin{matrix}-1&-2&2\\-3&4&1\\5&-1&4\end{matrix} \right) &= \left[
\begin{array}{ccc|cc} \cssId{i00}{-1} & \cssId{i01}{-2} & \cssId{i02}{2} & \cssId{i03}{-1} & \cssId{i04}{-2} \\
\cssId{i10}{-3} & \cssId{i11}{4} & \cssId{i12}{1} & \cssId{i13}{-3} & \cssId{i14}{4} \\
\cssId{i20}{5} & \cssId{i21}{-1} & \cssId{i22}{4} & \cssId{i23}{5} & \cssId{i24}{-1}
\end{array} \right | = \\\\
&= \cssId{u0}{-1 \cdot 4 \cdot 4} \cssId{u1}{+\left(-2\right) \cdot 1 \cdot 5} \cssId{u2}{+2 \cdot \left(-3\right) \cdot \left(-1\right)} \cssId{u3}{-5 \cdot 4 \cdot 2} \cssId{u4}{-\left(-1\right) \cdot 1 \cdot \left(-1\right)} \cssId{u5}{-4 \cdot \left(-3\right) \cdot \left(-2\right)} = \\\\
&= -16 + \left( -10\right) + 6 - 40 - 1 - 24 = -85
\end{aligned}
$$
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