Question
Evaluate the following determinant:
$$ det \left( \begin{matrix}1&-4&2\\0&-2&-2\\1&0&5\end{matrix} \right) $$Answer
$$ det \left( \begin{matrix}1&-4&2\\0&-2&-2\\1&0&5\end{matrix} \right) = 2 $$
Explanation
To find the 3x3 determinant, we can use the Rule of Sarrus. .
$$ \begin{aligned} det \left( \begin{matrix}1&-4&2\\0&-2&-2\\1&0&5\end{matrix} \right) &= \left[
\begin{array}{ccc|cc} \cssId{i00}{1} & \cssId{i01}{-4} & \cssId{i02}{2} & \cssId{i03}{1} & \cssId{i04}{-4} \\
\cssId{i10}{0} & \cssId{i11}{-2} & \cssId{i12}{-2} & \cssId{i13}{0} & \cssId{i14}{-2} \\
\cssId{i20}{1} & \cssId{i21}{0} & \cssId{i22}{5} & \cssId{i23}{1} & \cssId{i24}{0}
\end{array} \right | = \\\\
&= \cssId{u0}{1 \cdot \left(-2\right) \cdot 5} \cssId{u1}{+\left(-4\right) \cdot \left(-2\right) \cdot 1} \cssId{u2}{+2 \cdot 0 \cdot 0} \cssId{u3}{-1 \cdot \left(-2\right) \cdot 2} \cssId{u4}{-0 \cdot \left(-2\right) \cdot 1} \cssId{u5}{-5 \cdot 0 \cdot \left(-4\right)} = \\\\
&= -10 + 8 + 0 - \left( -4\right) - 0 - 0 = 2
\end{aligned}
$$
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