Question
Evaluate the following determinant:
$$ det \left( \begin{matrix}1&-2&-2\\-3&-1&1\\1&-4&-3\end{matrix} \right) $$Answer
$$ det \left( \begin{matrix}1&-2&-2\\-3&-1&1\\1&-4&-3\end{matrix} \right) = -3 $$
Explanation
To find the 3x3 determinant, we can use the Rule of Sarrus. .
$$ \begin{aligned} det \left( \begin{matrix}1&-2&-2\\-3&-1&1\\1&-4&-3\end{matrix} \right) &= \left[
\begin{array}{ccc|cc} \cssId{i00}{1} & \cssId{i01}{-2} & \cssId{i02}{-2} & \cssId{i03}{1} & \cssId{i04}{-2} \\
\cssId{i10}{-3} & \cssId{i11}{-1} & \cssId{i12}{1} & \cssId{i13}{-3} & \cssId{i14}{-1} \\
\cssId{i20}{1} & \cssId{i21}{-4} & \cssId{i22}{-3} & \cssId{i23}{1} & \cssId{i24}{-4}
\end{array} \right | = \\\\
&= \cssId{u0}{1 \cdot \left(-1\right) \cdot \left(-3\right)} \cssId{u1}{+\left(-2\right) \cdot 1 \cdot 1} \cssId{u2}{+\left(-2\right) \cdot \left(-3\right) \cdot \left(-4\right)} \cssId{u3}{-1 \cdot \left(-1\right) \cdot \left(-2\right)} \cssId{u4}{-\left(-4\right) \cdot 1 \cdot 1} \cssId{u5}{-\left(-3\right) \cdot \left(-3\right) \cdot \left(-2\right)} = \\\\
&= 3 + \left( -2\right) + \left( -24\right) - 2 - \left( -4\right) - \left( -18\right) = -3
\end{aligned}
$$
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