Question
Evaluate the following determinant:
$$ det \left( \begin{matrix}2&3&-1\\-4&-5&2\\6&1&3\end{matrix} \right) $$Answer
$$ det \left( \begin{matrix}2&3&-1\\-4&-5&2\\6&1&3\end{matrix} \right) = 12 $$
Explanation
To find the 3x3 determinant, we can use the Rule of Sarrus. .
$$ \begin{aligned} det \left( \begin{matrix}2&3&-1\\-4&-5&2\\6&1&3\end{matrix} \right) &= \left[
\begin{array}{ccc|cc} \cssId{i00}{2} & \cssId{i01}{3} & \cssId{i02}{-1} & \cssId{i03}{2} & \cssId{i04}{3} \\
\cssId{i10}{-4} & \cssId{i11}{-5} & \cssId{i12}{2} & \cssId{i13}{-4} & \cssId{i14}{-5} \\
\cssId{i20}{6} & \cssId{i21}{1} & \cssId{i22}{3} & \cssId{i23}{6} & \cssId{i24}{1}
\end{array} \right | = \\\\
&= \cssId{u0}{2 \cdot \left(-5\right) \cdot 3} \cssId{u1}{+3 \cdot 2 \cdot 6} \cssId{u2}{+\left(-1\right) \cdot \left(-4\right) \cdot 1} \cssId{u3}{-6 \cdot \left(-5\right) \cdot \left(-1\right)} \cssId{u4}{-1 \cdot 2 \cdot 2} \cssId{u5}{-3 \cdot \left(-4\right) \cdot 3} = \\\\
&= -30 + 36 + 4 - 30 - 4 - \left( -36\right) = 12
\end{aligned}
$$
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