Question
Evaluate the following determinant:
$$ det \left( \begin{matrix}1&2&3\\2&-3&-1\\-3&4&5\end{matrix} \right) $$Answer
$$ det \left( \begin{matrix}1&2&3\\2&-3&-1\\-3&4&5\end{matrix} \right) = -28 $$
Explanation
To find the 3x3 determinant, we can use the Rule of Sarrus. .
$$ \begin{aligned} det \left( \begin{matrix}1&2&3\\2&-3&-1\\-3&4&5\end{matrix} \right) &= \left[
\begin{array}{ccc|cc} \cssId{i00}{1} & \cssId{i01}{2} & \cssId{i02}{3} & \cssId{i03}{1} & \cssId{i04}{2} \\
\cssId{i10}{2} & \cssId{i11}{-3} & \cssId{i12}{-1} & \cssId{i13}{2} & \cssId{i14}{-3} \\
\cssId{i20}{-3} & \cssId{i21}{4} & \cssId{i22}{5} & \cssId{i23}{-3} & \cssId{i24}{4}
\end{array} \right | = \\\\
&= \cssId{u0}{1 \cdot \left(-3\right) \cdot 5} \cssId{u1}{+2 \cdot \left(-1\right) \cdot \left(-3\right)} \cssId{u2}{+3 \cdot 2 \cdot 4} \cssId{u3}{-\left(-3\right) \cdot \left(-3\right) \cdot 3} \cssId{u4}{-4 \cdot \left(-1\right) \cdot 1} \cssId{u5}{-5 \cdot 2 \cdot 2} = \\\\
&= -15 + 6 + 24 - 27 - \left( -4\right) - 20 = -28
\end{aligned}
$$
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