Question
Evaluate the following determinant:
$$ det \left( \begin{matrix}1&2&1\\3&-4&-2\\5&3&5\end{matrix} \right) $$Answer
$$ det \left( \begin{matrix}1&2&1\\3&-4&-2\\5&3&5\end{matrix} \right) = -35 $$
Explanation
To find the 3x3 determinant, we can use the Rule of Sarrus. .
$$ \begin{aligned} det \left( \begin{matrix}1&2&1\\3&-4&-2\\5&3&5\end{matrix} \right) &= \left[
\begin{array}{ccc|cc} \cssId{i00}{1} & \cssId{i01}{2} & \cssId{i02}{1} & \cssId{i03}{1} & \cssId{i04}{2} \\
\cssId{i10}{3} & \cssId{i11}{-4} & \cssId{i12}{-2} & \cssId{i13}{3} & \cssId{i14}{-4} \\
\cssId{i20}{5} & \cssId{i21}{3} & \cssId{i22}{5} & \cssId{i23}{5} & \cssId{i24}{3}
\end{array} \right | = \\\\
&= \cssId{u0}{1 \cdot \left(-4\right) \cdot 5} \cssId{u1}{+2 \cdot \left(-2\right) \cdot 5} \cssId{u2}{+1 \cdot 3 \cdot 3} \cssId{u3}{-5 \cdot \left(-4\right) \cdot 1} \cssId{u4}{-3 \cdot \left(-2\right) \cdot 1} \cssId{u5}{-5 \cdot 3 \cdot 2} = \\\\
&= -20 + \left( -20\right) + 9 - \left( -20\right) - \left( -6\right) - 30 = -35
\end{aligned}
$$
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