Question
Evaluate the following determinant:
$$ det \left( \begin{matrix}3&2&10\\-6&2&5\\4&0&1\end{matrix} \right) $$Answer
$$ det \left( \begin{matrix}3&2&10\\-6&2&5\\4&0&1\end{matrix} \right) = -22 $$
Explanation
To find the 3x3 determinant, we can use the Rule of Sarrus. .
$$ \begin{aligned} det \left( \begin{matrix}3&2&10\\-6&2&5\\4&0&1\end{matrix} \right) &= \left[
\begin{array}{ccc|cc} \cssId{i00}{3} & \cssId{i01}{2} & \cssId{i02}{10} & \cssId{i03}{3} & \cssId{i04}{2} \\
\cssId{i10}{-6} & \cssId{i11}{2} & \cssId{i12}{5} & \cssId{i13}{-6} & \cssId{i14}{2} \\
\cssId{i20}{4} & \cssId{i21}{0} & \cssId{i22}{1} & \cssId{i23}{4} & \cssId{i24}{0}
\end{array} \right | = \\\\
&= \cssId{u0}{3 \cdot 2 \cdot 1} \cssId{u1}{+2 \cdot 5 \cdot 4} \cssId{u2}{+10 \cdot \left(-6\right) \cdot 0} \cssId{u3}{-4 \cdot 2 \cdot 10} \cssId{u4}{-0 \cdot 5 \cdot 3} \cssId{u5}{-1 \cdot \left(-6\right) \cdot 2} = \\\\
&= 6 + 40 + 0 - 80 - 0 - \left( -12\right) = -22
\end{aligned}
$$
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