Question
Evaluate the following determinant:
$$ det \left( \begin{matrix}2&1&1\\0&3&-1\\1&0&2\end{matrix} \right) $$Answer
$$ det \left( \begin{matrix}2&1&1\\0&3&-1\\1&0&2\end{matrix} \right) = 8 $$
Explanation
To find the 3x3 determinant, we can use the Rule of Sarrus. .
$$ \begin{aligned} det \left( \begin{matrix}2&1&1\\0&3&-1\\1&0&2\end{matrix} \right) &= \left[
\begin{array}{ccc|cc} \cssId{i00}{2} & \cssId{i01}{1} & \cssId{i02}{1} & \cssId{i03}{2} & \cssId{i04}{1} \\
\cssId{i10}{0} & \cssId{i11}{3} & \cssId{i12}{-1} & \cssId{i13}{0} & \cssId{i14}{3} \\
\cssId{i20}{1} & \cssId{i21}{0} & \cssId{i22}{2} & \cssId{i23}{1} & \cssId{i24}{0}
\end{array} \right | = \\\\
&= \cssId{u0}{2 \cdot 3 \cdot 2} \cssId{u1}{+1 \cdot \left(-1\right) \cdot 1} \cssId{u2}{+1 \cdot 0 \cdot 0} \cssId{u3}{-1 \cdot 3 \cdot 1} \cssId{u4}{-0 \cdot \left(-1\right) \cdot 2} \cssId{u5}{-2 \cdot 0 \cdot 1} = \\\\
&= 12 + \left( -1\right) + 0 - 3 - 0 - 0 = 8
\end{aligned}
$$
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