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