Question
Evaluate the following determinant:
$$ det \left( \begin{matrix}4&2&-2\\-5&3&2\\-2&4&1\end{matrix} \right) $$Answer
$$ det \left( \begin{matrix}4&2&-2\\-5&3&2\\-2&4&1\end{matrix} \right) = 10 $$
Explanation
To find the 3x3 determinant, we can use the Rule of Sarrus. .
$$ \begin{aligned} det \left( \begin{matrix}4&2&-2\\-5&3&2\\-2&4&1\end{matrix} \right) &= \left[
\begin{array}{ccc|cc} \cssId{i00}{4} & \cssId{i01}{2} & \cssId{i02}{-2} & \cssId{i03}{4} & \cssId{i04}{2} \\
\cssId{i10}{-5} & \cssId{i11}{3} & \cssId{i12}{2} & \cssId{i13}{-5} & \cssId{i14}{3} \\
\cssId{i20}{-2} & \cssId{i21}{4} & \cssId{i22}{1} & \cssId{i23}{-2} & \cssId{i24}{4}
\end{array} \right | = \\\\
&= \cssId{u0}{4 \cdot 3 \cdot 1} \cssId{u1}{+2 \cdot 2 \cdot \left(-2\right)} \cssId{u2}{+\left(-2\right) \cdot \left(-5\right) \cdot 4} \cssId{u3}{-\left(-2\right) \cdot 3 \cdot \left(-2\right)} \cssId{u4}{-4 \cdot 2 \cdot 4} \cssId{u5}{-1 \cdot \left(-5\right) \cdot 2} = \\\\
&= 12 + \left( -8\right) + 40 - 12 - 32 - \left( -10\right) = 10
\end{aligned}
$$
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