Answer
The definition of a third order determinant is,
$\left| \begin{matrix}
{{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\
{{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\
{{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right|$ $={{a}_{1}}{{b}_{2}}{{c}_{3}}+{{b}_{1}}{{c}_{2}}{{a}_{3}}+{{c}_{1}}{{a}_{2}}{{b}_{3}}-{{a}_{3}}{{b}_{2}}{{c}_{1}}-{{b}_{3}}{{c}_{2}}{{a}_{1}}-{{c}_{3}}{{a}_{2}}{{b}_{1}}$
Work Step by Step
The definition of a third order determinant is,
$\left| \begin{matrix}
{{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\
{{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\
{{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right|={{a}_{1}}{{b}_{2}}{{c}_{3}}+{{b}_{1}}{{c}_{2}}{{a}_{3}}+{{c}_{1}}{{a}_{2}}{{b}_{3}}-{{a}_{3}}{{b}_{2}}{{c}_{1}}-{{b}_{3}}{{c}_{2}}{{a}_{1}}-{{c}_{3}}{{a}_{2}}{{b}_{1}}$
It can be further solved as:
$\begin{align}
& \left| \begin{matrix}
{{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\
{{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\
{{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right|={{a}_{1}}{{b}_{2}}{{c}_{3}}+{{b}_{1}}{{c}_{2}}{{a}_{3}}+{{c}_{1}}{{a}_{2}}{{b}_{3}}-{{a}_{3}}{{b}_{2}}{{c}_{1}}-{{b}_{3}}{{c}_{2}}{{a}_{1}}-{{c}_{3}}{{a}_{2}}{{b}_{1}} \\
& ={{a}_{1}}{{b}_{2}}{{c}_{3}}-{{b}_{3}}{{c}_{2}}{{a}_{1}}-{{c}_{3}}{{a}_{2}}{{b}_{1}}+{{c}_{1}}{{a}_{2}}{{b}_{3}}+{{b}_{1}}{{c}_{2}}{{a}_{3}}-{{a}_{3}}{{b}_{2}}{{c}_{1}} \\
& ={{a}_{1}}({{b}_{2}}{{c}_{3}}-{{b}_{3}}{{c}_{2}})-{{a}_{2}}({{c}_{3}}{{b}_{1}}-{{c}_{1}}{{b}_{3}})+{{a}_{3}}({{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}) \\
& ={{a}_{1}}\left| \begin{matrix}
{{b}_{2}} & {{c}_{2}} \\
{{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right|-{{a}_{2}}\left| \begin{matrix}
{{b}_{1}} & {{c}_{1}} \\
{{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right|+{{a}_{3}}\left| \begin{matrix}
{{b}_{1}} & {{c}_{1}} \\
{{b}_{2}} & {{c}_{2}} \\
\end{matrix} \right|\text{ (expanding along 1st column)}
\end{align}$
Similarly, expand it along any row or column of the determinant.
A third-order determinant can be evaluated with the help of any row or column.
