Precalculus (6th Edition) Blitzer

The definition of a third order determinant is, $\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\ \end{matrix} \right|$ $={{a}_{1}}{{b}_{2}}{{c}_{3}}+{{b}_{1}}{{c}_{2}}{{a}_{3}}+{{c}_{1}}{{a}_{2}}{{b}_{3}}-{{a}_{3}}{{b}_{2}}{{c}_{1}}-{{b}_{3}}{{c}_{2}}{{a}_{1}}-{{c}_{3}}{{a}_{2}}{{b}_{1}}$
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.