Compute the determinant in Exercise 3 using a cofactor expansion across the first row. Also, compute the determinant by a cofactor expansion down the second column.

3. \(\left| {\begin{aligned}{*{20}{c}}{\bf{2}}&{ - {\bf{2}}}&{\bf{3}}\\{\bf{3}}&{\bf{1}}&{\bf{2}}\\{\bf{1}}&{\bf{3}}&{ - {\bf{1}}}\end{aligned}} \right|\)

The determinant computed by acofactor expansion across the i^th row is

\(\det A = {a_{i1}}{C_{i1}} + {a_{i2}}{C_{i2}} + \cdots + {a_{in}}{C_{in}}\).

The determinant computed by a cofactor expansion down the j^th column is

\(\det A = {a_{1j}}{C_{1j}} + {a_{2j}}{C_{2j}} + \cdots + {a_{nj}}{C_{nj}}\).

Here, A is an \(n \times n\) matrix, and \({C_{ij}} = {\left( { - 1} \right)^{i + j}}{A_{ij}}\).

\(\begin{aligned}{c}\left| {\begin{aligned}{*{20}{c}}2&{ - 2}&3\\3&1&2\\1&3&{ - 1}\end{aligned}} \right| = {a_{11}}{C_{11}} + {a_{12}}{C_{12}} + {a_{13}}{C_{13}}\\ = {a_{11}}{\left( { - 1} \right)^{1 + 1}}\det {A_{11}} + {a_{12}}{\left( { - 1} \right)^{1 + 2}}\det {A_{12}} + {a_{13}}{\left( { - 1} \right)^{1 + 3}}\det {A_{13}}\\ = 2\left| {\begin{aligned}{*{20}{c}}1&2\\3&{ - 1}\end{aligned}} \right| - \left( { - 2} \right)\left| {\begin{aligned}{*{20}{c}}3&2\\1&{ - 1}\end{aligned}} \right| + 3\left| {\begin{aligned}{*{20}{c}}3&1\\1&3\end{aligned}} \right|\\ = 2\left( { - 7} \right) + 2\left( { - 5} \right) + 3\left( 8 \right)\\ = - 14 - 10 + 24\\ = 0\end{aligned}\)

\(\begin{aligned}{c}\left| {\begin{aligned}{*{20}{c}}2&{ - 2}&3\\3&1&2\\1&3&{ - 1}\end{aligned}} \right| = {a_{12}}{C_{12}} + {a_{22}}{C_{22}} + {a_{32}}{C_{32}}\\ = {a_{12}}{\left( { - 1} \right)^{1 + 2}}\det {A_{12}} + {a_{22}}{\left( { - 1} \right)^{2 + 2}}\det {A_{22}} + {a_{32}}{\left( { - 1} \right)^{3 + 2}}\det {A_{32}}\\ = - \left( { - 2} \right)\left| {\begin{aligned}{*{20}{c}}3&2\\1&{ - 1}\end{aligned}} \right| + 1\left| {\begin{aligned}{*{20}{c}}2&3\\1&{ - 1}\end{aligned}} \right| - 3\left| {\begin{aligned}{*{20}{c}}2&3\\3&2\end{aligned}} \right|\\ = 2\left( { - 5} \right) + 1\left( { - 5} \right) - 3\left( { - 5} \right)\\ = - 10 - 5 + 15\\ = 0\end{aligned}\)