Compute the determinants in Exercises 9-14 by cofactor expnasions. At each step, choose a row or column that involves the least amount of computation.

\(\left| {\begin{aligned}{*{20}{c}}{\bf{3}}&{\bf{5}}&{ - {\bf{6}}}&{\bf{4}}\\{\bf{0}}&{ - {\bf{2}}}&{\bf{3}}&{ - {\bf{3}}}\\{\bf{0}}&{\bf{0}}&{\bf{1}}&{\bf{5}}\\{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{3}}\end{aligned}} \right|\)

The determinant can be expanded as shown below:

\(\left| {\begin{aligned}{*{20}{c}}3&5&{ - 6}&4\\0&{ - 2}&3&{ - 3}\\0&0&1&5\\0&0&0&3\end{aligned}} \right| = {\left( { - 1} \right)^{1 + 1}} \cdot 3\left| {\begin{aligned}{*{20}{c}}{ - 2}&3&{ - 3}\\0&1&5\\0&0&3\end{aligned}} \right|\)

The determinant can be expanded as shown below:

\({\left( { - 1} \right)^{1 + 1}} \cdot 3\left| {\begin{aligned}{*{20}{c}}{ - 2}&3&{ - 3}\\0&1&5\\0&0&3\end{aligned}} \right| = 3{\left( { - 1} \right)^{1 + 1}} \cdot \left( { - 2} \right)\left| {\begin{aligned}{*{20}{c}}1&5\\0&3\end{aligned}} \right|\)

The determinant can be expanded as follows:

\(\begin{aligned}{c}3{\left( { - 1} \right)^{1 + 1}} \cdot \left( { - 2} \right)\left| {\begin{aligned}{*{20}{c}}1&5\\0&3\end{aligned}} \right| = \left( { - 6} \right){\left( { - 1} \right)^{1 + 1}}\left( 1 \right)\left( 3 \right)\\ = - 18\end{aligned}\)

So, the value of the determinant is \( - 18\).