Question

Combine the methods of row reduction and cofactor expansion to compute the determinant in Exercise 11.

11. \(\left| {\begin{aligned}{*{20}{c}}{\bf{3}}&{\bf{4}}&{ - {\bf{3}}}&{ - {\bf{1}}}\\{\bf{3}}&{\bf{0}}&{\bf{1}}&{ - {\bf{3}}}\\{ - {\bf{6}}}&{\bf{0}}&{ - {\bf{4}}}&{\bf{3}}\\{\bf{6}}&{\bf{8}}&{ - {\bf{4}}}&{ - {\bf{1}}}\end{aligned}} \right|\)

Short answer

\(\left| {\begin{aligned}{*{20}{c}}3&4&{ - 3}&{ - 1}\\3&0&1&{ - 3}\\{ - 6}&0&{ - 4}&3\\6&8&{ - 4}&{ - 1}\end{aligned}} \right| = - 48\)

Step-by-step solution

Create zero in the second column

Add \( - 2\) times row 1 to row 4 to obtain

\(\left| {\begin{aligned}{*{20}{c}}3&4&{ - 3}&{ - 1}\\3&0&1&{ - 3}\\{ - 6}&0&{ - 4}&3\\6&8&{ - 4}&{ - 1}\end{aligned}} \right| \sim \left| {\begin{aligned}{*{20}{c}}3&4&{ - 3}&{ - 1}\\3&0&1&{ - 3}\\{ - 6}&0&{ - 4}&3\\0&0&2&1\end{aligned}} \right|\).

Use cofactor expansion down the second column

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

Create zero in the first column

Add 2 times row 1 to row 2 to obtain

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

Use cofactor expansion down the first column

\(\begin{aligned}{c} - 4\left| {\begin{aligned}{*{20}{c}}3&1&{ - 3}\\0&{ - 2}&{ - 3}\\0&2&1\end{aligned}} \right| = - 4\left( {3\left| {\begin{aligned}{*{20}{c}}{ - 2}&{ - 3}\\2&1\end{aligned}} \right| + 0 + 0} \right)\\ = - 4\left( {3\left( 4 \right)} \right)\\ = - 48\end{aligned}\)

Hence, \(\left| {\begin{aligned}{*{20}{c}}3&4&{ - 3}&{ - 1}\\3&0&1&{ - 3}\\{ - 6}&0&{ - 4}&3\\6&8&{ - 4}&{ - 1}\end{aligned}} \right| = - 48\).