Question

Combine the methods of row reduction and cofactor expansion to compute the determinant in Exercise 12.

12. \(\left| {\begin{aligned}{*{20}{c}}{ - {\bf{1}}}&{\bf{2}}&{\bf{3}}&{\bf{0}}\\{\bf{3}}&{\bf{4}}&{\bf{3}}&{\bf{0}}\\{{\bf{11}}}&{\bf{4}}&{\bf{6}}&{\bf{6}}\\{\bf{4}}&{\bf{2}}&{\bf{4}}&{\bf{3}}\end{aligned}} \right|\)

Short answer

\(\left| {\begin{aligned}{*{20}{c}}{ - 1}&2&3&0\\3&4&3&0\\{11}&4&6&6\\4&2&4&3\end{aligned}} \right| = 6\)

Step-by-step solution

Create zero in the fourth column

Add \( - 2\) times row 4 to row 3 to obtain

\(\left| {\begin{aligned}{*{20}{c}}{ - 1}&2&3&0\\3&4&3&0\\{11}&4&6&6\\4&2&4&3\end{aligned}} \right| \sim \left| {\begin{aligned}{*{20}{c}}{ - 1}&2&3&0\\3&4&3&0\\3&0&{ - 2}&0\\4&2&4&3\end{aligned}} \right|\).

Use cofactor expansion down the fourth column

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

Create zero in the second column

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

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

Use cofactor expansion down the second column

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

Hence, \(\left| {\begin{aligned}{*{20}{c}}{ - 1}&2&3&0\\3&4&3&0\\{11}&4&6&6\\4&2&4&3\end{aligned}} \right| = 6\).