Question

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.

\[\left| {\begin{array}{*{20}{c}}{\bf{3}}&{\bf{0}}&{\bf{0}}&{\bf{0}}\\{\bf{7}}&{ - {\bf{2}}}&{\bf{0}}&{\bf{0}}\\{\bf{2}}&{\bf{6}}&{\bf{3}}&{\bf{0}}\\{\bf{3}}&{ - {\bf{8}}}&{\bf{4}}&{ - {\bf{3}}}\end{array}} \right|\]

Short answer

The value of the determinant is 54.

Step-by-step solution

Expand the determinant about the first row

The determinant can be expanded as shown below:

\(\left| {\begin{array}{*{20}{c}}3&0&0&0\\7&{ - 2}&0&0\\2&6&3&0\\3&{ - 8}&4&{ - 3}\end{array}} \right| = {\left( { - 1} \right)^{1 + 1}} \cdot 3\left| {\begin{array}{*{20}{c}}{ - 2}&0&0\\6&3&0\\{ - 8}&4&{ - 3}\end{array}} \right|\)

Expand the determinant about the first row

The determinant can be expanded as shown below:

\({\left( { - 1} \right)^{1 + 1}} \cdot 3\left| {\begin{array}{*{20}{c}}{ - 2}&0&0\\6&3&0\\{ - 8}&4&{ - 3}\end{array}} \right| = 3{\left( { - 1} \right)^{1 + 1}} \cdot \left( { - 2} \right)\left| {\begin{array}{*{20}{c}}3&0\\4&{ - 3}\end{array}} \right|\)

Expand the determinant about the first column

The determinant can be expanded as shown below:

\(\begin{array}{c}3{\left( { - 1} \right)^{1 + 1}} \cdot \left( { - 2} \right)\left| {\begin{array}{*{20}{c}}3&0\\4&{ - 3}\end{array}} \right| = \left( { - 6} \right){\left( { - 1} \right)^{1 + 1}}\left( 3 \right)\left( { - 3} \right)\\ = 54\end{array}\)

So, the value of the determinant is 54.