Question

Compute the determinant in Exercise 10 by cofactor expansions. At each step, choose a row or column that involves the least amount of computation.

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

Short answer

\(\left| {\begin{array}{*{20}{c}}1&{ - 2}&5&2\\0&0&3&0\\2&{ - 4}&{ - 3}&5\\2&0&3&5\end{array}} \right| = 12\)

Step-by-step solution

Write the determinant formula

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

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

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

For the least amount of computation, choose a row or column that has the maximum entries as zero.

Use cofactor expansion across the second row

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

Use cofactor expansion across the third row

\(\begin{array}{c}\left| {\begin{array}{*{20}{c}}1&{ - 2}&5&2\\0&0&3&0\\2&{ - 4}&{ - 3}&5\\2&0&3&5\end{array}} \right| = - 3\left| {\begin{array}{*{20}{c}}1&{ - 2}&2\\2&{ - 4}&5\\2&0&5\end{array}} \right|\\ = - 3\left( {2\left| {\begin{array}{*{20}{c}}{ - 2}&2\\{ - 4}&5\end{array}} \right| + 0 + 5\left| {\begin{array}{*{20}{c}}1&{ - 2}\\2&{ - 4}\end{array}} \right|} \right)\\ = - 3\left( {2\left( { - 2} \right) + 5\left( 0 \right)} \right)\\ = - 3\left( { - 4} \right)\\ = 12\end{array}\)