Question

Compute the determinant in Exercise 9 by cofactor expansions. At each step, choose a row or column that involves the least amount of computation.

9. \(\left| {\begin{array}{*{20}{c}}{\bf{4}}&{\bf{0}}&{\bf{0}}&{\bf{5}}\\{\bf{1}}&{\bf{7}}&{\bf{2}}&{ - {\bf{5}}}\\{\bf{3}}&{\bf{0}}&{\bf{0}}&{\bf{0}}\\{\bf{8}}&{\bf{3}}&{\bf{1}}&{\bf{7}}\end{array}} \right|\)

Short answer

\(\left| {\begin{array}{*{20}{c}}4&0&0&5\\1&7&2&{ - 5}\\3&0&0&0\\8&3&1&7\end{array}} \right| = 15\)

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 minimum computation, choose a row or column that has zero as maximum entries.

Use cofactor expansion across the third row

\(\begin{array}{c}\left| {\begin{array}{*{20}{c}}4&0&0&5\\1&7&2&{ - 5}\\3&0&0&0\\8&3&1&7\end{array}} \right| = 3\left| {\begin{array}{*{20}{c}}0&0&5\\7&2&{ - 5}\\3&1&7\end{array}} \right| + 0 + 0 + 0\\ = 3\left| {\begin{array}{*{20}{c}}0&0&5\\7&2&{ - 5}\\3&1&7\end{array}} \right|\end{array}\)

Use cofactor expansion across the first row

\(\begin{array}{c}\left| {\begin{array}{*{20}{c}}4&0&0&5\\1&7&2&{ - 5}\\3&0&0&0\\8&3&1&7\end{array}} \right| = 3\left( {0 + 0 + 5\left| {\begin{array}{*{20}{c}}7&2\\3&1\end{array}} \right|} \right)\\ = 3\left( {5\left( 1 \right)} \right)\\ = 15\end{array}\)