Compute the determinant in Exercise 2 using a cofactor expansion across the first row. Also compute the determinant by a cofactor expansion down the second column.

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

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

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

The determinant computed by a cofactor expansion down the j^th column is

\(\det A = {a_{1j}}{C_{1j}} + {a_{2j}}{C_{2j}} + \cdots + {a_{nj}}{C_{nj}}\).

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

\(\begin{aligned}{c}\left| {\begin{aligned}{*{20}{c}}0&4&1\\5&{ - 3}&0\\2&3&1\end{aligned}} \right| = {a_{11}}{C_{11}} + {a_{12}}{C_{12}} + {a_{13}}{C_{13}}\\ = {a_{11}}{\left( { - 1} \right)^{1 + 1}}\det {A_{11}} + {a_{12}}{\left( { - 1} \right)^{1 + 2}}\det {A_{12}} + {a_{13}}{\left( { - 1} \right)^{1 + 3}}\det {A_{13}}\\ = 0\left| {\begin{aligned}{*{20}{c}}{ - 3}&0\\3&1\end{aligned}} \right| - 4\left| {\begin{aligned}{*{20}{c}}5&0\\2&1\end{aligned}} \right| + 1\left| {\begin{aligned}{*{20}{c}}5&{ - 3}\\2&3\end{aligned}} \right|\\ = 0 - 4\left( 5 \right) + 1\left( {21} \right)\\ = - 20 + 21\\ = 1\end{aligned}\)

\(\begin{aligned}{c}\left| {\begin{aligned}{*{20}{c}}0&4&1\\5&{ - 3}&0\\2&3&1\end{aligned}} \right| = {a_{12}}{C_{12}} + {a_{22}}{C_{22}} + {a_{32}}{C_{32}}\\ = {a_{12}}{\left( { - 1} \right)^{1 + 2}}\det {A_{12}} + {a_{22}}{\left( { - 1} \right)^{2 + 2}}\det {A_{22}} + {a_{32}}{\left( { - 1} \right)^{3 + 2}}\det {A_{32}}\\ = - 4\left| {\begin{aligned}{*{20}{c}}5&0\\2&1\end{aligned}} \right| + \left( { - 3} \right)\left| {\begin{aligned}{*{20}{c}}0&1\\2&1\end{aligned}} \right| - 3\left| {\begin{aligned}{*{20}{c}}0&1\\5&0\end{aligned}} \right|\\ = - 4\left( 5 \right) - 3\left( { - 2} \right) - 3\left( { - 5} \right)\\ = - 20 + 6 + 15\\ = 1\end{aligned}\)