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Compute the determinant in Exercise 7 using a cofactor expansion across the first row.

7. \[\left| {\begin{array}{*{20}{c}}{\bf{4}}&{\bf{3}}&{\bf{0}}\\{\bf{6}}&{\bf{5}}&{\bf{2}}\\{\bf{9}}&{\bf{7}}&{\bf{3}}\end{array}} \right|\]

Short Answer

Expert verified

Thus, \[\left| {\begin{array}{*{20}{c}}4&3&0\\6&5&2\\9&7&3\end{array}} \right| = 4\].

Step by step solution

01

Write the determinant formula

The determinant computed by acofactor expansion across the ith 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 \[{C_{ij}} = {\left( { - 1} \right)^{i + j}}{A_{ij}}\].

02

Use the cofactor expansion across the first row

\[\begin{array}{c}\left| {\begin{array}{*{20}{c}}4&3&0\\6&5&2\\9&7&3\end{array}} \right| = {a_{11}}{C_{11}} + {a_{12}}{C_{12}} + {a_{13}}{C_{13}}\\ = {a_{11}}{\left( { - 1} \right)^{1 + 1}}\det {A_{11}} + {a_{12}}{\left( { - 1} \right)^{1 + 2}}\det {A_{12}} + {a_{13}}{\left( { - 1} \right)^{1 + 3}}\det {A_{13}}\\ = 4\left| {\begin{array}{*{20}{c}}5&2\\7&3\end{array}} \right| - 3\left| {\begin{array}{*{20}{c}}6&2\\9&3\end{array}} \right| + 0\left| {\begin{array}{*{20}{c}}6&5\\9&7\end{array}} \right|\\ = 4\left( 1 \right) - 3\left( 0 \right) + 0\\ = 4\end{array}\]

03

Conclusion

Hence, the given determinant using a cofactor expansion across the first row is 4.

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