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Q6E

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Linear Algebra and its Applications
Found in: Page 165
Linear Algebra and its Applications

Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

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Short Answer

Compute the determinant in Exercise 6 using a cofactor expansion across the first row.

6. \(\left| {\begin{aligned}{*{20}{c}}{\bf{5}}&{ - {\bf{2}}}&{\bf{2}}\\{\bf{0}}&{\bf{3}}&{ - {\bf{3}}}\\{\bf{2}}&{ - {\bf{4}}}&{\bf{7}}\end{aligned}} \right|\)

Thus, \(\left| {\begin{aligned}{*{20}{c}}5&{ - 2}&2\\0&3&{ - 3}\\2&{ - 4}&7\end{aligned}} \right| = 45\).

See the step by step solution

Step by Step Solution

Step 1: Write the determinant formula

The determinant computed by a cofactor 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 \({C_{ij}} = {\left( { - 1} \right)^{i + j}}{A_{ij}}\).

Step 2: Use the cofactor expansion across the first row

\(\begin{aligned}{c}\left| {\begin{aligned}{*{20}{c}}5&{ - 2}&2\\0&3&{ - 3}\\2&{ - 4}&7\end{aligned}} \right| = {a_{11}}{C_{11}} + {a_{12}}{C_{12}} + {a_{13}}{C_{13}}\\ = {a_{11}}{\left( { - 1} \right)^{1 + 1}}\det {A_{11}} + {a_{12}}{\left( { - 1} \right)^{1 + 2}}\det {A_{12}} + {a_{13}}{\left( { - 1} \right)^{1 + 3}}\det {A_{13}}\\ = 5\left| {\begin{aligned}{*{20}{c}}3&{ - 3}\\{ - 4}&7\end{aligned}} \right| - \left( { - 2} \right)\left| {\begin{aligned}{*{20}{c}}0&{ - 3}\\2&7\end{aligned}} \right| + 2\left| {\begin{aligned}{*{20}{c}}0&3\\2&{ - 4}\end{aligned}} \right|\\ = 5\left( 9 \right) + 2\left( 6 \right) + 2\left( { - 6} \right)\\ = 45 + 12 - 12\\ = 45\end{aligned}\)

Step 3: Conclusion

Hence, the given determinant using a cofactor expansion across the first row is 45.

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