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Q6E

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Found in: Page 165

### 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

# 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 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.