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Expert-verified 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 8 using a cofactor expansion across the first row. 8. $$\left| {\begin{array}{*{20}{c}}{\bf{4}}&{\bf{1}}&{\bf{2}}\\{\bf{4}}&{\bf{0}}&{\bf{3}}\\{\bf{3}}&{ - {\bf{2}}}&{\bf{5}}\end{array}} \right|$$

$$\left| {\begin{array}{*{20}{c}}4&1&2\\4&0&3\\3&{ - 2}&5\end{array}} \right| = - 3$$

See the step by step solution

## Step 1: Write the determinant formula

The determinant computed by 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 cofactor expansion across the first row

$$\begin{array}{c}\left| {\begin{array}{*{20}{c}}4&1&2\\4&0&3\\3&{ - 2}&5\end{array}} \right| = {a_{11}}{C_{11}} + {a_{12}}{C_{12}} + {a_{13}}{C_{13}}\\ = {a_{11}}{\left( { - 1} \right)^{1 + 1}}\det {A_{11}} + {a_{12}}{\left( { - 1} \right)^{1 + 2}}\det {A_{12}} + {a_{13}}{\left( { - 1} \right)^{1 + 3}}\det {A_{13}}\\ = 4\left| {\begin{array}{*{20}{c}}0&3\\{ - 2}&5\end{array}} \right| - 1\left| {\begin{array}{*{20}{c}}4&3\\3&5\end{array}} \right| + 2\left| {\begin{array}{*{20}{c}}4&0\\3&{ - 2}\end{array}} \right|\\ = 4\left( 6 \right) - 1\left( {11} \right) + 2\left( { - 8} \right)\\ = 24 - 11 - 16\\ = - 3\end{array}$$

## Step 3: Draw a conclusion

The determinant obtained by cofactor expansion across the first row is $$- 3$$. ### Want to see more solutions like these? 