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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 2 using a cofactor expansion across the first row. Also compute the determinant by a cofactor expansion down the second column.\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|

Thus, \left| {\begin{aligned}{*{20}{c}}0&4&1\\5&{ - 3}&0\\2&3&1\end{aligned}} \right| = 1.

See the step by step solution

## Step 1: Write the determinant formula

The determinant computed by a cofactor expansion across the ith row is

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

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

## Step 2: Use the cofactor expansion across the first row

\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}

## Step 3: Use the cofactor expansion down the second column

\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} ### Want to see more solutions like these? 