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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 4 using a cofactor expansion across the first row. Also compute the determinant by a cofactor expansion down the second column.4. \left| {\begin{aligned}{*{20}{c}}{\bf{1}}&{\bf{2}}&{\bf{4}}\\{\bf{3}}&{\bf{1}}&{\bf{1}}\\{\bf{2}}&{\bf{4}}&{\bf{2}}\end{aligned}} \right|

Thus \left| {\begin{aligned}{*{20}{c}}1&2&4\\3&1&1\\2&4&2\end{aligned}} \right| = 30.

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}}1&2&4\\3&1&1\\2&4&2\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}}\\ = 1\left| {\begin{aligned}{*{20}{c}}1&1\\4&2\end{aligned}} \right| - 2\left| {\begin{aligned}{*{20}{c}}3&1\\2&2\end{aligned}} \right| + 4\left| {\begin{aligned}{*{20}{c}}3&1\\2&4\end{aligned}} \right|\\ = 1\left( { - 2} \right) - 2\left( 4 \right) + 4\left( {10} \right)\\ = - 2 - 8 + 40\\ = 30\end{aligned}

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

\begin{aligned}{c}\left| {\begin{aligned}{*{20}{c}}1&2&4\\3&1&1\\2&4&2\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}}\\ = - 2\left| {\begin{aligned}{*{20}{c}}3&1\\2&2\end{aligned}} \right| + 1\left| {\begin{aligned}{*{20}{c}}1&4\\2&2\end{aligned}} \right| - 4\left| {\begin{aligned}{*{20}{c}}1&4\\3&1\end{aligned}} \right|\\ = - 2\left( 4 \right) + 1\left( { - 6} \right) - 4\left( { - 11} \right)\\ = - 8 - 6 + 44\\ = 30\end{aligned}

Thus, the value of provided determinant is 30.