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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

# In Exercise 33-36, verify that $$\det EA = \left( {\det E} \right)\left( {\det A} \right)$$where E is the elementary matrix shown and $$A = \left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right]$$.34. $$\left[ {\begin{array}{*{20}{c}}1&0\\k&1\end{array}} \right]$$

It is verified that $$\det EA = \left( {\det E} \right)\left( {\det A} \right)$$.

See the step by step solution

## Step 1: Determine matrix $$EA$$

It is given that $$A = \left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right],{\rm{ }}E = \left[ {\begin{array}{*{20}{c}}1&0\\k&1\end{array}} \right]$$.

Compute matrix $$EA$$ as shown below:

$$\begin{array}{c}EA = \left[ {\begin{array}{*{20}{c}}1&0\\k&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{a + 0}&{b + 0}\\{ka + c}&{kb + d}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}a&b\\{ka + c}&{kb + d}\end{array}} \right]\end{array}$$

## Step 2: Verify that $$\det EA = \left( {\det E} \right)\left( {\det A} \right)$$

The determinants of matrices E and A are shown below:

$\begin{array}{c}\det E = \left| {\begin{array}{*{20}{c}}1&0\\k&1\end{array}} \right|\\ = 1\\\det A = \left| {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right|\\ = ad - bc\end{array}$

The determinant of matrix $$EA$$ is shown below:

$$\begin{array}{c}\det EA = \left| {\begin{array}{*{20}{c}}a&b\\{ka + c}&{kb + d}\end{array}} \right|\\ = a\left( {kb + d} \right) - \left( {ka + c} \right)b\\ = kab + ad - kab - bc\\ = 1\left( {ad - bc} \right)\\ = \left( {\det E} \right)\left( {\det A} \right)\end{array}$$

Thus, it is verified that $$\det EA = \left( {\det E} \right)\left( {\det A} \right)$$.