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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{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right].33. \left[ {\begin{aligned}{*{20}{c}}1&k\\0&1\end{aligned}} \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{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right],{\rm{ }}E = \left[ {\begin{aligned}{*{20}{c}}1&k\\0&1\end{aligned}} \right].

Compute matrix $$EA$$ as shown below:

\begin{aligned}{c}EA = \left[ {\begin{aligned}{*{20}{c}}1&k\\0&1\end{aligned}} \right]\left[ {\begin{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right]\\ = \left[ {\begin{aligned}{*{20}{c}}{a + kc}&{b + kd}\\{0 + c}&{0 + d}\end{aligned}} \right]\\ = \left[ {\begin{aligned}{*{20}{c}}{a + kc}&{b + kd}\\c&d\end{aligned}} \right]\end{aligned}

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

The determinant of matrices E and A are shown below:

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

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

\begin{aligned}{c}\det EA = \left| {\begin{aligned}{*{20}{c}}{a + kc}&{b + kd}\\c&d\end{aligned}} \right|\\ = \left( {a + kc} \right)d - c\left( {b + kd} \right)\\ = ad + kcd - bc - kcd\\ = 1\left( {ad - bc} \right)\\ = \left( {\det E} \right)\left( {\det A} \right)\end{aligned}

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