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Linear Algebra and its Applications
Found in: Page 165
Linear Algebra and its Applications

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

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Short Answer

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 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 \(\det EA = \left( {\det E} \right)\left( {\det A} \right)\).

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