Question

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. \(\left[ {\begin{aligned}{*{20}{c}}1&k\\0&1\end{aligned}} \right]\)

Short answer

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

Step-by-step solution

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:

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

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

The determinant of matrices Eand 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 \(\det EA = \left( {\det E} \right)\left( {\det A} \right)\).