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 \(A = \left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right]\).

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

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 \(A = \left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right],{\rm{ }}E = \left[ {\begin{array}{*{20}{c}}0&1\\1&0\end{array}} \right]\).

Compute matrix \(EA\) as shown below:

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

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

The determinants of matrices Eand A are shown below:

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

The determinant of matrix \(EA\) is shown below:

\(\begin{array}{c}\det EA = \left| {\begin{array}{*{20}{c}}c&d\\a&b\end{array}} \right|\\ = cb - da\\ = - 1\left( {ad - bc} \right)\\ = \left( {\det E} \right)\left( {\det A} \right)\end{array}\)

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