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Q34Q

Expert-verifiedFound in: Page 165

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

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:

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

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

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