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Q24E

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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 19-24, explore the effect of an elementary row operation on the determinant of a matrix. In each case, state the row operation and describe how it affects the determinant.

\[\left[ {\begin{array}{*{20}{c}}{\bf{1}}&{\bf{0}}&{\bf{1}}\\{ - {\bf{3}}}&{\bf{4}}&{ - {\bf{4}}}\\{\bf{2}}&{ - {\bf{3}}}&{\bf{1}}\end{array}} \right],\left[ {\begin{array}{*{20}{c}}k&{\bf{0}}&k\\{ - {\bf{3}}}&{\bf{4}}&{ - {\bf{4}}}\\{\bf{2}}&{ - {\bf{3}}}&{\bf{1}}\end{array}} \right]\]

The row operation scales row 1 by k and multiplies the determinant by k.

See the step by step solution

Step by Step Solution

Step 1: Find the determinant of the first matrix

The determinant of the matrix \(\left[ {\begin{array}{*{20}{c}}1&0&1\\{ - 3}&4&{ - 4}\\2&{ - 3}&1\end{array}} \right]\) can be calculated as shown below:

\(\begin{array}{c}\left| {\begin{array}{*{20}{c}}1&0&1\\{ - 3}&4&{ - 4}\\2&{ - 3}&1\end{array}} \right| = 1\left| {\begin{array}{*{20}{c}}4&{ - 4}\\{ - 3}&1\end{array}} \right| - 0 + 1\left| {\begin{array}{*{20}{c}}{ - 3}&4\\2&{ - 3}\end{array}} \right|\\ = \left( {4 - 12} \right) + \left( {9 - 8} \right)\\ = - 7\end{array}\)

Step 2: Find the determinant of the second matrix

The determinant of the matrix \(\left[ {\begin{array}{*{20}{c}}k&0&k\\{ - 3}&4&{ - 4}\\2&{ - 3}&1\end{array}} \right]\) can be calculated as shown below:

\(\begin{array}{c}\left| {\begin{array}{*{20}{c}}k&0&k\\{ - 3}&4&{ - 4}\\2&{ - 3}&1\end{array}} \right| = k\left| {\begin{array}{*{20}{c}}4&{ - 4}\\{ - 3}&1\end{array}} \right| - 0 + k\left| {\begin{array}{*{20}{c}}{ - 3}&4\\2&{ - 3}\end{array}} \right|\\ = k\left( {4 - 12} \right) + k\left( {9 - 8} \right)\\ = - 7k\end{array}\)

So, the row operation scales row 1 by k and multiplies the determinant by k.

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