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Found in: Page 165

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

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 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.