Linear Algebra and Its Applications, 5th Edition
Authors: David C. Lay, Steven R. Lay, Judi J. McDonald
ISBN-13: 978-0321982384
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See our solution for Question 12E from Chapter 3.2 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 12E
Chapter:
Problem:
Combine the methods of row reduction and cofactor expansion to compute the determinants in Exercises 11–14.
Step-by-Step Solution
Given Information
We have to find the following determinant by using methods of row reduction and cofactor expansion: \[M = \left| {\begin{array}{*{20}{c}} { - 1}&2&3&0\\ 3&4&3&0\\ {11}&4&6&6\\ 4&2&4&3 \end{array}} \right|\]
Step-1: Expand the determinant from fourth column
\[\begin{array}{l} \left| {\begin{array}{*{20}{c}} { - 1}&2&3&0\\ 3&4&3&0\\ {11}&4&6&6\\ 4&2&4&3 \end{array}} \right| = \left| {\begin{array}{*{20}{c}} { - 1}&2&3&0\\ 3&4&3&0\\ 3&0&{ - 2}&0\\ 4&2&4&3 \end{array}} \right|\\ = 3\left| {\begin{array}{*{20}{c}} { - 1}&2&3\\ 3&4&3\\ 3&0&{ - 2} \end{array}} \right| \end{array}\]
Step-2:
Apply Row-2 minus two times Row-1 in Row-2 \[ 3 \left| \begin{array} { c c c } { - 1 } & { 2 } & { 3 } \\ { 3 } & { 4 } & { 3 } \\ { 3 } & { 0 } & { - 2 } \end{array} \right| = 3 \left| \begin{array} { c c c } { - 1 } & { 2 } & { 3 } \\ { 5 } & { 0 } & { - 3 } \\ { 3 } & { 0 } & { - 2 } \end{array} \right| R _ { 2 } \rightarrow R _ { 2 } - 2 R _ { 1 } \] Expand along second column: \[\begin{array}{l} 3\left| {\begin{array}{*{20}{c}} { - 1}&2&3\\ 5&0&{ - 3}\\ { - 3}&0&{ - 2} \end{array}} \right| = (3)(2){( - 1)^{1 + 2}}\left| {\begin{array}{*{20}{c}} 5&{ - 3}\\ 3&{ - 2} \end{array}} \right|\\ = (3)( - 2)( - 10 + 9)\\ = ( - 6)( - 1)\\ = 6 \end{array}\] Therefore,
The determinant is 6
We have to find the following determinant by using methods of row reduction and cofactor expansion: \[M = \left| {\begin{array}{*{20}{c}} { - 1}&2&3&0\\ 3&4&3&0\\ {11}&4&6&6\\ 4&2&4&3 \end{array}} \right|\]
Step-1: Expand the determinant from fourth column
\[\begin{array}{l} \left| {\begin{array}{*{20}{c}} { - 1}&2&3&0\\ 3&4&3&0\\ {11}&4&6&6\\ 4&2&4&3 \end{array}} \right| = \left| {\begin{array}{*{20}{c}} { - 1}&2&3&0\\ 3&4&3&0\\ 3&0&{ - 2}&0\\ 4&2&4&3 \end{array}} \right|\\ = 3\left| {\begin{array}{*{20}{c}} { - 1}&2&3\\ 3&4&3\\ 3&0&{ - 2} \end{array}} \right| \end{array}\]
Step-2:
Apply Row-2 minus two times Row-1 in Row-2 \[ 3 \left| \begin{array} { c c c } { - 1 } & { 2 } & { 3 } \\ { 3 } & { 4 } & { 3 } \\ { 3 } & { 0 } & { - 2 } \end{array} \right| = 3 \left| \begin{array} { c c c } { - 1 } & { 2 } & { 3 } \\ { 5 } & { 0 } & { - 3 } \\ { 3 } & { 0 } & { - 2 } \end{array} \right| R _ { 2 } \rightarrow R _ { 2 } - 2 R _ { 1 } \] Expand along second column: \[\begin{array}{l} 3\left| {\begin{array}{*{20}{c}} { - 1}&2&3\\ 5&0&{ - 3}\\ { - 3}&0&{ - 2} \end{array}} \right| = (3)(2){( - 1)^{1 + 2}}\left| {\begin{array}{*{20}{c}} 5&{ - 3}\\ 3&{ - 2} \end{array}} \right|\\ = (3)( - 2)( - 10 + 9)\\ = ( - 6)( - 1)\\ = 6 \end{array}\] Therefore,
The determinant is 6