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 5E from Chapter 3.2 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 5E
Chapter:
Problem:
Find the determinants in Exercises by row reduction to echelon form.
Step-by-Step Solution
Given Information
We are given that a matrix \[A = \left| {\begin{array}{*{20}{c}}1&5&{ - 4}\\{ - 1}&{ - 4}&5\\{ - 2}&{ - 8}&7\end{array}} \right|\]We have to find the determinant of the matrix using row reduced echelon form.
Step-1: Echelon Form
\[\begin{array}{l}A = \left| {\begin{array}{*{20}{c}}1&5&{ - 4}\\{ - 1}&{ - 4}&5\\{ - 2}&{ - 8}&7\end{array}} \right|\\ = \left| {\begin{array}{*{20}{c}}1&5&{ - 4}\\0&1&1\\0&2&{ - 1}\end{array}} \right|::\,\,\left\{ \begin{array}{l}{R_2} = {R_2} + {R_1}\\{R_3} = {R_3} + 2{R_1}\end{array} \right\}\\ = \left| {\begin{array}{*{20}{c}}1&5&{ - 4}\\0&1&1\\0&0&{ - 3}\end{array}} \right|::\,\,\left\{ {{R_3} = {R_3} - 2{R_1}} \right\}\end{array}\]
Step-2: The determinant
Since the matrix is in echelon form (Triangular Matrix), the determinant can be found by product of its diagonal entries \[\begin{array}{l}D = 1 \times 1 \times \left( { - 3} \right)\\ = - 3\end{array}\]Therefore, the determinant is
\[D = - 3\]
We are given that a matrix \[A = \left| {\begin{array}{*{20}{c}}1&5&{ - 4}\\{ - 1}&{ - 4}&5\\{ - 2}&{ - 8}&7\end{array}} \right|\]We have to find the determinant of the matrix using row reduced echelon form.
Step-1: Echelon Form
\[\begin{array}{l}A = \left| {\begin{array}{*{20}{c}}1&5&{ - 4}\\{ - 1}&{ - 4}&5\\{ - 2}&{ - 8}&7\end{array}} \right|\\ = \left| {\begin{array}{*{20}{c}}1&5&{ - 4}\\0&1&1\\0&2&{ - 1}\end{array}} \right|::\,\,\left\{ \begin{array}{l}{R_2} = {R_2} + {R_1}\\{R_3} = {R_3} + 2{R_1}\end{array} \right\}\\ = \left| {\begin{array}{*{20}{c}}1&5&{ - 4}\\0&1&1\\0&0&{ - 3}\end{array}} \right|::\,\,\left\{ {{R_3} = {R_3} - 2{R_1}} \right\}\end{array}\]
Step-2: The determinant
Since the matrix is in echelon form (Triangular Matrix), the determinant can be found by product of its diagonal entries \[\begin{array}{l}D = 1 \times 1 \times \left( { - 3} \right)\\ = - 3\end{array}\]Therefore, the determinant is
\[D = - 3\]