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 27E from Chapter 3.2 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 27E
Chapter:
Problem:
In Exercises 27 and 28, A and B are n × n matrices. Mark each statement True or False. Justify each answer. a. A row replacement operation does not affect the determinant of a matrix.
Step-by-Step Solution
Given Information
We are given with some statements, we have to find whether they are True or False
Step-1: (a) .
Statement: A row replacement operation does not affect the determinant of a matrix
The statement is True. From Statement (a) of theorem-3, If a multiple of one row of A is added to another row to produce a matrix B, the \[\det A = \det B\].
True
Step-2: (b) .
Statement: The determinant of A is the product of the pivots in any echelon form U of A, multiplied by $(-1)^r$, where $r$ is the number of row interchanges made during row reduction from A to U
The statement is False. From Example-2 of the textbook, the given statement is true only for invertible matrix. \[\det A = \left\{ {\begin{array}{*{20}{c}}{{{\left( { - 1} \right)}^r} \times \left( {{\rm{Product}}\,\,{\rm{of}}\,\,{\rm{pivots}}\,\,{\rm{in}}\,{\rm{U}}} \right)}&{{\rm{A}}\,\,{\rm{is}}\,\,{\rm{invertible}}}\\0&{{\rm{A}}\,\,{\rm{is}}\,\,{\rm{not}}\,\,{\rm{invertible}}}\end{array}} \right\}\] False
Step-3: (c) .
Statement: If the columns of A are linearly dependent, then$\det A = 0$
The statement is True. Refer to paragraph below Theorem-4, when th columns of A are linearly dependent, the determinant of the matrix is zero.
True
Step-4: (d) .
Statement: $\det \left( {A + B} \right) = \det A + \det B$
The statement is False. Take an example of two matrices A and B
\[\begin{array}{l}A = \left[ {\begin{array}{*{20}{c}}{{a_1}}&{{a_2}}\\{{a_3}}&{{a_4}}\end{array}} \right]\\B = \left[ {\begin{array}{*{20}{c}}{{b_1}}&{{b_2}}\\{{b_3}}&{{b_4}}\end{array}} \right]\end{array}\]Their Determinants:\[\begin{array}{l}\det \left( A \right) = {a_1}{a_4} - {a_2}{a_3}\\\det \left( B \right) = {b_1}{b_4} - {b_2}{b_3}\end{array}\]Sum of the determinents : \[\det \left( A \right) + \det \left( B \right) = {a_1}{a_4} + {b_1}{b_4} - {a_2}{a_3} - {b_2}{b_3}\]Sum of the matrices: \[A + B = \left[ {\begin{array}{*{20}{c}}{{a_1} + {b_1}}&{{a_2} + {b_2}}\\{{a_3} + {b_3}}&{{a_4} + {b_4}}\end{array}} \right]\]The determinant of the sum of matrices:\[\begin{array}{l}\det \left( {A + B} \right) = \left( {{a_1} + {b_1}} \right) \times \left( {{a_4} + {b_4}} \right) - \left( {{a_2} + {b_2}} \right) \times \left( {{a_3} + {b_3}} \right)\\ = {a_1}{a_4} + {b_1}{b_4} + {a_1}{b_4} + {b_1}{a_4} - \left( {{a_2}{a_3} + {b_2}{b_3} + {a_3}{b_2} + {b_3}{a_2}} \right)\end{array}\]We can see that sum of determinants is not equal to determinant of sum of matrices.
False
We are given with some statements, we have to find whether they are True or False
Step-1: (a) .
Statement: A row replacement operation does not affect the determinant of a matrix
The statement is True. From Statement (a) of theorem-3, If a multiple of one row of A is added to another row to produce a matrix B, the \[\det A = \det B\].
True
Step-2: (b) .
Statement: The determinant of A is the product of the pivots in any echelon form U of A, multiplied by $(-1)^r$, where $r$ is the number of row interchanges made during row reduction from A to U
The statement is False. From Example-2 of the textbook, the given statement is true only for invertible matrix. \[\det A = \left\{ {\begin{array}{*{20}{c}}{{{\left( { - 1} \right)}^r} \times \left( {{\rm{Product}}\,\,{\rm{of}}\,\,{\rm{pivots}}\,\,{\rm{in}}\,{\rm{U}}} \right)}&{{\rm{A}}\,\,{\rm{is}}\,\,{\rm{invertible}}}\\0&{{\rm{A}}\,\,{\rm{is}}\,\,{\rm{not}}\,\,{\rm{invertible}}}\end{array}} \right\}\] False
Step-3: (c) .
Statement: If the columns of A are linearly dependent, then$\det A = 0$
The statement is True. Refer to paragraph below Theorem-4, when th columns of A are linearly dependent, the determinant of the matrix is zero.
True
Step-4: (d) .
Statement: $\det \left( {A + B} \right) = \det A + \det B$
The statement is False. Take an example of two matrices A and B
\[\begin{array}{l}A = \left[ {\begin{array}{*{20}{c}}{{a_1}}&{{a_2}}\\{{a_3}}&{{a_4}}\end{array}} \right]\\B = \left[ {\begin{array}{*{20}{c}}{{b_1}}&{{b_2}}\\{{b_3}}&{{b_4}}\end{array}} \right]\end{array}\]Their Determinants:\[\begin{array}{l}\det \left( A \right) = {a_1}{a_4} - {a_2}{a_3}\\\det \left( B \right) = {b_1}{b_4} - {b_2}{b_3}\end{array}\]Sum of the determinents : \[\det \left( A \right) + \det \left( B \right) = {a_1}{a_4} + {b_1}{b_4} - {a_2}{a_3} - {b_2}{b_3}\]Sum of the matrices: \[A + B = \left[ {\begin{array}{*{20}{c}}{{a_1} + {b_1}}&{{a_2} + {b_2}}\\{{a_3} + {b_3}}&{{a_4} + {b_4}}\end{array}} \right]\]The determinant of the sum of matrices:\[\begin{array}{l}\det \left( {A + B} \right) = \left( {{a_1} + {b_1}} \right) \times \left( {{a_4} + {b_4}} \right) - \left( {{a_2} + {b_2}} \right) \times \left( {{a_3} + {b_3}} \right)\\ = {a_1}{a_4} + {b_1}{b_4} + {a_1}{b_4} + {b_1}{a_4} - \left( {{a_2}{a_3} + {b_2}{b_3} + {a_3}{b_2} + {b_3}{a_2}} \right)\end{array}\]We can see that sum of determinants is not equal to determinant of sum of matrices.
False