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 29E from Chapter 2.8 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 29E
Chapter:
Problem:
Construct a nonzero 3 × 3 matrix A and a nonzero vector b such that b is in Nul A.
Step-by-Step Solution
Given Information
We have to construct a nonzero $3 \times 3$ matrix A and a vector b such that b is in Nul A.
Step-1:
The Null space of matrix A is the set of solutions that satisfy the homogeneous system $Ax=0$
Step-2:
Let the nonzero matrix and vector are: \[ A = \left( \begin{array} { l l l } { 1 } & { 0 } & { 0 } \\ { 0 } & { 1 } & { 0 } \\ { 0 } & { 0 } & { 0 } \end{array} \right) \text { and } \mathbf { b } = \left( \begin{array} { l } { 0 } \\ { 0 } \\ { 1 } \end{array} \right) \] Compute their product: \[\begin{array}{l} A{\bf{b}} = \left( {\begin{array}{*{20}{l}} 1&0&0\\ 0&1&0\\ 0&0&0 \end{array}} \right)\left( {\begin{array}{*{20}{l}} 0\\ 0\\ 1 \end{array}} \right)\\ = \left( {\begin{array}{*{20}{l}} {1(0) + 0(0) + 0(1)}\\ {0(0) + 1(0) + 0(0)}\\ {0(0) + 0(0) + 0(1)} \end{array}} \right)\\ = \left( {\begin{array}{*{20}{l}} {0 + 0 + 0}\\ {0 + 0 + 0}\\ {0 + 0 + 0} \end{array}} \right)\\ = \left( {\begin{array}{*{20}{l}} 0\\ 0\\ 0 \end{array}} \right) \end{array}\] Therefore,
Hence, the vector b is Null space of A
We have to construct a nonzero $3 \times 3$ matrix A and a vector b such that b is in Nul A.
Step-1:
The Null space of matrix A is the set of solutions that satisfy the homogeneous system $Ax=0$
Step-2:
Let the nonzero matrix and vector are: \[ A = \left( \begin{array} { l l l } { 1 } & { 0 } & { 0 } \\ { 0 } & { 1 } & { 0 } \\ { 0 } & { 0 } & { 0 } \end{array} \right) \text { and } \mathbf { b } = \left( \begin{array} { l } { 0 } \\ { 0 } \\ { 1 } \end{array} \right) \] Compute their product: \[\begin{array}{l} A{\bf{b}} = \left( {\begin{array}{*{20}{l}} 1&0&0\\ 0&1&0\\ 0&0&0 \end{array}} \right)\left( {\begin{array}{*{20}{l}} 0\\ 0\\ 1 \end{array}} \right)\\ = \left( {\begin{array}{*{20}{l}} {1(0) + 0(0) + 0(1)}\\ {0(0) + 1(0) + 0(0)}\\ {0(0) + 0(0) + 0(1)} \end{array}} \right)\\ = \left( {\begin{array}{*{20}{l}} {0 + 0 + 0}\\ {0 + 0 + 0}\\ {0 + 0 + 0} \end{array}} \right)\\ = \left( {\begin{array}{*{20}{l}} 0\\ 0\\ 0 \end{array}} \right) \end{array}\] Therefore,
Hence, the vector b is Null space of A