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 26E from Chapter 1.7 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 26E
Chapter:
Problem:
In Exercises 23–26, describe the possible echelon forms of the matrix. Use the notation of Example 1 in Section 1.2. A is a 4 × 3 matrix, A = [a1 a2 a3] such that {a1 a2} is linearly independent and a3 is not in Span {a1 a2}. Example 1: a. Determine if the set { v1, v2, v3,} is linearly independent. b. If possible, find a linear dependence relation among v1, v2 and v3,
Step-by-Step Solution
Given Information
For the given $4 \times 3$ matrix, we have to find a possible form of the echelon matrix.
Step-1:
It is given that the 3 columns of the matrix are such that first two columns ($a_1, a_2$) are linearly independent but the third column ($a_3$) is not in Span of first two columns. This implies that the set ($a_1, a_2, a_3$) is linearly independent.
Step-2:
Since the matrix has 4 rows and 3 columns, it can have a maximum of 3 pivot elements. Moreover, all the columns are linearly independent, the matrix must have 3 pivot columns. So a possible form of the matrix is given below, with $p$ as pivot element and $*$ as any nonzero value
\[ \left[ \begin{array} { l l l } { \mathbf { p } } & { * } & { * } \\ { 0 } & { \mathbf { p} } & { * } \\ { 0 } & { 0 } & { \mathbf { p } } \\ { 0 } & { 0 } & { 0 } \end{array} \right] \]
For the given $4 \times 3$ matrix, we have to find a possible form of the echelon matrix.
Step-1:
It is given that the 3 columns of the matrix are such that first two columns ($a_1, a_2$) are linearly independent but the third column ($a_3$) is not in Span of first two columns. This implies that the set ($a_1, a_2, a_3$) is linearly independent.
Step-2:
Since the matrix has 4 rows and 3 columns, it can have a maximum of 3 pivot elements. Moreover, all the columns are linearly independent, the matrix must have 3 pivot columns. So a possible form of the matrix is given below, with $p$ as pivot element and $*$ as any nonzero value
\[ \left[ \begin{array} { l l l } { \mathbf { p } } & { * } & { * } \\ { 0 } & { \mathbf { p} } & { * } \\ { 0 } & { 0 } & { \mathbf { p } } \\ { 0 } & { 0 } & { 0 } \end{array} \right] \]