Find bases for the column space, the row space, and the null space of matrix A. You should verify that the Rank-Nullity Theorem holds. An equivalent echelon form of matrix A is given to make your work easier. 3 2 A = -5 3 -5 Basis for the column space of A is 19 3 1 0 2 3 4 -1 9 4 - [1 0 0 0 1 0 0 01 (0.00)) 000 [000] Basis for the row space of A is Note that since the only solution to Ax is the
Find bases for the column space, the row space, and the null space of matrix A. You should verify that the Rank-Nullity Theorem holds. An equivalent echelon form of matrix A is given to make your work easier. 3 2 A = -5 3 -5 Basis for the column space of A is 19 3 1 0 2 3 4 -1 9 4 - [1 0 0 0 1 0 0 01 (0.00)) 000 [000] Basis for the row space of A is Note that since the only solution to Ax is the
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Matrix Spaces and Bases**
**Objective:**
Find bases for the column space, the row space, and the null space of matrix \(A\). Verify the Rank-Nullity Theorem.
**Matrix \(A\):**
\[
A = \begin{bmatrix}
3 & 19 & 3 \\
2 & 1 & 0 \\
-5 & 2 & 3 \\
3 & 4 & -1 \\
-5 & 9 & 4 \\
\end{bmatrix}
\]
**Equivalent Echelon Form of Matrix \(A\):**
\[
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{bmatrix}
\]
**1. Basis for the Column Space of \(A\):**
- The pivot columns of the echelon form indicate which columns in the original matrix form a basis. Typically, these would align with the corresponding non-zero columns in the original matrix.
**2. Basis for the Row Space of \(A\):**
- The non-zero rows of the echelon form matrix represent a basis for the row space.
**3. Basis for the Null Space of \(A\):**
- Since the only solution to \(A\mathbf{x} = 0\) is the zero vector, there is no non-trivial basis for the null space.
**Verification:**
- Verify the Rank-Nullity Theorem:
- Rank(\(A\)) = Number of linearly independent rows = 3
- Nullity(\(A\)) = Number of free variables = 0
- Total columns = Rank + Nullity, which should hold true.
This explanation helps learners visualize and comprehend the role of matrix transformations in determining spaces and how to verify these with the Rank-Nullity Theorem.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc167378a-6772-4355-bdb0-a5fec1af8d0e%2F5d4702dd-222c-4d45-b6a8-2854aaf043e3%2Fj9ect1_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Matrix Spaces and Bases**
**Objective:**
Find bases for the column space, the row space, and the null space of matrix \(A\). Verify the Rank-Nullity Theorem.
**Matrix \(A\):**
\[
A = \begin{bmatrix}
3 & 19 & 3 \\
2 & 1 & 0 \\
-5 & 2 & 3 \\
3 & 4 & -1 \\
-5 & 9 & 4 \\
\end{bmatrix}
\]
**Equivalent Echelon Form of Matrix \(A\):**
\[
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{bmatrix}
\]
**1. Basis for the Column Space of \(A\):**
- The pivot columns of the echelon form indicate which columns in the original matrix form a basis. Typically, these would align with the corresponding non-zero columns in the original matrix.
**2. Basis for the Row Space of \(A\):**
- The non-zero rows of the echelon form matrix represent a basis for the row space.
**3. Basis for the Null Space of \(A\):**
- Since the only solution to \(A\mathbf{x} = 0\) is the zero vector, there is no non-trivial basis for the null space.
**Verification:**
- Verify the Rank-Nullity Theorem:
- Rank(\(A\)) = Number of linearly independent rows = 3
- Nullity(\(A\)) = Number of free variables = 0
- Total columns = Rank + Nullity, which should hold true.
This explanation helps learners visualize and comprehend the role of matrix transformations in determining spaces and how to verify these with the Rank-Nullity Theorem.
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