d. Find a basis for the column space of of A. If necessary, enter a1 for a, etc., or enter coordinate vectors of the form <1,2,3> or <1,2,3,4,5>. Enter your answer as a comma separated list of vectors. A basis for the column space of A is { } e. The dimension of the null space of A is , and the null space of A is a subspace of f. If x1 = (5, 2, –1, 1,0), then Ax1 = .Is x1 in the null space of A? choose v g. If x2 = (-4, –2, –1,0, 1), then Ax2 = Is x2 in the null space of A? choose v h. If xg = 3x2 – 4x1 = , then Axg = Is xg in the null space of A? choose v

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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Suppose **a₁**, **a₂**, **a₃**, **a₄**, and **a₅** are vectors in ℝ³. Let \( A = (\mathbf{a₁} \mid \mathbf{a₂} \mid \mathbf{a₃} \mid \mathbf{a₄} \mid \mathbf{a₅}) \), and

\[
\text{rref}(A) =
\begin{bmatrix}
1 & 0 & 0 & -5 & 4 \\
0 & 1 & 0 & -2 & 2 \\
0 & 0 & 1 & 1 & 1
\end{bmatrix}
\]

### Explanation of the Diagram:

The matrix shown is in reduced row echelon form (rref), which is a form of a matrix used to solve linear equations. Each row corresponds to a linear equation, and the matrix representation helps identify the relationships and dependencies among the vectors \( \mathbf{a₁}, \mathbf{a₂}, \mathbf{a₃}, \mathbf{a₄}, \) and \( \mathbf{a₅} \).

- The first row indicates \( 1x₁ + 0x₂ + 0x₃ = -5x₄ + 4x₅ \).
- The second row shows \( 0x₁ + 1x₂ + 0x₃ = -2x₄ + 2x₅ \).
- The third row presents \( 0x₁ + 0x₂ + 1x₃ = 1x₄ + 1x₅ \).

The zeros in each row highlight the pivots, and the non-zero entries on the right specify relationships between the respective vectors \( \mathbf{a₄} \) and \( \mathbf{a₅} \).
Transcribed Image Text:Suppose **a₁**, **a₂**, **a₃**, **a₄**, and **a₅** are vectors in ℝ³. Let \( A = (\mathbf{a₁} \mid \mathbf{a₂} \mid \mathbf{a₃} \mid \mathbf{a₄} \mid \mathbf{a₅}) \), and \[ \text{rref}(A) = \begin{bmatrix} 1 & 0 & 0 & -5 & 4 \\ 0 & 1 & 0 & -2 & 2 \\ 0 & 0 & 1 & 1 & 1 \end{bmatrix} \] ### Explanation of the Diagram: The matrix shown is in reduced row echelon form (rref), which is a form of a matrix used to solve linear equations. Each row corresponds to a linear equation, and the matrix representation helps identify the relationships and dependencies among the vectors \( \mathbf{a₁}, \mathbf{a₂}, \mathbf{a₃}, \mathbf{a₄}, \) and \( \mathbf{a₅} \). - The first row indicates \( 1x₁ + 0x₂ + 0x₃ = -5x₄ + 4x₅ \). - The second row shows \( 0x₁ + 1x₂ + 0x₃ = -2x₄ + 2x₅ \). - The third row presents \( 0x₁ + 0x₂ + 1x₃ = 1x₄ + 1x₅ \). The zeros in each row highlight the pivots, and the non-zero entries on the right specify relationships between the respective vectors \( \mathbf{a₄} \) and \( \mathbf{a₅} \).
d. Find a basis for the column space of \( A \). If necessary, enter \(\mathbf{a_1}\) for \(\mathbf{a_1}\), etc., or enter coordinate vectors of the form \(\langle 1, 2, 3 \rangle\) or \(\langle 1, 2, 3, 4, 5 \rangle\). Enter your answer as a comma-separated list of vectors.

A basis for the column space of \( A \) is \(\{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\}\).

e. The dimension of the null space of \( A \) is \(\_\_\_\_\_\), and the null space of \( A \) is a subspace of \(\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\).
 
f. If \(\mathbf{x_1} = \langle 5, 2, -1, 1, 0 \rangle\), then \(A\mathbf{x_1} = \_\_\_\_\_\_\_\_\_\_\_\_\_\). Is \(\mathbf{x_1}\) in the null space of \( A \)? \(\text{choose}\)

g. If \(\mathbf{x_2} = \langle -4, -2, -1, 0, 1 \rangle\), then \(A\mathbf{x_2} = \_\_\_\_\_\_\_\_\_\_\_\_\_\). Is \(\mathbf{x_2}\) in the null space of \( A \)? \(\text{choose}\)

h. If \(\mathbf{x_3} = 3\mathbf{x_2} - 4\mathbf{x_1} = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_,\) then \(A\mathbf{x_3} = \_\_\_\_\_\_\_\_\_\_\_\_\_\). Is \(\mathbf{x_3}\) in the null space of \( A \)? \(\text{choose}\)

i. Find a basis for the null space of \( A \). If necessary, enter \(\mathbf{a_1}\) for \(\
Transcribed Image Text:d. Find a basis for the column space of \( A \). If necessary, enter \(\mathbf{a_1}\) for \(\mathbf{a_1}\), etc., or enter coordinate vectors of the form \(\langle 1, 2, 3 \rangle\) or \(\langle 1, 2, 3, 4, 5 \rangle\). Enter your answer as a comma-separated list of vectors. A basis for the column space of \( A \) is \(\{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\}\). e. The dimension of the null space of \( A \) is \(\_\_\_\_\_\), and the null space of \( A \) is a subspace of \(\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\). f. If \(\mathbf{x_1} = \langle 5, 2, -1, 1, 0 \rangle\), then \(A\mathbf{x_1} = \_\_\_\_\_\_\_\_\_\_\_\_\_\). Is \(\mathbf{x_1}\) in the null space of \( A \)? \(\text{choose}\) g. If \(\mathbf{x_2} = \langle -4, -2, -1, 0, 1 \rangle\), then \(A\mathbf{x_2} = \_\_\_\_\_\_\_\_\_\_\_\_\_\). Is \(\mathbf{x_2}\) in the null space of \( A \)? \(\text{choose}\) h. If \(\mathbf{x_3} = 3\mathbf{x_2} - 4\mathbf{x_1} = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_,\) then \(A\mathbf{x_3} = \_\_\_\_\_\_\_\_\_\_\_\_\_\). Is \(\mathbf{x_3}\) in the null space of \( A \)? \(\text{choose}\) i. Find a basis for the null space of \( A \). If necessary, enter \(\mathbf{a_1}\) for \(\
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