A matrix A= -10 -5 -1 40 37 -12] 20 20 -7 4 1 1 Find a basis for Nul A: 1 reduces to 0 0 -4 0 -4 3 -32 002
A matrix A= -10 -5 -1 40 37 -12] 20 20 -7 4 1 1 Find a basis for Nul A: 1 reduces to 0 0 -4 0 -4 3 -32 002
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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![The image shows a problem about finding the basis for the null space (Nul A) of a given matrix. The original matrix \( A \) and its row-reduced echelon form are displayed:
**Matrix A:**
\[
A = \begin{bmatrix}
-10 & 40 & 37 & -12 \\
-5 & 20 & 20 & -7 \\
-1 & 4 & 1 & 1
\end{bmatrix}
\]
**Row-Reduced Echelon Form of A:**
\[
\begin{bmatrix}
1 & -4 & -4 & 3 \\
0 & 0 & -3 & 2 \\
0 & 0 & 0 & 2
\end{bmatrix}
\]
The task is to find a basis for the null space of matrix \( A \). There is an empty template with four spaces, likely for listing the basis vectors for the null space.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9c49f70f-02c0-4076-b0a5-564b9a6f1d8d%2F14b58c00-19b3-4665-af14-60d8c59192ff%2Fuhfejbh_processed.png&w=3840&q=75)
Transcribed Image Text:The image shows a problem about finding the basis for the null space (Nul A) of a given matrix. The original matrix \( A \) and its row-reduced echelon form are displayed:
**Matrix A:**
\[
A = \begin{bmatrix}
-10 & 40 & 37 & -12 \\
-5 & 20 & 20 & -7 \\
-1 & 4 & 1 & 1
\end{bmatrix}
\]
**Row-Reduced Echelon Form of A:**
\[
\begin{bmatrix}
1 & -4 & -4 & 3 \\
0 & 0 & -3 & 2 \\
0 & 0 & 0 & 2
\end{bmatrix}
\]
The task is to find a basis for the null space of matrix \( A \). There is an empty template with four spaces, likely for listing the basis vectors for the null space.
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