2 - 2 8 - 20 0 -2 A matrix A= - 2 1 1 reduces to 2 0 – 4 6 1 -9 0 1 3 Find a basis for Nul A: 2.
2 - 2 8 - 20 0 -2 A matrix A= - 2 1 1 reduces to 2 0 – 4 6 1 -9 0 1 3 Find a basis for Nul A: 2.
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 Reduction and Null Space Basis
#### Matrix A
The matrix \( A \) is given by:
\[
A = \begin{bmatrix}
-2 & -2 & 2 & 8 \\
-6 & -2 & 1 & 1 \\
-6 & 0 & -1 & -9
\end{bmatrix}
\]
#### Row Reduction
The matrix \( A \) reduces to its row-echelon form:
\[
\begin{bmatrix}
-2 & 0 & 0 & -2 \\
0 & 2 & 0 & -4 \\
0 & 0 & 1 & 3
\end{bmatrix}
\]
#### Null Space
To find a basis for the null space of matrix \( A \) (Nul \( A \)), follow these steps:
1. **Identify Free Variables**: In the row-echelon form, any column without a leading entry is a free variable.
2. **Express Basic Variables in Terms of Free Variables**: Solve the system for the basic variables in terms of the free variables.
#### Basis for Nul A
The null space basis is represented as a set of vectors. The placeholders are provided to input the solutions:
\[
\begin{bmatrix}
\text{[ \ \ ]} \\
\text{[ \ \ ]} \\
\text{[ \ \ ]} \\
\text{[ \ \ ]}
\end{bmatrix}
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd2a435ab-e32d-4bb5-91c7-35592295b90b%2F20b75deb-e066-43ee-82cf-c49008ab7d2c%2Flxs8l27_processed.png&w=3840&q=75)
Transcribed Image Text:### Matrix Reduction and Null Space Basis
#### Matrix A
The matrix \( A \) is given by:
\[
A = \begin{bmatrix}
-2 & -2 & 2 & 8 \\
-6 & -2 & 1 & 1 \\
-6 & 0 & -1 & -9
\end{bmatrix}
\]
#### Row Reduction
The matrix \( A \) reduces to its row-echelon form:
\[
\begin{bmatrix}
-2 & 0 & 0 & -2 \\
0 & 2 & 0 & -4 \\
0 & 0 & 1 & 3
\end{bmatrix}
\]
#### Null Space
To find a basis for the null space of matrix \( A \) (Nul \( A \)), follow these steps:
1. **Identify Free Variables**: In the row-echelon form, any column without a leading entry is a free variable.
2. **Express Basic Variables in Terms of Free Variables**: Solve the system for the basic variables in terms of the free variables.
#### Basis for Nul A
The null space basis is represented as a set of vectors. The placeholders are provided to input the solutions:
\[
\begin{bmatrix}
\text{[ \ \ ]} \\
\text{[ \ \ ]} \\
\text{[ \ \ ]} \\
\text{[ \ \ ]}
\end{bmatrix}
\]
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