3. If AA'GAA' = AA' and A'AHA'A = A'A, then A'GAHA' is the Penrose inverse of A. Use this principle to find the Penrose inverse of A = 0 1 1 011 1 1 0
3. If AA'GAA' = AA' and A'AHA'A = A'A, then A'GAHA' is the Penrose inverse of A. Use this principle to find the Penrose inverse of A = 0 1 1 011 1 1 0
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
This covers the topic of finding the penrose inverse of a given matrix, specifically in the linear models course. Attached is the given problem, and also attached is my attempt at the problem. My professor marked my answer wrong, but I do not see why I got it wrong, I thought my approach was correct.
![**Problem 3**
If \( AA'GAA' = AA' \) and \( A'AH'A'A = A'A \), then \( A'GAHA' \) is the Penrose inverse of \( A \). Use this principle to find the Penrose inverse of
\[
A = \begin{pmatrix}
0 & 1 & 1 \\
0 & 1 & 1 \\
1 & 1 & 0
\end{pmatrix}.
\]
**Explanation:**
This problem involves finding the Penrose (or Moore-Penrose) inverse of a matrix using a given principle. Here, a matrix \( A \) is provided, and you are tasked to apply the given identities to obtain its Penrose inverse. The provided matrix is a \( 3 \times 3 \) matrix, containing both zero and one entries.
To solve this problem, you need to check and verify if the given conditions are satisfied for a candidate matrix \( G \) related to \( A \), then determine and compute \( A'GAHA' \) to find the Penrose inverse.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F99690567-bdca-4632-b99f-493e44b366e1%2F818bbc85-17bd-4d2a-924e-79451578cc0d%2Fij0tgcb_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem 3**
If \( AA'GAA' = AA' \) and \( A'AH'A'A = A'A \), then \( A'GAHA' \) is the Penrose inverse of \( A \). Use this principle to find the Penrose inverse of
\[
A = \begin{pmatrix}
0 & 1 & 1 \\
0 & 1 & 1 \\
1 & 1 & 0
\end{pmatrix}.
\]
**Explanation:**
This problem involves finding the Penrose (or Moore-Penrose) inverse of a matrix using a given principle. Here, a matrix \( A \) is provided, and you are tasked to apply the given identities to obtain its Penrose inverse. The provided matrix is a \( 3 \times 3 \) matrix, containing both zero and one entries.
To solve this problem, you need to check and verify if the given conditions are satisfied for a candidate matrix \( G \) related to \( A \), then determine and compute \( A'GAHA' \) to find the Penrose inverse.
![Certainly! Below is the transcription suitable for an educational website:
---
### Gaussian Reduction Method
#### Initial Matrices
\( A = \begin{pmatrix} 0 & 2 & 1 \\ 0 & 1 & 2 \\ 1 & 2 & 3 \end{pmatrix} \)
\( A^{-1} = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 2 \end{pmatrix} \)
#### Calculation Steps
1. **Equation Formation:**
\( S = A \cdot A^{-1} - I = \begin{pmatrix} 0 & 2 & 1 \\ 0 & 1 & 2 \\ 1 & 2 & 3 \end{pmatrix} \cdot \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 2 \end{pmatrix} - \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \)
2. **Vector Conversion:**
\( \begin{pmatrix} 2 & 1 \end{pmatrix} - 1 \cdot \begin{pmatrix} 2 \cdot \dfrac{1}{3} \\ \dfrac{1}{3} \end{pmatrix} = \begin{pmatrix} -213 & -313 \\ -13 & 213 \end{pmatrix} \)
#### Intermediate Matrices
\[
\begin{pmatrix} -219 & 219 & 219 \\ 113 & 113 & 113 \end{pmatrix}
\xrightarrow{}
\text{Final Answer!}
\]
#### Final Calculation:
\[
\begin{pmatrix} -219 & 219 & 219 \\ 113 & 113 & 113 \end{pmatrix}
\]
**Note:** The final matrix encapsulated in a box is the result of the transformations applied during the Gaussian reduction.
---
The exercise demonstrates problem-solving steps utilizing matrix operations, specifically geared for linear algebra methodology in solving systems of equations.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F99690567-bdca-4632-b99f-493e44b366e1%2F818bbc85-17bd-4d2a-924e-79451578cc0d%2Fl0sueig_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Certainly! Below is the transcription suitable for an educational website:
---
### Gaussian Reduction Method
#### Initial Matrices
\( A = \begin{pmatrix} 0 & 2 & 1 \\ 0 & 1 & 2 \\ 1 & 2 & 3 \end{pmatrix} \)
\( A^{-1} = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 2 \end{pmatrix} \)
#### Calculation Steps
1. **Equation Formation:**
\( S = A \cdot A^{-1} - I = \begin{pmatrix} 0 & 2 & 1 \\ 0 & 1 & 2 \\ 1 & 2 & 3 \end{pmatrix} \cdot \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 2 \end{pmatrix} - \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \)
2. **Vector Conversion:**
\( \begin{pmatrix} 2 & 1 \end{pmatrix} - 1 \cdot \begin{pmatrix} 2 \cdot \dfrac{1}{3} \\ \dfrac{1}{3} \end{pmatrix} = \begin{pmatrix} -213 & -313 \\ -13 & 213 \end{pmatrix} \)
#### Intermediate Matrices
\[
\begin{pmatrix} -219 & 219 & 219 \\ 113 & 113 & 113 \end{pmatrix}
\xrightarrow{}
\text{Final Answer!}
\]
#### Final Calculation:
\[
\begin{pmatrix} -219 & 219 & 219 \\ 113 & 113 & 113 \end{pmatrix}
\]
**Note:** The final matrix encapsulated in a box is the result of the transformations applied during the Gaussian reduction.
---
The exercise demonstrates problem-solving steps utilizing matrix operations, specifically geared for linear algebra methodology in solving systems of equations.
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