B: Suppose T(x, y, z) = (3x − 6y +3z, x+y+z, -x). b) Using the columns of T, prove that T is bijective. Then find the matrix of the inverse of T.
B: Suppose T(x, y, z) = (3x − 6y +3z, x+y+z, -x). b) Using the columns of T, prove that T is bijective. Then find the matrix of the inverse of T.
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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![### Educational Problem: Linear Transformation and Inverse Matrix
#### Problem Statement:
**B:** Suppose \( T(x, y, z) = (3x - 6y + 3z, x + y + z, -x) \).
**b)** Using the columns of \( T \), prove that \( T \) is bijective. Then find the matrix of the inverse of \( T \).
#### Explanation:
1. **Define the Linear Transformation \( T(x, y, z) \):**
The given transformation \( T \) maps a 3-dimensional vector \((x, y, z)\) to another 3-dimensional vector \((3x - 6y + 3z, x + y + z, -x) \).
2. **Matrix Representation of \( T \):**
To represent the transformation as a matrix \( A \), express the transformed coordinates in terms of the original coordinates \((x, y, z) \):
\[
A \begin{bmatrix}
x \\
y \\
z
\end{bmatrix}
= \begin{bmatrix}
3 & -6 & 3 \\
1 & 1 & 1 \\
-1 & 0 & 0
\end{bmatrix} \begin{bmatrix}
x \\
y \\
z
\end{bmatrix}
\]
3. **Prove Bijectivity:**
- A transformation is bijective if it is both injective (one-to-one) and surjective (onto).
- For a linear transformation described by a matrix, it is bijective if and only if the matrix is invertible.
- Check for invertibility by finding the determinant of matrix \( A \) and ensuring it is non-zero.
\[
\text{det}(A) = \begin{vmatrix}
3 & -6 & 3 \\
1 & 1 & 1 \\
-1 & 0 & 0
\end{vmatrix} = 3 \cdot \begin{vmatrix}
1 & 1 \\
0 & 0
\end{vmatrix} - (-6) \cdot \begin{vmatrix}
1 & 1 \\
-](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F544a9f38-3e68-4113-87a7-993914a696cb%2F4b59f247-ecb7-4426-9e28-f58672d7f07c%2Fibsvvyo_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Educational Problem: Linear Transformation and Inverse Matrix
#### Problem Statement:
**B:** Suppose \( T(x, y, z) = (3x - 6y + 3z, x + y + z, -x) \).
**b)** Using the columns of \( T \), prove that \( T \) is bijective. Then find the matrix of the inverse of \( T \).
#### Explanation:
1. **Define the Linear Transformation \( T(x, y, z) \):**
The given transformation \( T \) maps a 3-dimensional vector \((x, y, z)\) to another 3-dimensional vector \((3x - 6y + 3z, x + y + z, -x) \).
2. **Matrix Representation of \( T \):**
To represent the transformation as a matrix \( A \), express the transformed coordinates in terms of the original coordinates \((x, y, z) \):
\[
A \begin{bmatrix}
x \\
y \\
z
\end{bmatrix}
= \begin{bmatrix}
3 & -6 & 3 \\
1 & 1 & 1 \\
-1 & 0 & 0
\end{bmatrix} \begin{bmatrix}
x \\
y \\
z
\end{bmatrix}
\]
3. **Prove Bijectivity:**
- A transformation is bijective if it is both injective (one-to-one) and surjective (onto).
- For a linear transformation described by a matrix, it is bijective if and only if the matrix is invertible.
- Check for invertibility by finding the determinant of matrix \( A \) and ensuring it is non-zero.
\[
\text{det}(A) = \begin{vmatrix}
3 & -6 & 3 \\
1 & 1 & 1 \\
-1 & 0 & 0
\end{vmatrix} = 3 \cdot \begin{vmatrix}
1 & 1 \\
0 & 0
\end{vmatrix} - (-6) \cdot \begin{vmatrix}
1 & 1 \\
-
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