Determine whether the linear transformation T is one-to-one and whether it maps as specified Let T be the linear transformation whose standard matrix is 1-2 3 -1 3-4 A- -2 -9 5 Determine whether the linear transformation T is one-to-one and whether it maps R³ onto R³. O Not one-to-one; not onto ³ O One-to-one; not onto O Not one-to-one; onto ³ O One-to-one; onto ³
Determine whether the linear transformation T is one-to-one and whether it maps as specified Let T be the linear transformation whose standard matrix is 1-2 3 -1 3-4 A- -2 -9 5 Determine whether the linear transformation T is one-to-one and whether it maps R³ onto R³. O Not one-to-one; not onto ³ O One-to-one; not onto O Not one-to-one; onto ³ O One-to-one; onto ³
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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Question
![**Determine Whether the Linear Transformation T is One-to-One and Onto**
To determine if the linear transformation \( T \) is one-to-one and onto, we need to analyze the given standard matrix \( A \).
### Given Data:
#### Matrix \( A \):
\[ A = \begin{pmatrix}
1 & -2 & 3 \\
-1 & 3 & -4 \\
-2 & -9 & 5
\end{pmatrix} \]
### Objective:
Determine whether the linear transformation \( T \) is:
- One-to-one (injective)
- Onto (surjective)
### Options to Choose From:
1. Not one-to-one; not onto \(\mathbb{R}^3\)
2. One-to-one; not onto \(\mathbb{R}^3\)
3. Not one-to-one; onto \(\mathbb{R}^3\)
4. One-to-one; onto \(\mathbb{R}^3\)
### Explanation of Terms:
**One-to-one (Injective):** A transformation \( T \) is one-to-one if different vectors in the domain map to different vectors in the codomain. This is often checked by ensuring that the kernel of \( T \) (the set of all vectors that map to the zero vector) contains only the zero vector.
**Onto (Surjective):** A transformation \( T \) is onto if every vector in the codomain is mapped to by at least one vector in the domain. This means that the range of \( T \) is the entire codomain.
You must determine the rank of the matrix \( A \). If the rank is equal to the number of columns (and rows, since it is a square matrix), then \( A \) defines a transformation that is both injective and surjective.
### Steps to Determine Injectivity and Surjectivity:
1. **Calculate the determinant of \( A \).**
2. **Evaluate the rank of matrix \( A \).**
3. **Analyze the options based on the determinant and rank results.**
By following these steps, you can determine the correct answer among the provided choices.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F157e66f5-e794-4af2-988e-1885b818a80a%2F373fc7ee-2a37-4fe4-a90e-8144ed244fe3%2Fwyvlnyg4_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Determine Whether the Linear Transformation T is One-to-One and Onto**
To determine if the linear transformation \( T \) is one-to-one and onto, we need to analyze the given standard matrix \( A \).
### Given Data:
#### Matrix \( A \):
\[ A = \begin{pmatrix}
1 & -2 & 3 \\
-1 & 3 & -4 \\
-2 & -9 & 5
\end{pmatrix} \]
### Objective:
Determine whether the linear transformation \( T \) is:
- One-to-one (injective)
- Onto (surjective)
### Options to Choose From:
1. Not one-to-one; not onto \(\mathbb{R}^3\)
2. One-to-one; not onto \(\mathbb{R}^3\)
3. Not one-to-one; onto \(\mathbb{R}^3\)
4. One-to-one; onto \(\mathbb{R}^3\)
### Explanation of Terms:
**One-to-one (Injective):** A transformation \( T \) is one-to-one if different vectors in the domain map to different vectors in the codomain. This is often checked by ensuring that the kernel of \( T \) (the set of all vectors that map to the zero vector) contains only the zero vector.
**Onto (Surjective):** A transformation \( T \) is onto if every vector in the codomain is mapped to by at least one vector in the domain. This means that the range of \( T \) is the entire codomain.
You must determine the rank of the matrix \( A \). If the rank is equal to the number of columns (and rows, since it is a square matrix), then \( A \) defines a transformation that is both injective and surjective.
### Steps to Determine Injectivity and Surjectivity:
1. **Calculate the determinant of \( A \).**
2. **Evaluate the rank of matrix \( A \).**
3. **Analyze the options based on the determinant and rank results.**
By following these steps, you can determine the correct answer among the provided choices.
![**Problem Statement:**
Solve the problem.
The columns of \(I_3 = \begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix}\) are \(e_1 = \begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix}, e_2 = \begin{bmatrix}
0 \\
1 \\
0
\end{bmatrix}, e_3 = \begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix}\).
Suppose that \( T \) is a linear transformation from \( \mathbb{R}^3 \) into \( \mathbb{R}^2 \) such that
\[
T(e_1) = \begin{bmatrix}
3 \\
-2
\end{bmatrix}, T(e_2) = \begin{bmatrix}
5 \\
0
\end{bmatrix}, \text{ and } T(e_3) = \begin{bmatrix}
-5 \\
1
\end{bmatrix}.
\]
Find a formula for the image of an arbitrary \( x = \begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix}\) in \( \mathbb{R}^3 \).
**Answer Options:**
a) \( T \left( \begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
\end{bmatrix} \right) = \begin{bmatrix}
3x_1 + 5x_2 - 5x_3 \\
-2x_1 + x_3 \\
\end{bmatrix} \)
b) \( T \left( \begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
\end{bmatrix} \right) = \begin{bmatrix}
3x_1 - 2x_2 \\
5x_1 \\
5x_2 + x_3 \\
\end{bmatrix} \)
c) \( T \left( \begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
\end{bmatrix](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F157e66f5-e794-4af2-988e-1885b818a80a%2F373fc7ee-2a37-4fe4-a90e-8144ed244fe3%2Fscqqpos_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Solve the problem.
The columns of \(I_3 = \begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix}\) are \(e_1 = \begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix}, e_2 = \begin{bmatrix}
0 \\
1 \\
0
\end{bmatrix}, e_3 = \begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix}\).
Suppose that \( T \) is a linear transformation from \( \mathbb{R}^3 \) into \( \mathbb{R}^2 \) such that
\[
T(e_1) = \begin{bmatrix}
3 \\
-2
\end{bmatrix}, T(e_2) = \begin{bmatrix}
5 \\
0
\end{bmatrix}, \text{ and } T(e_3) = \begin{bmatrix}
-5 \\
1
\end{bmatrix}.
\]
Find a formula for the image of an arbitrary \( x = \begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix}\) in \( \mathbb{R}^3 \).
**Answer Options:**
a) \( T \left( \begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
\end{bmatrix} \right) = \begin{bmatrix}
3x_1 + 5x_2 - 5x_3 \\
-2x_1 + x_3 \\
\end{bmatrix} \)
b) \( T \left( \begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
\end{bmatrix} \right) = \begin{bmatrix}
3x_1 - 2x_2 \\
5x_1 \\
5x_2 + x_3 \\
\end{bmatrix} \)
c) \( T \left( \begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
\end{bmatrix
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