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To find a basis for the kernel of the linear transformation \( T \), we need to determine the null space (or kernel) of the matrix \( A \) given by:

\[
A = \begin{pmatrix}
1 & 2 & 1 \\
0 & 1 & 3 \\
-1 & 3 & 4
\end{pmatrix}
\]

The kernel of \( T \) consists of all vectors \( \mathbf{x} \in \mathbb{R}^3 \) such that \( A\mathbf{x} = \mathbf{0} \).

Let's solve the equation \( A\mathbf{x} = \mathbf{0} \), where \( \mathbf{x} = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} \).

This leads to the following system of linear equations:

\[
\begin{aligned}
1. & \quad x_1 + 2x_2 + x_3 = 0, \\
2. & \quad x_2 + 3x_3 = 0, \\
3. & \quad -x_1 + 3x_2 + 4x_3 = 0.
\end{aligned}
\]

To find the solution set, we'll solve this system using Gaussian elimination or row reduction.

### Step 1: Write the augmented matrix
\[
\left[\begin{array}{ccc|c}
1 & 2 & 1 & 0 \\
0 & 1 & 3 & 0 \\
-1 & 3 & 4 & 0 \\
\end{array}\right]
\]

### Step 2: Row reduction
1. Add row 1 to row 3 to eliminate the first entry in the third row:

\[
\left[\begin{array}{ccc|c}
1 & 2 & 1 & 0 \\
0 & 1 & 3 & 0 \\
0 & 5 & 5 & 0 \\
\end{array}\right]
\]

2. Subtract 5 times row 2 from row 3 to eliminate the second entry in the third row:

\[
\left[\begin{array}{ccc|c}
1 & 2 & 1 & 0 \\
0 & 1 & 3 & 0 \\
0 & 0 & -10 & 0 \\
\end{array}\right]
\]

3. Divide the third row by -10:

\[
\left[\begin{array}{ccc|c}
1 & 2 & 1 & 0 \\
0 & 1 & 3 & 0 \\
0 & 0 & 1 & 0 \\
\end{array}\right]
\]

### Step 3: Back-substitution
From the reduced matrix, we have the following equations:
\[
\begin{aligned}
x_3 & = 0, \\
x_2 + 3x_3 & = 0 \quad \Rightarrow \quad x_2 = 0, \\
x_1 + 2x_2 + x_3 & = 0 \quad \Rightarrow \quad x_1 = 0.
\end{aligned}
\]

This implies that the only solution to the system is the trivial solution \( \mathbf{x} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \).

### Conclusion
Since the kernel consists only of the zero vector, the kernel is trivial, and therefore, the basis for the kernel of \( T \) is the empty set. This indicates that the linear transformation \( T \) is injective (one-to-one), meaning there are no non-trivial solutions to \( A\mathbf{x} = \mathbf{0} \).

Would you like any additional details or explanations?

Here are five follow-up questions to explore this topic further:
1. What does it mean for a linear transformation to be injective?
2. How would the analysis change if there were non-trivial solutions in the kernel?
3. What is the rank-nullity theorem and how does it apply here?
4. Can you find the image (or range) of the transformation \( T \)?
5. How would you find a basis for the image of \( T \)?

**Tip:** Understanding the relationship between the rank of a matrix and the dimension of its kernel is key in linear algebra.

Explore how to find a basis for the kernel of the linear transformation T using matrix operations and the rank-nullity theorem. This detailed solution covers Gaussian elimination, null space determination, and the implications of a trivial kernel. Suitable for advanced undergraduate students and beyond.

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