We are given the linear transformation $T: \mathbb{R}^3 \to \mathbb{R}^2$ defined by:

$T(x_1, x_2, x_3) = (x_1, x_2)$

To find the kernel of $T$:

The kernel of $T$, denoted $\text{Ker}(T)$, consists of all vectors $(x_1, x_2, x_3) \in \mathbb{R}^3$ that are mapped to the zero vector in $\mathbb{R}^2$, i.e.,

$T(x_1, x_2, x_3) = (0, 0)$

This means that:

$T(x_1, x_2, x_3) = (x_1, x_2) = (0, 0)$

Thus, for a vector $(x_1, x_2, x_3)$ to be in the kernel, it must satisfy:

$x_1 = 0 \quad \text{and} \quad x_2 = 0$

However, there is no condition on $x_3$. Therefore, the kernel consists of all vectors of the form $(0, 0, x_3)$, where $x_3 \in \mathbb{R}$.

Conclusion:

The kernel of $T$ is:

$\text{Ker}(T) = \{(0, 0, x_3) : x_3 \in \mathbb{R}\}$

This matches option b:

$\boxed{b. \{(0, 0, x) : x \in \mathbb{R}\}}$

Would you like further details or have any questions?

Here are 5 related questions:

1. What is the definition of the image (range) of a linear transformation?
2. How do we compute the rank of a linear transformation like $T$?
3. What is the rank-nullity theorem and how does it apply to this problem?
4. Can the kernel of a linear transformation be the whole domain?
5. How does the matrix representation of $T$ help in finding its kernel?

Tip: The kernel of a transformation tells you about the solutions to $T(v) = 0$ and reveals which inputs get "collapsed" to zero.