Can someone check my working please. Thanks.
Let $T:V \to R$ be a linear transformation, $B_V = \{v_1,v_2,v_3,v_4 \}$ a basis for $V$ and $B_R = \{r_1,r_2,r_3 \}$ a basis for $R$.
We have:
$T(v_1)= r_1 + r_2$ $T(v_2)= r_2 + r_3$ $T(v_3)= r_3 + 2r_2 - r_1$ $T(v_4)= r_1 - r_3$
(a) Find the matrix that represents $T$ with respect to the bases $B_V$ and $B_R$.
I got the matrix
$$T=\begin{bmatrix} 1 & 0 & -1 & 1 \\ 1 & 1 & 2 & 0\\ 0 & 1 & 1 & -1 \end{bmatrix}$$
(b) What is the range of $T$?
$$ \mathrm{ran}(T) = \mathrm{span}\left( \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}, \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix},\begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} \right) \\ = \mathrm{span}\left( \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \right)$$
(c) What is the kernel of $T$?
$$ \mathrm{ker}(T) = \mathrm{span}\left( \begin{bmatrix} -1 \\ 1 \\ 0 \\ 1 \end{bmatrix}\right) $$
Question (a) looks fine.
For (b), normally what you want to do is to find a computationally useful formula for the range. The definition is of course the span of the columns. But the range is given as the set of all linear combinations of the columns, so we can choose better linear combinations than the ones given in front of us; to do so, we can put the vectors in column echelon form. Using the notation where we replace column $C_i$ with the column $a C_i + b C_j$ (where $a, b$ are real numbers) by writing $C_i \to a C_i + b C_j$, it looks like this: \begin{gather*} \begin{aligned} \begin{bmatrix} 1 & 0 & -1 & 1 \\ 1 & 1 & 2 & 0 \\ 0 & 1 & 1 & -1 \end{bmatrix} & \to \begin{bmatrix} 1 & 0 & 0 & 0 \\ 1 & 1 & 3 & -1 \\ 0 & 1 & 1 & -1 \end{bmatrix} \quad (C_3 \to C_3 + C_1, \quad C_4 \to C_4 - C_1) \\ & \to \begin{bmatrix} 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 0 & 1 & -3 & 0 \end{bmatrix} \quad (C_3 \to C_3 - 3 C_2, \quad C_4 \to C_4 + C_2) \\ & \to \begin{bmatrix} 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \end{bmatrix} \quad (C_3 \to (-1/3)C_3) \\ & \to \begin{bmatrix} 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{bmatrix} \quad (C_2 \to C_2 - C_3) \\ & \to \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{bmatrix} \quad (C_1 \to C_1 - C_2) \end{aligned} \end{gather*} Why did we do this? Well, if in the range I can contain my initial set of columns, I can contain also linear combinations of them (because the range is a subspace of the codomain). So I change my set of columns with invertible linear transformations as to completely determine the range but with a different set of columns that is easier to observe.
In this case, we can see that the range is equal to $R$; something that was not immediately visible from your original answer. Although it was technically correct, it was not exactly the answer that we were looking for.
For (c), you need to play the same game, but this time with the rows. Why? Well, the kernel is the set of vectors $v \in V$ with $T(v) = 0$. But this is the same if we take an invertible transformation $P$ and solve for $P(T(v)) = 0$; such transformations in matrix form correspond to manipulations of the rows (because they correspond to multiplying by invertible matrices on the left). So we simplify our matrix and put it in row echelon form to have a better description of the kernel. Here it goes:
\begin{gather*} \begin{aligned} \begin{bmatrix} 1 & 0 & -1 & 1 \\ 1 & 1 & 2 & 0 \\ 0 & 1 & 1 & -1 \end{bmatrix} & \to \begin{bmatrix} 1 & 0 & -1 & 1 \\ 0 & 1 & 3 & -1 \\ 0 & 1 & 1 & -1 \end{bmatrix} \quad (R_2 \to R_2 - R_1) \\ & \to \begin{bmatrix} 1 & 0 & -1 & 1 \\ 0 & 1 & 3 & -1 \\ 0 & 0 & -2 & 0 \end{bmatrix} \quad (R_3 \to R_3 - R_2) \\ & \to \begin{bmatrix} 1 & 0 & -1 & 1 \\ 0 & 1 & 3 & -1 \\ 0 & 0 & 1 & 0 \end{bmatrix} \quad (R_3 \to (-1/2)R_3) \\ & \to \begin{bmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 0 \end{bmatrix} \quad (R_1 \to R_1 + R_3, \quad R_2 \to R_2 - 3 R_3) \end{aligned} \end{gather*} Note that in all those matrix computations, I only allow myself to do two operations simultaneously if they can be computed in parallel. I never skip a single manipulation. That allows me to double-check after the fact without having to rely on memory, which can lead to trouble.
Now that we have the matrix in row echelon form, we can parametrize the kernel: if $T(a_1 v_1 + a_2 v_2 + a_3 v_3 + a_4 v_4) = 0$, the row echelon form tells us the following: $$ (1) a_1 + (1) a_4 = 0, \quad (1) a_2 + (-1) a_4 = 0, \quad (1) a_3 = 0. $$ We can use $a_4 = a$ as a free parameter, which shows $a_2 = a$ and $a_1 = -a$, so that the kernel is generated by a single vector, namely $$ \begin{bmatrix} -1 \\ 1 \\ 0 \\ 1 \end{bmatrix}. $$ So your answer was correct.
Hope that helps,