$$ A = \begin{bmatrix} 1 & 2 & 0 & -1 & 5 \\ 2 & 0 & 2 & 0 & 1 \\ 1 & 1 & -1 & 3 & 2 \\ 0 & 3 & -3 & 2 & 6 \end{bmatrix} $$
I put that matrix into reduced row echelon form https://www.emathhelp.net/calculators/linear-algebra/reduced-row-echelon-form-rref-calculator/?i=%5B%5B1%2C2%2C0%2C-1%2C5%5D%2C%5B2%2C0%2C2%2C0%2C1%5D%2C%5B1%2C1%2C-1%2C3%2C2%5D%2C%5B0%2C3%2C-3%2C2%2C6%5D%5D&reduced=on
and get $$ A = \begin{bmatrix} 1 & 0 & 0 & 7/3 & 0 \\ 0 & 1 & 0 & -5/3 & 5/2 \\ 0 & 0 & 1 & -7/3 & 1/2 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix} $$
$a + 7/3d = 0$
$b - 5/3d + 5/2e = 0$
$c - 7/3d + 1/2e = 0$
$0 = 0$
So the kernel is
$ \begin{bmatrix} -7/3 \\ 5/3 \\ 7/3 \\ 1 \\ 0 \end{bmatrix} $ and $ \begin{bmatrix} 0 \\ -5/2 \\ -1/2 \\ 0 \\ 1 \end{bmatrix} $
For the image, we simply look at the pivots 1 of the reduced row echelon form, and take the corresponding columns of the original matrix, so we have
$ \begin{bmatrix} 1 \\ 2 \\ 1 \\ 0 \end{bmatrix} $, $ \begin{bmatrix} 2 \\ 0 \\ 1 \\ 3 \end{bmatrix} $ and $ \begin{bmatrix} 0 \\ 2 \\ -1 \\ -3 \end{bmatrix} $
Those are the basis for the kernel and image of this matrix, is this correct ? I'm quite new to the subject so I'm quite unsure, if I forgot or missed something. Also, I don't really understand why they talk about $\phi$ in the title, is it important to know that it is given by left multiplication with the matrix ?
At the very end of the exercise, they ask what the dimensions of the kernel and image are. It seems to me that dimension of the image is 3, because the original matrix has three linearly independent vectors, but what is the dimension of the kernel ?
Thanks for your help !