I'm having a bit of trouble with a question from a previous years undergraduate algebra exam.
Consider the matrix $$A=\begin{bmatrix} 0 & 0 & 22 & 0 \\ -2 & 2 & -6 & -4 \\ 2 & 2 & 6 & 8 \end{bmatrix}\in M_{3\times 4}(\mathbb Z)$$ Let $ϕ :\mathbb Z^4 →\mathbb Z^3$ be the module homomorphism whose matrix with respect to the standard bases is $A$. Find a basis $\{b_1, b_2, b_3\}$ for $\mathbb Z^3$ and integers $d_1, d_2, d_3$ such that $\{d_1b_1, d_2b_2, d_3b_3\}$ is a basis for the image of $ϕ$.
This is part c) of a three part question, the earlier parts consisting of
a) finding a diagonal matrix $D$ such that $D=XAY$ for some invertible integer matrices $X$ and $Y$, and
(b) Find the invertible integer matrices $X$ and $Y$ such that $D=XAY$.
This is relatively straightforward, and I ended up calculating $D$ to be the matrix
\begin{bmatrix} 2 & 0 & 0 & 0 \\ 0 & 4 & 0 & 0 \\ 0 & 0 & 22 & 0 \end{bmatrix}
since the question did not explicitly require $d_1 \mid d_2 \mid d_3$ as is required for invariant factors (unless I've goofed the calculations which is a distinct possibility), as well as the matrices $X$ and $Y$ by applying the relevant row and column operations to the 3x3 and 4x4 identity matrices respectively.
I am stuck however as to how to complete the final calculation asked in the indented section above and my course notes only briefly mention that such a calculation is possible. Any insight would be greatly appreciated.
It seems that the image of $\phi$ is the column space. So $b_1=\begin{bmatrix} 0 & -2&2 \end{bmatrix}$, $b_2=\begin{bmatrix} 0 & 2&2\end{bmatrix}$ $b_3=\begin{bmatrix} 22&-6&6\end{bmatrix}$. And we can eliminate the fourth column. Then we get a basis $B=\{b_1,b_2,b_3\}$ of the image of $\phi$ with $d_1=d_2=d_3=1$.
I am wondering if the question have some mistake or I misunderstand something...
Or another possibility is that: it is supposed that $d_1=2$, $d_2=4$, $d_3=22$ and we want to find a basis of the image of $\phi$. Then since $\operatorname{rank}A=3$ the basis is $b_1=\begin{bmatrix} 1 & 0&0 \end{bmatrix}$, $b_2=\begin{bmatrix} 0 & 1&0\end{bmatrix}$ $b_3=\begin{bmatrix} 0&0&1\end{bmatrix}$.