<s>[INST] Find the last column of a matrix. Find the matrix. $A\left[\begin{matrix} 1  & 0 \\ 0 & 0 \\ 1 & 1 \end{matrix} \right]$ =  $\left[\begin{matrix} 2  & 3 \\ -1 & 0 \\ 5 & -7 \\ 0 & 6 \end{matrix} \right]$
(1)Find the last column of $A$.
(2)Find the matrix $A$.
I figured that the size of $A$ has to be $4 \times 3$. Therefore there are 4 entries in any column of $A$.
The answer to (1) is: $\left[\begin{matrix} 3 \\ 0 \\ -7 \\ 6 \end{matrix} \right]$
and the answer to (2) is $\left[\begin{matrix} -1  & * & 3 \\ -1  & * & 0  \\ 12  & * & -7  \\ -6  & * & 6  \end{matrix} \right]$ where * is any  number.
However I have no idea how to get this answer.
 [/INST] Multiplying $A$ by $\begin{bmatrix} 1 \\ 0 \\ 1\end{bmatrix}$ on the right sums the first column of $A$ with its last column.
Multiplying $A$ by $\begin{bmatrix} 0 \\ 0 \\ 1\end{bmatrix}$ on the right maintains the last column.
Using the last column and the first line of this answer you're able to obtain the first column of the product.
It also follows that the second column of $A$ is irrelevant.
This can and should all be confirmed by writing $A=\begin{bmatrix} a_{11} & a_{12} & a_{13}\\  a_{21} & a_{22} & a_{23}\\  a_{31} & a_{32} & a_{33}\\  a_{41} & a_{42} & a_{42}\\\end{bmatrix}$ and carrying out the computation.
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