I took a test today and ran across this question:$$Are\space the\space vectors\space \pmatrix{
1\\
1\\} and\space \pmatrix{
1\\
20\\} in\space the\space image\space of\space the\space matrix\space \pmatrix{
-1&1\\
1&2\\} $$
First, the answer is yes. I gave my professor two reasons:
Reason #$1$: $$$$The matrix $A=\pmatrix{
-1&1\\
1&2\\}$ is invertible, since it has a nonzero determinant, thus all of $\mathbb{R}^2$ is in the image of $A$. In other words, the row vectors form a basis for $\mathbb{R}^2$ since they are linearly independent and span $\mathbb{R}^2.$
She marked, "explain."
Reason #$2$:$$$$ The systems $A\vec{x_1}=\vec{b_1}$ and $A\vec{x_2}=\vec{b_2}$ are consistent where $A=\pmatrix{
-1&1\\
1&2\\}$ and $\vec{b_1}$ and $\vec{b_2}$ are $\pmatrix{
1\\
1\\} and\space \pmatrix{
1\\
20\\}$ respectively.
$$$$I feel as if this is ample reason and explanation for the question in absence of a proof which was not required. Is there anything wrong with my answer? Even after asking my prof, I still could not see my error or where there was an ambiguity in my first reason.
Best Answer
I don't see anything wrong, but it is a bit of a laundry list. One challenge of grading someones math is trying to figure out if they actually understand the words they are using.
Unless she asked for two reasons, you should only give one correct answer. You could say, for example, that since the determinant is $-3$, the linear map is surjective. There might be some theorem she specifically wanted you to state.
One guaranteed correct answer is to just demonstrate that they are in the image. It is easy to calculate the inverse of a $2\times 2$ matrix, so you should be able to find a preimage for $(1, 1)$ and $(1,20)$ directly.