Are these vectors in the image of the matrix?

Question

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.

Answer 1

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.

Answer 2

Though it is true that the rows of the matrix span $\mathbb R^2$, since the matrix is invertible, hence the rank is $2$, the dimension of $\mathbb R^2$... she may have been looking for something about the columns; since in general the image is the span of the columns... the part about the rows might be considered something of a coincidence (it isn't true for a nonsquare matrix, generally...).

Answer 3

Your reason #2 is unsatisfying. It's tautological - it's as if you said "$b_1$ and $b_2$ are in the image, because $b_1$ and $b_2$ are in the image". It's technically true but you haven't demonstrated that the systems are indeed consistent. You've basically asked the professor to do the work for you and verify themselves that the systems are consistent, and hence, the vectors are in the image.

Reason #1 is better, since it's easier to be convinced that the determinant is indeed nonzero. It might fly in an advanced course, where it's clear that anyone can compute the determinant in their head (of course, in this case you wouldn't have such an easy exercise - it will only be a part of a bigger derivation).

But in a basic course, you are supposed to show that you actually know how to calculate the determinant. Again you've asked the grader to do the work for you. You should actually show the calculation for the determinant, show what number it results in, and only then assert that it's nonzero, and hence the matrix is invertible and every vector in the codomain is in the image.