I am studying vector calculus, and I encountered a question, which I find the solution to this question somewhat colliding with my knowledge of linear algebra.

The question is this.

Consider the function $f:R^2rightarrow R^3$ given by $f(x)=Ax$, where A=begin{bmatrix}

2 & -1 \

5 & 0 \

-6 &3

end{bmatrix} and the vector $x$ in $R^2$ is written as the 2×1 column matrix x=begin{bmatrix}

x_1 \

x_2

end{bmatrix}

Describe the range of f.

From what I learned in my linear algebra course, f is a linear transformation with a standard matrix as A, and therefore the range of A is a span of the column vectors of A. So I thought that since the column vectors are linearly independent, the range is a plane in $R^3$ spanned by two vectors (2, 5, -6) and (-1, 0, 3).

However, the solution manual had a different answer, it said that $$f(x)=(2x_1-x_2, 5x_1, -6x_1+3x_2)$$

and since $$f_3(x_1, x_2)=-6x_1+3x_2=-3(2x_1-x_2)=-3f_1(x_1, x_2)$$

The range of f would be all of $y=(y_1, y_2, y_3)$ in $R^3$ such that $y_3=-3y_1$.

I am pretty much confused right now…can anybody explain to me about this result?