Given $f: \mathbb R^2\to \mathbb R^2$ where $f((2,7)) = (7,5)$ and $f((1,3)) = (4,1)$ find $f(3,5)$ I'm trying to solve this problem, I took the basis $B=\langle(2,7)^T, (1,3)^T\rangle$ and then build the matrix of $f$ application in $B$:
$$
M_b(f)=\left(\begin{array}{cc}
7&4\\5&1
\end{array}\right).
$$
I've thought that now I can get $f(3,5)$ just saying that $f: x\rightarrow M_b(f) x$ but I'm wrong because my answer is not the correct one. Could you explain me where I'm wrong?
 A: I assume $f:\Bbb{R^2}\to \Bbb{R^2}$ be a linear map.
Given $f(2, 7) =(7, 5) $ and $f(1, 3) =(4, 1) $
To find $f(3, 5) $ , first we have to write $(3, 5) $ as a linear combination of $(2, 7) $ and $(1, 3)$.
$\begin{bmatrix}2 & 4 \\ 7 & 1 \\\end{bmatrix}\begin{bmatrix}c_1 \\ c_2\\\end{bmatrix}=\begin{bmatrix}3 \\ 5\\\end{bmatrix}$
Solving the above equation, we get $c_1={-4}$, $c_2=11$
Hence, $(3, 5)={-4}(2, 7) +11(1, 3) $
Now, \begin{align}f(3, 5)&={-4}\space f\space (2, 7) +11 \space f\space (1, 3) \\&= 
{-4}(7, 5) +11(4, 1) \\ &=(16,-9)\end{align}
A: If $f$ is a linear map which I am assuming is the case then you can (as you tried) express the linear map in terms of a matrix. The "mistake" you did is that you wrote the images of the basis $B$ as columns so your matrix is the matrix representation with resprect to the bases $B$ and $((1,0),(0,1))$ meaning you wrote down the images of $B$ in terms of the standard basis. If you want to keep working with that matrix you need to add a base change in order to use that $f(x) = M_b(f)x$ because you are putting in the vector $(3,5)$ in terms of the standard basis and not in terms of $B$. You have to multiply from the right the base change matrix from the standard basis to $B$. That means that we have to multiply by the matrix $A$ that has the columns describing the standard basis in terms of $B$, so:
$$(1,0) = a(2,7) + b(1,3) \Rightarrow a=-3, b=7$$
so the first column of the matrix $A$ is $(-3,7)^T$. For the second vector
$$(0,1) = a(2,7)+b(1,3) \Rightarrow b=-2,a=1$$
That gives us $$A= \begin{pmatrix}-3&1\\7&-2 \end{pmatrix}$$
and for the matrix $M_b(f)A$ it is indeed true that $$f(x)= M_b(f)Ax=\begin{pmatrix}7&-1\\-8&3 \end{pmatrix}x$$
and therefore $f(3,5)=(16,-9)^T$
A little background: If you want to describe a linear map with a matrix there are always two bases involved: An "input base" and an "output base". The $i$-th column of the matrix then has entries that are the coordinates of the image of the $i$-th vector of the "input base" with respect to the "output base". So if $(v_1, \dots, v_n)$ is the input base and $(w_1, \dots, w_m)$ is the output base and the representation of $f(v_i)$ in the output base reads
$$f(v_i) = a_1w_1 + \dots + a_mw_m$$
with some numbers $a_j$, $j=1, \dots, m$ then the $i$-th column of the matrix is precisely $(a_1, \dots, a_m)^T$.
If you want to evaluate $f(x)$ by multiplying $x$ with a matrix you need to make sure that the matrix and the vector are given with respect to the same basis. So if you want to find $f(3,5)$ then the vector $(3,5)$ is implicitely given as coordinates in the standard basis (because $(3,5) = 3 \cdot (1,0) + 5 \cdot (0,1)$) and therefore your matrix needs to have the standard basis as the input basis (which was not the case at first). Therefore we multiply the matrix describing the identity transformation with the standard base as input base and the base $B$ as output base.
