Quick doubt about linear application and its matrix

[Felafel]

Homework Statement

Let $f:\mathbb{R}^3\to \mathbb{R}^3$ such that $v_1=(1,0,1) , v_2=(0,1,-1), v_3=(0,0,2)$ and $f(v_1)=(3,1,0), f(v_2)=(-1,0,2), f(v_3)=(0,2,0)$
find $M^{E,E}_f$ where $E=(e_1,e_2,e_3)$ is the canonical basis.

The Attempt at a Solution

i see
$v_1=e_1+e_3$
$v_2=e_2-e_3$
$v_3=2e_3$
thus
$f(e_1)+f(e_3)=(3,1,0)$
$f(e_2)-f(e_3)=(-1, 0 ,2)$
$2f(e_3)=(0,2,0)$
solving the system i get
$f(e_3)=(0,1,0)$
$f(e_1)=(3,0,0)$
$f(e_2)=(-1,1,2)$

and so I thought the matrix was:

(3 -1 0)
(0 1 1)
(0 2 0)

but according to my book the solution should be:
(3 0 0)
(-1 1 2)
(0 1 0)

why? shouldn't the image of the vector form the columns instead of the rows?
thank you in advance :)

 

[Dick]

Homework Statement

Let $f:\mathbb{R}^3\to \mathbb{R}^3$ such that $v_1=(1,0,1) , v_2=(0,1,-1), v_3=(0,0,2)$ and $f(v_1)=(3,1,0), f(v_2)=(-1,0,2), f(v_3)=(0,2,0)$
find $M^{E,E}_f$ where $E=(e_1,e_2,e_3)$ is the canonical basis.

The Attempt at a Solution

i see
$v_1=e_1+e_3$
$v_2=e_2-e_3$
$v_3=2e_3$
thus
$f(e_1)+f(e_3)=(3,1,0)$
$f(e_2)-f(e_3)=(-1, 0 ,2)$
$2f(e_3)=(0,2,0)$
solving the system i get
$f(e_3)=(0,1,0)$
$f(e_1)=(3,0,0)$
$f(e_2)=(-1,1,2)$

and so I thought the matrix was:

(3 -1 0)
(0 1 1)
(0 2 0)

but according to my book the solution should be:
(3 0 0)
(-1 1 2)
(0 1 0)

why? shouldn't the image of the vector form the columns instead of the rows?
thank you in advance :)

Your answer looks right assuming the vectors are column vectors. Perhaps the book is treating them as row vectors and multiplying the matrix on the right?

 

[Felafel]

i don't see why it would do that though.
it's just wrong, probably
thank you :)