Let T: R3 →R3 be a Linear transformation such that T (2e1 – e2 = 5e3 ) =[3 -1] T( -e1+e2+5e3) = [ 2 -3] T( e1 – e2 + e3) = [ 2 1] Find the matrix A of transformation such that T(x) = A x
The columns of A are T(e1), T(e2), T(e3).
tx
ok, I don't get it how you find A inverse?
the first matrix?
I think you made a mistake loser66. If we add the 2nd and the third equation we have \[ T(6e3)=[4, -2]\Leftrightarrow T(e3)=[2/3 ,-1/3] \]
Go ahead :)
tx, yes I got the first part
right
right
ok
- T (e1) + T (e2) + 5 Te(3) = [ 2 -3]
T e(1) - T (e2) + T(e3) = [2 1]
I'll be right back,,,thank you
Be careful loser66. The map is from R^3 to R^2. Hence A will be a 2 by 3 matrix
You should write [3 -1], [2 -3] and [2 1] on your matrix as columns
We don't have to find the invers to solve this problem. The idea is to write e1,e2,e3 as linear combination of 2e1 – e2 = 5e3, -e1+e2+5e3 and e1 – e2 + e3
yes, you are right that should be a plus
I misscopied it
Before you leave, your mistake is to treat T(e1), (e2), T(e3) as number and you put them as one column vector
I'm here
yes, they are vector so in this case we can't put it as \[ \begin{pmatrix}T(e_1)\\T(e_2)\\T(e_3)\end{pmatrix} \]
you guys are great
To find \(T(e_1)\) first we need to find \(a,b,c\in R\) such that \[ e_1=a(2e1 – e2 + 5e3)+b( -e1+e2+5e3)+c( e1 – e2 + e3) \] once we know \(a,b,c\) then by linearity we can compute T(e1). The same goes gor T(e2) and T(e3)
I am afraid I am not as generous as you are :). I think that is enough hint to do the problem.
alittle bit more plz
if we simplify the right hand side by collecting similar term we have \[ e1=(2a-b+c)e1+(-a-b-c)e2+(5a+5b+c)e3 \] Now comparing the left and the right we have 2a-b+c=1, -a-b-c=0 and 5a+5b+c=0. So you have a system of equation and you can solve it using gaussian elimination.
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