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Mathematics 15 Online
OpenStudy (anonymous):

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

OpenStudy (watchmath):

The columns of A are T(e1), T(e2), T(e3).

OpenStudy (anonymous):

tx

OpenStudy (anonymous):

ok, I don't get it how you find A inverse?

OpenStudy (anonymous):

the first matrix?

OpenStudy (watchmath):

I think you made a mistake loser66. If we add the 2nd and the third equation we have \[ T(6e3)=[4, -2]\Leftrightarrow T(e3)=[2/3 ,-1/3] \]

OpenStudy (watchmath):

Go ahead :)

OpenStudy (anonymous):

tx, yes I got the first part

OpenStudy (anonymous):

right

OpenStudy (anonymous):

right

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

- T (e1) + T (e2) + 5 Te(3) = [ 2 -3]

OpenStudy (anonymous):

T e(1) - T (e2) + T(e3) = [2 1]

OpenStudy (anonymous):

I'll be right back,,,thank you

OpenStudy (watchmath):

Be careful loser66. The map is from R^3 to R^2. Hence A will be a 2 by 3 matrix

OpenStudy (watchmath):

You should write [3 -1], [2 -3] and [2 1] on your matrix as columns

OpenStudy (watchmath):

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

OpenStudy (watchmath):

yes, you are right that should be a plus

OpenStudy (watchmath):

I misscopied it

OpenStudy (watchmath):

Before you leave, your mistake is to treat T(e1), (e2), T(e3) as number and you put them as one column vector

OpenStudy (anonymous):

I'm here

OpenStudy (watchmath):

yes, they are vector so in this case we can't put it as \[ \begin{pmatrix}T(e_1)\\T(e_2)\\T(e_3)\end{pmatrix} \]

OpenStudy (anonymous):

you guys are great

OpenStudy (watchmath):

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)

OpenStudy (watchmath):

I am afraid I am not as generous as you are :). I think that is enough hint to do the problem.

OpenStudy (anonymous):

alittle bit more plz

OpenStudy (watchmath):

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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