Question

T: R^3 --> R^3 such that
T(1,1,1)=(2,0,1)
T(0,1,0)=(-3,2-1)
T(1,0,1)=(1,1,0)

find T(2,-1,1)

Answer 1

You want to write (2,-1,1) as a linear combination of the other three 'input' vectors that you are given:
(2,-1,1) = a(1,1,1) + b(0,1,0) + c(1,0,1)

Then T(2,-1,1) = aT(1,1,1) + bT(0,1,0) + cT(1,0,1)

If you just look at this for a moment, you should be able to figure a,b,c out. But if you're really really stuck, solve the system of equations.

Answer 2

hey, the values of a, b and c are not really necessary to be equal to (2,-1,1) right ?

Answer 3

I don't quite understand what you're asking.

If you're asking "(a,b,c) is not necessarily equal to (2,-1,1)" the answer is no, they are in fact not.

But if you're asking, do we need to find the constants a,b,c such that
(2,-1,1) = a(1,1,1) + b(0,1,0) + c(1,0,1)
the answer is an definite yes.

Answer 4

yeah, im talking about the first one, okay sir ! thanks a lot !!

Answer 5

Hang on a second: are you sure these equations are correct?
T(1,1,1)=(2,0,1)
T(0,1,0)=(-3,2-1)
T(1,0,1)=(1,1,0)

In particular, I think one of the vectors on the left side is not correct.

Answer 6

and that's because the three vectors: (1,1,1), (0,1,0) and (1,0,1) are not linearly independent and do not span R^3.

In particular, there are no constants a,b,c such that:
(2,-1,1) = a(1,1,1) + b(0,1,0) + c(1,0,1)

Answer 7

okok ! wait sir, how can i know if a S spans R^3 ?

Answer 8

coz i just know that if the determinant of S in non zero then it spans R^3

Answer 9

Yes, if you took the determinant of the matrix comprising of these three vectors as columns or rows, it would be zero.

In any case, double check the question.

Answer 10

ok, so therefore if determinant is zero, is does not span?

Answer 11

yes