suppose we want to find out how T transforms vectors in B'-coordinates.

well, all we know at the moment, is what T does to B-coordinates (the standard ones).

but whatever [T]_{B'}is, we know that the first column of [T]_{B'}is:

[T]_{B'}[b_{1}]_{B'}.

so let's find [T(b_{1})]_{B}= [T]_{B}[b_{1}]_{B}, and then figure out how to express this in B'-coordinates.

in B-coordinates, b_{1}is (1,1,1). and T(1,1,1) = (5,7,5). since this is 5(1,1,1) + 2(1,1,0) - 2(1,0,0), in B'-coordinates, this is [5,2,-2]_{B'}.

ok, now let's do the same with the 2nd B' basis vector (1,1,0). computing T(1,1,0) = (4,5,1) = 1(1,1,1) + 4(1,1,0) - 1(1,0,0), so this is [1,4,-1]_{B'}.

finally, we have T(0,0,1) = (1,2,4) = 4(1,1,1) - 2(1,1,0) - 1(1,0,0), so this is [4,-2,-1]_{B'}.

therefore, it would seem that [T]_{B'}=

.

well, let's see if this checks out.

we should have [T]_{B'}= P^{-1}[T]_{B}P, where:

now [T]_{B}P =

(these numbers should look familiar, they are the images of the B'-basis under T in the standard basis).

and P^{-1}[T]_{B}P =

i have NO idea what you are doing with the row-reduction in part (b).