First, a little theory:
Let T be a linear transformation from Vector space U to vector space V. It can be written as a matrix by choosing specific
bases for U and V. In particular, if
is a basis for U and
is a basis for V, then every vector in U can be written as "
" for some numbers a and b, and every vector in V can be written as "
" (I have assumed in this example that U has dimension 2 and V has dimension 3 because that is the case in your problem.)
When we have agreed to use a particular ordered basis, we can drop the basis vectors themselves and and write
as (a, b). Similarly, we can write
as (x, y, z).
That is how we can then write the linear transformation as a matrix: T is applied to members of U which can be represented by two numbers so the matrix for T must have two columns. The result is a vector in V which can be represented by 3 numbers so the matrix for T must have 3 rows. That much tells us it must be of the form
.
Now, a very nice thing happens! The basis vectors pf U, themselves,
and
are equal to
and
and so are represented by (1, 0) and (0, 1). Look what happens when we multiply our matrix by those:
and
That is, applying the linear transformation to the basis vectors gives us the columns of the matrix in turn!
If T is a linear transformation from U to V, and we are given bases for U and V, we find the matrix representing T by applying T to each basis vector of U in turn, writing the result in terms of the basis for V. Each of those gives a column for T.
Now, for this particular problem, T is from
to
so the matrix form for T will have 2 columns and 3 rows just like the example above. We are told to find the
standard matrix which means using the
standard bases: {(1,0), (0,1)} for
and {(1,0,0), (0,1,0), (0,0,1)} for
.
Now, what are T(1,0) and T(0,1)? Well, we are told that T(2,-1)= T(2(1,0)-1(0,1))= 2T(1,0)-1T(0,1)= (1,0,2) and that T(-3,4)= T(-3(1,0)+ 4(0,1))= -3T(1,0)+ 4T(0,1)= (3,1,-1).
Writing A= T(1,0) and B= T(0,1) just to simplify the notation, we have 2A- B= (1,0,2) and -3A+ 4B= (3,1,-1). Mutiply the fist equation by 4 to get 8A- 4B= (4,0,8) and add that to -3A+ 4B= (3,1,-1). The "B" cancel leaving 5A= (7,1,7) so A= (7/5,1/5,7/5). Putting that back into 2A- B= (1,0,2) gives (14/5,2/5,14/5)- B= (1,0,2) so B= (14/5,2/5,14/5)- (1,0,2)= (9/5, 2/5,4/5).
Now we know that T(1,0)= (7/5,1/5,7/5) and T(0,1)= (9/5,2/5, 4/5), the two columns of the matix, so the matrix representation of T is
(Assuming I haven't made any arithmetic mistakes- better check that!)