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 specificbasesfor 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).Thatis 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 thestandardmatrix which means using thestandardbases: {(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!)