1. Understand the Goal
The problem states that the image (range space) of the linear transformation T is spanned or generated by two vectors:
v1=(1,2,0,−4)
v2=(2,0,−1,−3)
Any vector in the image of T must be a linear combination of these two generating vectors. We can find the general form of the transformation by setting:
T(x,y,z)=x⋅v1+y⋅v2
Note: Since the domain is R3, the transformation depends on three variables (x,y,z). However, because the image is only 2-dimensional (generated by 2 vectors), one of the standard basis components (like z) will map to the zero vector, or the variables can be rearranged to match the options. Let's test the standard basis alignment with the options.
2. Testing the Columns / Basis Vectors
Let's find the transformation by evaluating the linear combination:
T(x,y,z)=x(1,2,0,−4)+y(2,0,−1,−3)
Now, distribute the variables scalar-wise into each coordinate:
T(x,y,z)=(x,2x,0,−4x)+(2y,0,−y,−3y)
3. Simplify Component-by-Component
Add the corresponding components together:
Putting it all together into a single vector expression:
T(x,y,z)=(x+2y,2x,−y,−4x−3y)
