Final answer:
After analyzing, transformation (a) T([x y])=[-y -x] is both one-to-one and onto; transformation (b) T([x y z])=[x+2y-z -x+y+z] is also both one-to-one and onto; and transformation (c) T([x y])=[x+y -x+2y 3y] is neither one-to-one nor onto.
Step-by-step explanation:
The student's question revolves around determining whether the given linear transformations are one-to-one, onto, both, or neither. Let's analyze each transformation separately:
- Transformation (a): T([x y])=[-y -x]. To check if T is one-to-one, we see if two different inputs result in different outputs. Since swapping x and y and changing their signs will always result in a unique output for each unique input, T is one-to-one. To check if T is onto, every possible output vector in the codomain should be achievable. Since we can find an input that results in any given output, T is also onto. Hence, T is both one-to-one and onto.
- Transformation (b): T([x y z])=[x+2y-z -x+y+z]. To test for one-to-one, we set x+2y-z = a and -x+y+z = b for variables a and b representing any vector in the codomain. Solving these equations simultaneously, we can see that we can obtain unique values of x, y, z for unique values of a and b, indicating that T is one-to-one. For onto, we prove similarly that for every a and b in the output, we can find x, y, and z to satisfy the transformation, hence T is onto as well. Therefore, T is both one-to-one and onto.
- Transformation (c): T([x y])=[x+y -x+2y 3y]. To verify if T is one-to-one, we check for different inputs leading to unique outputs. However, if we set x=0 and y=0, we get the zero vector, and if we set x=2 and y=-2, we also get the zero vector, so T is not one-to-one. To check if T is onto, we see that since the third component is 3y, if we wanted an output vector with a third component that is not a multiple of 3, the transformation cannot provide it, so T is not onto. Hence, T is neither one-to-one nor onto.