Actually, there is an error here: with a= 0, b= 1, c= -6, not -3. A basis for the image is {(1, 0, -3); (0, 1, -6)}

Yes, that is perfectly valid. Any you can check that a= 1, b= -1, c= 3 and a= -2, b= 1, c= 0 also satisfy c+3a+6b = 0.I wonder if I get an equivalent result if I do it the following manner:

of course, in this case I'll need to check if are them linearly independent if I want to get a base.

thanks in advice

VIKKO

(If they were not independent that linear transformation would be singular and map [itex]R^2[/itex] into aonedimensional subspace of [itex]R^3[/itex].)