The three nonzero vectors in are , and . An invertible linear map permutes this three-element set, so part (b) gives a group homomorphism
If a matrix acts trivially on all three vectors, it fixes both basis vectors , so it is the identity matrix. Thus the kernel of a group homomorphism is trivial and is injective.
To count the domain, its first column may be any of the three nonzero vectors. The second column must be outside the two-element span of the first, leaving two choices. Hence
An injection between these finite sets of equal size is surjective, proving

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