Both are invertible, so each is P·(the other)·Q for suitable invertible P, Q — every invertible n×n matrix is equivalent to every other, because equivalence sees nothing finer than rank. But A has eigenvalue 1 and B has eigenvalue 2, and similarity preserves eigenvalues. Equivalence allows an independent change of basis in the domain and the codomain; similarity forces the same basis on both sides.
Counterexample bank
Part C is won by knowing which tempting claims are false. 149 counterexamples; 86 free. The rest come with the Notes pack.
Knowing the claim is false is half of it. The mistakes that cost marks shows the wrong options candidates actually pick, and the reasoning slip behind each one.
#1 · Analysis & Linear Algebra › Vector Spaces and Linear Maps
“Equivalent matrices are similar” — false
Counterexample: A = I₂ and B = 2I₂
linear algebra change of basis