Proof.
Assume (2). It suffices to show that B \otimes _ k k' is a matrix algebra. We know that B \otimes _ k B^{op} \cong \text{End}_ k(B). Since k' is the centralizer of k' in B^{op} by Lemma 11.7.3 we see that B \otimes _ k k' is the centralizer of k \otimes k' in B \otimes _ k B^{op} = \text{End}_ k(B). Of course this centralizer is just \text{End}_{k'}(B) where we view B as a k' vector space via the embedding k' \to B. Thus the result.
Assume (1). This means that we have an isomorphism A \otimes _ k k' \cong \text{End}_{k'}(V) for some k'-vector space V. Let B be the commutant of A in \text{End}_ k(V). Note that k' sits in B. By Lemma 11.7.2 the classes of A and B add up to zero in \text{Br}(k). From the dimension formula in Theorem 11.7.1 we see that
[B : k] [A : k] = \dim _ k(V)^2 = [k' : k]^2 \dim _{k'}(V)^2 = [k' : k]^2 [A : k].
Hence [B : k] = [k' : k]^2. Thus we have proved the result for the opposite to the Brauer class of A. However, k' splits the Brauer class of A if and only if it splits the Brauer class of the opposite algebra, so we win anyway.
\square
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