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Use the conjugacy class order of solution. The defining representation has characterIts character inner product with itself is $(4+4)/8=1$, proving that it is an irreducible representation. There is also a direct invariant subspace argument: $B$ has two distinct coordinate eigenspaces, and $A$ exchanges them, so no line is invariant under both.

Let $L_{st}$ denote the one-dimensional representation with $L_{st}(A)=s$ and $L_{st}(B)=t$. Then $L_{st}(-I)=1$ and $L_{st}(C)=st$. The complete character table isThe four scalar rows obey character orthogonality, and the dimensions exhaust the group order as established in solution.