<s>[INST] For all square matrices $A$ and $B$ of the same size, it is true that $(A+B)^2 = A^2 + 2AB + B^2$? The below statement is a true/false exercise.
Statement:
For all square matrices A and B of the same size, it is true that
$(A + B)2 = A^2 + 2AB + B^2$.
My thought process: Since it is not a proof, I figure I can show by example and come to a valid conclusion based on such example.
My work:
Come up with a square matrix A and B let both be a 2 by 2 matrix(rows and cols must be same).
Matrix $A$:
$A = \begin{array}{ccc}
3 & 5 \\
4 & 6 \\
\end{array} $
Matrix $B$:
$B =  \begin{array}{ccc}
5 & 8 \\
9 & 4 \\
\end{array} $
$A + B =  \begin{array}{ccc}
8 & 13 \\
13 & 10 \\
\end{array}$
$(A + B)^2 =  \begin{array}{ccc}
233 & 234 \\
234 & 264 \\
\end{array}$
$A^2 =  \begin{array}{ccc}
29 & 45 \\
36 & 56 \\
\end{array}$
$(AB) =  \begin{array}{ccc}
60 & 44 \\
74 & 56 \\
\end{array}$
$2(AB) =  \begin{array}{ccc}
120 & 88 \\
234 & 112 \\
\end{array}$
$B^2 =  \begin{array}{ccc}
97 & 72 \\
81 & 88 \\
\end{array}$
$A^2 + 2AB + B^2 =  \begin{array}{ccc}
246 & 205 \\
265 & 256 \\
\end{array}$
Based my above work, the answer is false. 
Is there another way to approach the problem? It seems like a lot of work needed to be done for a true/false question which raised my suspicion about whether there is a better way to look at the problem.
 [/INST] A counterexample is all you need, so you're done. You probably could have picked a simpler counterexample, say with most entries of $A$ and $B$ being $0,$ but yours works just fine.
As an alternative, note that if $A$ and $B$ are square matrices of the same size, then $$(A+B)^2=A(A+B)+B(A+B)=A^2+AB+BA+B^2.$$ From this, it follows that $$(A+B)^2=A^2+2AB+B^2$$ if and only if $AB=BA.$ So, any two square matrices $A,B$ of the same size such that $AB\neq BA$ will yield a counterexample.
</s>