Product of 3 Matrices Is this identity true for all possible values of $A,B$ and $C$?
$$A^TBC = C^TBA$$
where either 


*

*$A,B,C$ are square matrices of same size

*$A$ and $C$ are vectors of size $n$ and $B$ is square matrix of size $n\times n$

 A: Recall, to prove that a statement is not an identity (an equality that holds for all $A, B, C$), you need only find one counterexample that satisfies the premise, but fails to satisfy the equality/identity. 

$(1)$ Does the identity hold for all $A, B, C$ where $A, B, C$ are $n\times n$ square matrices for some $n$? For example, consider the following matrices:
Let $$A = 
\begin{bmatrix}\
1 &2\\
1 & 0\\
\end{bmatrix}
,\quad B=
\begin{bmatrix}\
\,\,1 & 1 \\
-1 & 2\\
\end{bmatrix},\quad
C=
\begin{bmatrix}\
2 & -1\\
\,\,0 & 1\\
\end{bmatrix}.\
$$



*

*You'll first need to find $A^T$ and $C^T$.

*Then compute $A^TBC$ and $C^TBA$.  

*Does $A^TBC = C^TBA$?

$(2)$ Does the identity hold for all $A, B, C$ when $A$ and $C$ are each an $n\times 1$ column vector and $B$ is an $n\times n$ square matrix? ($A$ and $C$ must be $n\times 1$ column vectors if $A^TBC$ and $C^TBA$ are to be defined.) 


*

*Note that $A^TBC$ and $C^TBA$ each evaluate to a scalar. 

*Try experimenting with different possibilities for $A, B, C$. Start simple; for example, try working with $n = 2$.

*Again, if you can find $A, B, C$ such that $A^TBC \neq C^TBA$, then $(2)$ cannot hold for all such $A, B, C$.



