I would list the stem first and then give my two solutions, but the answers are not the same, something must be wrong here.
question：
matrix A is $ \begin{pmatrix} 1 & 0& 0 &1 \\ 0& 1 & 1 & 0\\ 0& 1& -1 &0 \\ 1& 0& 0&-1 \end{pmatrix} $ and matrix B is $ \begin{pmatrix} 0& 0& 0 &1 \\ 0& 0 & 1 & 0\\ 0& 1& 0 &0 \\ 1& 0& 0&0 \end{pmatrix} $
Then,what is the value of matrix ABA?
1. My method one:
Use block matrix:
matrix A = $ \begin{pmatrix} 1 & 0& 0 &1 \\ 0& 1 & 1 & 0\\ 0& 1& -1 &0 \\ 1& 0& 0&-1 \end{pmatrix} =\begin{pmatrix} E & C\\ C&-E \end{pmatrix} $ and matrix B = $ \begin{pmatrix} 0& 0& 0 &1 \\ 0& 0 & 1 & 0\\ 0& 1& 0 &0 \\ 1& 0& 0&0 \end{pmatrix} =\begin{pmatrix} 0 & C\\ C&0 \end{pmatrix}$
so matric ABA = $\begin{pmatrix} E & C\\ C&-E \end{pmatrix} \begin{pmatrix} 0 & C\\ C&0 \end{pmatrix} \begin{pmatrix} E & C\\ C&-E \end{pmatrix} = \begin{pmatrix} 2E&0 \\ 0&-2E \end{pmatrix} =\begin{pmatrix} 2& 0& 0 &0 \\ 0& 2 & 0 & 0\\ 0& 0& -2 &0 \\ 0& 0& 0&-2 \end{pmatrix}$
2. My method two:
There should be something wrong with this method, but I can't find it.
matrix A = $ \begin{pmatrix} 1 & 0& 0 &1 \\ 0& 1 & 1 & 0\\ 0& 1& -1 &0 \\ 1& 0& 0&-1 \end{pmatrix} =\begin{pmatrix} 1 & 1& 0 &0 \\ 1& -1 & 0 & 0\\ 0& 0& 1 &1 \\ 0& 0& 1&-1 \end{pmatrix}= \begin{pmatrix} D & 0\\ 0& D \end{pmatrix} $
and matrix B = $ \begin{pmatrix} 0& 0& 0 &1 \\ 0& 0 & 1 & 0\\ 0& 1& 0 &0 \\ 1& 0& 0&0 \end{pmatrix}= \begin{pmatrix} 1& 0& 0 &0 \\ 0& 1 & 0 & 0\\ 0& 0& 1 &0 \\ 0& 0& 0&1 \end{pmatrix}=E $
so matric ABA =AEA= $A^2= \begin{pmatrix} D^2 & 0\\ 0& D^2 \end{pmatrix} =\begin{pmatrix} 2& 0& 0 &0 \\ 0& 2 & 0 & 0\\ 0& 0& 2 &0 \\ 0& 0& 0&2 \end{pmatrix}$
You can notice that the two methods get different answers, the problem should be with method 2, but I can't find it.
thanks for your help!
I will compile my comments into an answer.
If you do elementary transformations on your matrices, the resulting matrices are not equal to the starting ones, and you can't expect when you multiply different matrices that you'll get the same result. Let me show you why that can't work using $1×1$ matrices, aka numbers.
Let $A=1$ and $B=2$. Then, $ABA=2$. Let me now do elementary transformations on $A$ and $B$, let's say multiply its rows by $3$ to get $A′=3$, $B′=6$. Note that $A′B′A′=54≠2$.
It seems like you'd like to do something like this: $A′=S^{−1}AS$, $B′=S^{−1}BS$. Then, $$A′B′A′=(S^{−1}AS)(S^{−1}BS)(S^{−1}AS)=S^{−1}(ABA)S,$$ so $ABA=S(A′B′A′)S^{−1}.$ Note that result is still not necessarily equal, but by writing transformations explicitly, you can extract the product you want.
Also note that I do same transformations both to $A$ and $B$, so that $S^{−1}$ and $S$ cancel when we expand the product $A'B'A'$.
Let's say that $S$ is matrix corresponding to switching 2nd and 4th row/column (depending on which side we multiply). It's clear that $S^{−1} = S$, since if we do the switch twice, we get back to the same matrix.
Now,
$$A'\! =\! S^{-1}AS\! =\!\! \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ \end{array}\right)\!\! \left( \begin{array}{cccc} 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & -1 & 0 \\ 1 & 0 & 0 & -1 \\ \end{array} \right)\!\! \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ \end{array} \right)\!\! =\!\! \left( \begin{array}{cccc} 1 & 1 & 0 & 0 \\ 1 & -1 & 0 & 0 \\ 0 & 0 & -1 & 1 \\ 0 & 0 & 1 & 1 \\ \end{array} \right)$$ (notice that you have a mistake in your calculation).
Let's do the same for $B$:
$$ B' = S^{-1}BS = \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ \end{array}\right) \left( \begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ \end{array} \right) = \left( \begin{array}{cccc} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ \end{array} \right) $$
You can verify that
$$A'B'A' = \left( \begin{array}{cccc} 2 & 0 & 0 & 0 \\ 0 & -2 & 0 & 0 \\ 0 & 0 & -2 & 0 \\ 0 & 0 & 0 & 2 \\ \end{array} \right)$$ and that $$ABA = S(A'B'A')S^{-1} = \left( \begin{array}{cccc} 2 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & -2 & 0 \\ 0 & 0 & 0 & -2 \\ \end{array} \right).$$