Given a matrix A, how to find  B such that AB=BA Let $A = \begin{pmatrix}
1 &  1& 1\\ 
1 & 2 &3 \\ 
1 &4  & 5
\end{pmatrix}$ and  $D = \begin{pmatrix}
2 &  0& 0\\ 
0 & 3 &0 \\ 
0 &0  & 5
\end{pmatrix}$.
It is found that right-multiplication by D multiplies each column of A by the corresponding diagonal entry of D, whereas left-multiplication by D multiplies each row of A by the corresponding diagonal entry of D.
Construct a 3 x 3 matrix B, not the identity matrix or zero matrix, such that $AB=BA$.
 A: Note that the determinant of $A$ is $-2$, so your matrix is invertible i.e. there exists some matrix $B=A^{-1}$ such that $AB=BA=I$. If I calculated this correctly 
$$B = \begin{pmatrix}
1 &  1/2& -1/2\\ 
1 & -2 &1 \\ 
-1 & 3/2  & -1/2
\end{pmatrix}.$$
A: As Alex Becker said, we could use any scalar multiple of $I$ or $A^n$ where $A^n = A\cdot A \dots A$ where $A$ is multiplied by itself $A$ times.  We could also use $A^{-1}$ or $(A^{-1})^n$.
Suppose we have any matrices $B$ and $C$ that satisfy $AB=BA$ and $AC=CA$.  Then, $A(c_1 B + c_2 C) = c_1 A B + c_2 A C = c_1 BA + c_2 CA = (c_1 B + c_2 C)A$.
From this we conclude that the collection of all matrices that commute with $A$ (under matrix multiplication) forms a vector space, since any linear combinations of matrices that commute with $A$ will also commute with $A$.
A: Notice that $A$ is diagonalizable, so any matirx with the same eigenvectors will commute with $A$, for example
$$B = \begin{pmatrix}
0.2481 & 0.31385 & -0.034765 \\ 
0.48816 & -0.65821 & 0.76725 \\ 
-0.20907 & 1.0811 & 0.28335
\end{pmatrix}$$
