Finding a matrix from equation we've got the following 4x4 Matrix 
$$\begin{pmatrix}
4 & -2 & 3 & 2\\ 
3 & 5 & 1 & -4\\
-1 & 6 & -4 & -7\\
-2 & 0 & -2 & 4
\end{pmatrix}$$
and I need to find $B$ from the equation: $(A-3I)B=0$.
i started to solve it by finding first $A-3I$. and I got:
$$\begin{pmatrix}
1 & -2 & 3 & 2\\ 
3 & 2 & 1 & -4\\
-1 & 6 & -7 & -7\\
-2 & 0 & -2 & 1
\end{pmatrix}$$
now I know that every column $[AB]^j$ [$j$ represents column number] can be calculated by $A[B]^j$ [$j$ represents column number].
and I was trying to solve it by multiplying $A$ with a specific column in $B$ but I wasn't able to reach the zero matrix.
EDIT:
almost forgot to mention that B needs to be 4x4 matrix and different form 0!

$$\begin{pmatrix}
1 & -2 & 3 & 2\\ 
3 & 2 & 1 & -4\\
-1 & 6 & -7 & -7\\
-2 & 0 & -2 & 1
\end{pmatrix}$$
is a zero divisor since : column$(3)=$column$(1)-$column$(2)$.
So the matrix: 
$$B=\begin{pmatrix}
0 & 0 & -a & 0\\ 
0 & 0 & a & 0\\
0 & 0 & a & 0\\
0 & 0 & 0 & 0
\end{pmatrix}$$
gives $AB=0$
