If $$B^t=\left( \begin{array}{cccc} 1 & 3 & 5 & 6 \\ 0 & 1 & 0 & 2 \\ 0 & 5 & 4 & 1 \end{array} \right)$$
calculate the original matrix $$B$$.
See development and solution
Development:
Now we just have to swap rows with columns again:
$$B=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 3 & 1 & 5 \\ 5 & 0 & 4 \\ 6 & 2 & 1 \end{array} \right)$$
If we now calculate the transpose of this matrix we see that it is $$B^t$$.
Solution:
$$B=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 3 & 1 & 5 \\ 5 & 0 & 4 \\ 6 & 2 & 1 \end{array} \right)$$