Find the determinant of a 5x5 matrix Find the determinant of the following matrix: 
$$\begin{bmatrix}
1& 1& 1& 1& 1\\
3 & 3 &3 &3 &2\\
4& 4& 4& 3& 3\\
5& 5& 4& 4&  4\\
 6& 5& 5& 5 &5\end{bmatrix}$$
Laplace doesn't seem like the best method here, can we somehow turn this into a triangular matrix so that the determinant is the product of the elements on the main diagonal? 
I multiplied the first row by $(-3)$ and added it to he second one, then by $(-4)$ and added it to the third one, by $(-5)$ and added it to the fourth one, and by $(-6)$ and added it to last one. 
$$\begin{vmatrix}
1& 1& 1& 1& 1\\
3 & 3 &3 &3 &2\\
4& 4& 4& 3& 3\\
5& 5& 4& 4&  4\\
 6& 5& 5& 5 &5
\end{vmatrix}=\begin{vmatrix}
1& 1& 1& 1& 1\\
0& 0 &0 &0 &-1\\
0& 0& 0& -1& -1\\
0& 0& -1& -1&  -1\\
 0& -1& -1& -1 &-1
\end{vmatrix}$$
What should I do now?

Now you can expand by the first collumn, and you get
$$\begin{vmatrix} 0 & 0 &0 &-1 \\ 0 & 0 &-1 &-1 \\ 0 &-1 &-1 &-1 \\ -1 &-1&-1&-1  \end{vmatrix}$$
which is triangular, so the determinant equals $(-1)^4=1$.
