Solving $n\times n$ determinant using triangular shape I have just started learning to solve nth order determinants by getting it into the triangular shape ( in this way the determinant is equal to the multiple of main or additional diagonal + the determination of the sign ). I have solved a couple of easy ones, but got stuck on this one ( which seems easy, but however I manipulate the rows or columns I can't get it into the triangular shape ).
$$
        \begin{vmatrix}
        5 & 3 & 3 & \cdots & 3 & 3 \\
        3 & 6 & 3 & \cdots & 3 & 3 \\
        3 & 3 & 6 & \cdots & 3 & 3 \\
        \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
        3 & 3 & 3 & \cdots & 6 & 3 \\
        3 & 3 & 3 & \cdots & 3 & 6 \\
        \end{vmatrix}
$$
If someone could give the solution and explain the crucial steps to solving this, it would be very appreciated.

First, you can subtract the last line from the others which gives you :$$\begin{vmatrix}
   2 & 0 &\ldots&0&-3 \\
   0 & 3 &\ddots&\vdots& \vdots \\ 
   \vdots &\Large{0}&\ddots&0& \vdots \\
   0&\ldots&0&3&-3\\
  3&\ldots&\ldots&3&6
\end{vmatrix}$$
Then subtract $\frac{3}{2} L_{1}$ from $L_{n}$:
$$\begin{vmatrix}
   2 & 0 &\ldots&0&-3 \\
   0 & 3 &\ddots&\vdots& \vdots \\ 
   \vdots &\Large{0}&\ddots&0& \vdots \\
   0&\ldots&0&3&-3\\
  0&3&\ldots&3&\frac{3}{2}
\end{vmatrix}$$
Finally, subtract $L_{i}$ for every $i \in [2,n-1]$ from $L_{n}$:
$$\begin{vmatrix}
   2 & 0 &\ldots&0&-3 \\
   0 & 3 &\ddots&\vdots& \vdots \\ 
   \vdots &\Large{0}&\ddots&0& \vdots \\
   0&\ldots&0&3&-3\\
  0&\ldots&\ldots&0&\frac{3}{2}+3(n-2)
\end{vmatrix}=\begin{vmatrix}
   2 & 0 &\ldots&0&-3 \\
   0 & 3 &\ddots&\vdots& \vdots \\ 
   \vdots &\Large{0}&\ddots&0& \vdots \\
   0&\ldots&0&3&-3\\
  0&\ldots&\ldots&0&3(n-\frac{3}{2})
\end{vmatrix}$$
These operations don't affect the determinant so you can apply the formula for a triangular shaped matrix.
