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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.

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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.

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  • $\begingroup$ I’ve never seen “subtract A to B”, but always “subtract A from B”. Do you mind if I ask where you’re from? Is that British English? $\endgroup$ – G Tony Jacobs Oct 6 '17 at 2:24
  • $\begingroup$ Sorry, it was a mistake. I am from France, it was definitely not British English. Thank you for the correction. $\endgroup$ – Théo D. Oct 6 '17 at 5:16
  • $\begingroup$ Thanks, I didn't think about subtracting fractions. $\endgroup$ – Jonas Petraška Oct 6 '17 at 6:28

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