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I have a determinant of nth order that I am not able to convert into a triangular shape. I believe that this determinant is quite easy, but I can't find a way to fully convert one of the corners into zeros. My other idea was to use the Laplace principle, but that didn't work as well. $$ \begin{vmatrix} 4 & 4 & 0 & 0 & \cdots & 0 & 0 \\ 1 & 4 & 4 & 0 & \cdots & 0 & 0 \\ 0 & 1 & 4 & 4 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & 4 & 4 \\ 0 & 0 & 0 & 0 & \cdots & 1 & 4 \\ \end{vmatrix} $$

If someone could present a detailed way of converting this determinant into a triangular shape, it would be much appreciated. In addition to that, maybe someone could give some tips for solving nth order determinant by converting it into a triangular shape, using the Laplace principle or any other more or less basic methods.

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    $\begingroup$ Use Laplace's formula for the first column. $\endgroup$ – mfl Oct 11 '17 at 14:21
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If you use Laplace's formula for the first column, as mfl suggests, you arrive at the recurrence relation $D_n=4(D_{n-1}-D_{n-2})$ with $D_1=4$ and $D_2=12$.

The solution is $D_n=(n+1)2^n$.

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  • $\begingroup$ Maybe you could explain this idea in depth ? I have been taught that the Laplace's principle is used when there are two or more rows/columns with zeros in corresponding places. Then those columns/rows can be removed and the minors of those elements could be calculated. Maybe this application of Laplace's theorem differs from yours. $\endgroup$ – Jonas Petraška Oct 11 '17 at 14:46
  • $\begingroup$ I don't understand what you say. "Corresponding places"? Laplace expansion is when you write the determinant as a sum over determinants of minors. This can always be used. $\endgroup$ – Marcel Oct 11 '17 at 14:51
  • $\begingroup$ Here's an example of how I was taught to use Laplace's principle : $$ \begin{vmatrix} 1 & 3 & 5 & -8 \\ 2 & 0 & 0 & 6 \\ -1 & 2 & 5 & 4 \\ -11 & 0 & 0 & 3 \\ \end{vmatrix} $$ Find minors of the removed rows and columns : $$ \begin{vmatrix} 2 & 6 \\ -11 & 3 \\ \end{vmatrix} $$ * $$ \begin{vmatrix} 3 & 5 \\ 2 & 5 \\ \end{vmatrix} $$ * $(-1)^{11}$ Then we solve the second order determinants and get the answer : -360 $\endgroup$ – Jonas Petraška Oct 11 '17 at 15:07
  • $\begingroup$ Just look up "Laplace expansion" in the Wikipedia, man $\endgroup$ – Marcel Oct 11 '17 at 15:12
  • $\begingroup$ Ok, thanks for the help ! $\endgroup$ – Jonas Petraška Oct 11 '17 at 15:17
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Alternatively: doing row operations to get a diagonal matrix: $$ R1\cdot (-\frac14)+R2\to R2; \ R2\cdot (-\frac13)+R3\to R3; \ etc$$ $$\begin{vmatrix} 4 & 4 & 0 & 0 & \cdots & 0 & 0 \\ 1 & 4 & 4 & 0 & \cdots & 0 & 0 \\ 0 & 1 & 4 & 4 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & 4 & 4 \\ 0 & 0 & 0 & 0 & \cdots & 1 & 4 \\ \end{vmatrix}=\begin{vmatrix} 4 & 4 & 0 & 0 & \cdots & 0 & 0 \\ 0 & \frac{12}{4} & 4 & 0 & \cdots & 0 & 0 \\ 0 & 0 & \frac{32}{12} & 4 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & \frac{n\cdot 2^{n-1}}{(n-1)\cdot 2^{n-2}} & 4 \\ 0 & 0 & 0 & 0 & \cdots & 0 & \frac{(n+1)\cdot 2^n}{n\cdot 2^{n-1}} \\ \end{vmatrix}= \\ \prod_{k=1}^n \frac{(k+1)\cdot 2^k}{k\cdot 2^{k-1}}=(n+1)\cdot 2^n.$$

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