# Solving $n$th order determinant

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.

• Use Laplace's formula for the first column.
– mfl
Oct 11 '17 at 14:21

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 $D_n=(n+1)2^n$.
• 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 Oct 11 '17 at 15:07
Alternatively: doing row operations to get a diagonal matrix: $$R1\cdot (-\frac14)+R2\to R2; \ R2\cdot (-\frac13)+R3\to R3; \ etc$$ $$\begin{vmatrix} 4 & 4 & 0 & 0 & \cdots & 0 & 0 \\ 1 & 4 & 4 & 0 & \cdots & 0 & 0 \\ 0 & 1 & 4 & 4 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & 4 & 4 \\ 0 & 0 & 0 & 0 & \cdots & 1 & 4 \\ \end{vmatrix}=\begin{vmatrix} 4 & 4 & 0 & 0 & \cdots & 0 & 0 \\ 0 & \frac{12}{4} & 4 & 0 & \cdots & 0 & 0 \\ 0 & 0 & \frac{32}{12} & 4 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & \frac{n\cdot 2^{n-1}}{(n-1)\cdot 2^{n-2}} & 4 \\ 0 & 0 & 0 & 0 & \cdots & 0 & \frac{(n+1)\cdot 2^n}{n\cdot 2^{n-1}} \\ \end{vmatrix}= \\ \prod_{k=1}^n \frac{(k+1)\cdot 2^k}{k\cdot 2^{k-1}}=(n+1)\cdot 2^n.$$