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The matrix $D_n(2,3,1)$ is to be written in the form

$$\pmatrix{3 & 1 & 0 & 0 & ... & 0 \\ 2 & 3 & 1 & 0 & ... & 0 \\ 0 & 2 & 3 & 1 &... &0\\ : & : & : & : & ... & : \\0 & 0 & 0 & 0 &... & 3}$$

I need to workout the determinant of this matrix $( d_n = \det(D_n))$. So I said to expand by row $1$. Looking the element at $a_{11}$, I can cancel that row and column out and get the determinant of that bit to be $3d_{n-1}$. Now, by expanding with element $a_{12}$, I get the determinant of that bit to be $-2d_{n-1}$ and so I get

$$d_n = 3d_{n-1} - 2d_{n-2}$$

The charateristic equation of this gives me $x^2 - 3x +2 = 0 = (x - 2)(x - 1)$ and so I get the roots to be $x_{1,2} = 1,2$. As we have two distinct roots, (here's where my theory has gone a bit funny) we get that it can be written in the form $d_n = C_1x_1^n + C_2 x_2^n$. I solved this and got that $C_1 = 2$ and $C_2 = -1$, but I'm not sure how to complete the question.

Does this just mean that the determinant is $2(1)^n - 2^n = 2 - 2^n$?

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From your recursive formula $d_n-d_{n-1}=2(d_{n-1}-d_{n-2})$. Using $d_{1}=3$ and $d_{2}=7$ we get $d_{n}-d_{n-1}=2^n$. And thus $d_n=d_1+\sum_{k=2}^n{2^k}=3+2(2^n-2)=2^{n+1}-1$. –  Sebastien B Feb 11 '13 at 22:10

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up vote 2 down vote accepted

All seems right.

Verifying for $n=1,2$, I suspect that when you calculated $C_1$ and $C_2$, the order of $x_{1,2}$ was mixed: $x_1=2,\ x_2=1$, and $$2\cdot 2^n-1$$ seems OK for small $n$'s.

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Oh yeah, sorry I did. I used weird notation in my working so I tried to keep it consistent in here. So would I write the determinant as $2^{n+1} - 1$? –  Kaish Feb 11 '13 at 22:03
    
Yes, it seems right. –  Berci Feb 11 '13 at 22:06

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