# Determinant of a matrix with variables in it

Assuming that $z \neq 0$, compute the determinant $d_n(z) = \det D_n \left(1, z, 1 - \frac{1}{z^2} \right)$, where $z$ is a complex variable. In particular, compute the value $d_n(\sqrt{2})$.

This matrix looks like this:

$$\pmatrix{z& 1 - \frac{1}{z^2} & 0 & 0 & ... & 0 \\ 1 & z & 1 - \frac{1}{z^2} & 0 & ... & 0 \\ 0 & 1 & z & 1 - \frac{1}{z^2} &... &0\\ 0 & 0 & 1 & z &... &0\\ : & : & : & : & ... & : \\ \\0 & 0 & 0 & 0 &... & z}.$$

Now to solve it, blah blah blah, you do the whole decomposition thing and I end up with

$$d_n = zd_{n-1} - \left(1 - \frac{1}{z^2} \right) d_{n-2},$$

and so from here, I get the characteristic polynomial to be

$$x^2 - z x + \left( 1 - \frac{1}{z^2} \right) = 0,$$

which, according to the answers, is correct. Now, I need to solve this and so I get two roots and then I can use the formulas to find the determinant. So to solve it, I said let's first multiply through by $z^2$ to get rid of the denominator and then use the quadratic formula to solve. So multiplying through gives me

$$z^2 x^2 - z^3x + (z^2 - 1) = 0,$$

and then putting this into the quadratic formula gives me

$$x_{1,2} = \frac{z^3 \pm \sqrt{z^6 - (4 \cdot z^2 \cdot(z^2 - 1))}}{2z^2} = \frac{z^3 \pm \sqrt{z^6 - 4z^4 + 4z^2}}{2z^2}$$

which can be simplified to give me

$$\frac{z \pm \sqrt{z^4 - 4z^2 + 4}}{2} = \frac{z \pm \sqrt{(z^2 - 2)^2}}{2} = \frac{z \pm z^2 - 2}{2}.$$

This then ends up giving me two more qudratics to solve, when in the answers it says the roots to that characteristic polynomial should be

$$\frac{1}{z} \hspace{1cm} \mathrm{and} \hspace{1cm} z - \frac{1}{z}.$$

Where have I gone wrong?

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@julien It is $d_{n-2}$ because I decompose with the $1 - \frac{1}{z}$ and then in my matrix of minors I decompose with respect to the $1$ in row $2$ column $1$ and that gives me $d_{n-2}$. EDIT: Sorry, I think you misunderstood the matrix. Let me write in one more row. – Kaish May 13 '13 at 19:45
@julien Yeah sorry, I was lazy with the matrix. I've added in an extra row. – Kaish May 13 '13 at 19:48
No they don't. Because I get my exact same matrix again, with just two rows and columns gone, hence $d_{n-2}$. Literally just cover up rows $1$ and $2$ and columns $1$ and $2$ and you'll see there's no triangular matrix. – Kaish May 13 '13 at 19:56
The mistake isn't in that bit. The mistake is in solving for the roots. – Kaish May 13 '13 at 19:56
Sorry, I did not check your edit carefully enough. The recurrence formula is correct. – 1015 May 13 '13 at 19:59