Determinant of a matrix

**joll** · July 9th 2011, 06:36 AM

Hello everyone,
Please help me with finding the determinant of
$\begin{vmatrix}x&-1&0&\cdots &0&0\\0&x&-1&\codts &0&0\\0&0&x&\ddots&\vdots&\vdots\\ \vdots&\vdot&\vdots &\ddots&-1&0\\0&0&0&\cdots&x&-1\\a_n&a_{n-1}&a_{n-2}&\cdots&a_1&a_0\end{vmatrix}$

I've tried to solve by myself, but I coudln't get anywhere. I do wish to improve my skill in mathematics; I am not asking for the complete answer. Just give me what to start, then I will try again. I do appreciate your help. Thank you.

**FernandoRevilla** · July 9th 2011, 07:23 AM

$D_2=\begin{vmatrix}{x}&{-1}\\{a_1}&{a_0}\end{vmatrix}=a_0x+a_1$

$D_3=\begin{vmatrix}{x}&{-1}&{0}\\{0}&{x}&{-1}\\{a_2}&{a_1}&{a_0}\end{vmatrix}=xD_2+a_2\begin{ vmatrix}{-1}&{\;\;0}\\{\;\;x}&{-1}\end{vmatrix}=a_0x^2+a_1x+a_2$

Conjecture $D_{n+1}=a_0x^n+a_1x^{n-1}+\ldots +a_n$ and use induction. The way we have found $D_3$ will help you.

**joll** · July 9th 2011, 04:11 PM

Thank you very much!
Assuming $D_{k+1} = a_0x^k+a_1x^{k-1}+\cdots +a_k$ ,
$D_{k+2}=xD_{k+1}+(-1)^{k+3}a_{k+1}\begin{vmatrix}-1&0&\cdots&\cdots&0\\x&-1&0&\cdots&0\\ \vdots&\vdots&\ddots&\ddots&0\\0&\cdots&\cdots&x&-1\end{vmatrix}$ .

and as $\begin{vmatrix}-1&0&\cdots&\cdots&0\\x&-1&0&\cdots&0\\ \vdots&\vdots&\ddots&\ddots&0\\0&\cdots&\cdots&x&-1\end{vmatrix}=(-1)^{k+1}$ , we get

$D_{k+2}=xD_{k+1}+a_{k+1}$ .

Therefore

$D_{k+2}=a_0^{k+1}+a_1x^{k}\cdots+a_{k+1}$

$D_2$ satisfies the first assumption, therefore via mathematical induction, $D_{n+1}=a_0x^n+a_1^{n-1}+\cdots+a_n\ \ \ ^\forall n\in \mathbb{Z},n\geq 0.$

Am I correct? Thank you very much for your help, and I will work even harder.

**FernandoRevilla** · July 9th 2011, 11:33 PM

Originally Posted by joll

Am I correct?

Yes, you are. Only one thing: it should be $n\geq 1$ instead of $n\geq 0$ . If $n=0$ , we have $D_{n+1}=D_1=\det [x]=x$ i.e. we have no $a_i$ and the rule defining the given determinant has no sense.

Thank you very much for your help,

You are welcome!

and I will work even harder.

Good. I am pretty sure.

**joll** · July 9th 2011, 11:51 PM

Thank you very much.

**NonCommAlg** · July 10th 2011, 12:59 AM

Originally Posted by joll

Hello everyone,
Please help me with finding the determinant of
$\begin{vmatrix}x&-1&0&\cdots &0&0\\0&x&-1&\codts &0&0\\0&0&x&\ddots&\vdots&\vdots\\ \vdots&\vdot&\vdots &\ddots&-1&0\\0&0&0&\cdots&x&-1\\a_n&a_{n-1}&a_{n-2}&\cdots&a_1&a_0\end{vmatrix}$

I've tried to solve by myself, but I coudln't get anywhere. I do wish to improve my skill in mathematics; I am not asking for the complete answer. Just give me what to start, then I will try again. I do appreciate your help. Thank you.

it can also be done without using induction. just expand the determinant along the last row and you'll quickly get the answer.

**joll** · July 10th 2011, 02:18 AM

I see... that way, the determinant can be written

$(-1)^{n+1}a_n\begin{vmatrix}-1&0&\cdots&0&0\\x&-1&\cdots&0&0\\0&x&\ddots&\vdots&\vdots\\ \vdots&&\ddots&-1&0\\0&0&\cdots&x&-1\end{vmatrix}+ (-1)^{n+2}a_{n-1}\begin{vmatrix} x&0&\cdots&0&0\\0&-1&\cdots&0&0\\0&x&\ddots&\vdots&\vdots\\ \vdots &&\ddots&-1&0\\0&0&\cdots&x&-1\end{vmatrix}\dots$ .

The determinent term in (n,i)cofactor (I mean, the \vmatrix part of the equation above) of the matrix is given by

$(-1)^{n-i}x^{i-1}$

Therefore

$D_{n+1}=a_0x^n+\dots a_n$ (am I correct?).

I haven't been learning linear algebra for long yet, so knowing many ways to solve one problem really helps me. Thank you very much.

**NonCommAlg** · July 10th 2011, 04:09 PM

Originally Posted by joll

I see... that way, the determinant can be written

$(-1)^{n+1}a_n\begin{vmatrix}-1&0&\cdots&0&0\\x&-1&\cdots&0&0\\0&x&\ddots&\vdots&\vdots\\ \vdots&&\ddots&-1&0\\0&0&\cdots&x&-1\end{vmatrix}+ (-1)^{n+2}a_{n-1}\begin{vmatrix} x&0&\cdots&0&0\\0&-1&\cdots&0&0\\0&x&\ddots&\vdots&\vdots\\ \vdots &&\ddots&-1&0\\0&0&\cdots&x&-1\end{vmatrix}\dots$ .

The determinent term in (n,i)cofactor (I mean, the \vmatrix part of the equation above) of the matrix is given by

$(-1)^{n-i}x^{i-1}$

Therefore

$D_{n+1}=a_0x^n+\dots a_n$ (am I correct?).

I haven't been learning linear algebra for long yet, so knowing many ways to solve one problem really helps me. Thank you very much.

yes, you've got the idea but you forgot that the original determinant is $(n+1) \times (n+1)$ not $n \times n$ .

so, after expanding along the last row, you'll get that the determinant is equal to

$\sum_{i=1}^{n+1}(-1)^{n+1+i}a_{n+1-i}(-1)^{n+1-i}x^{i-1}=\sum_{i=1}^{n+1}a_{n+1-i}x^{i-1}=a_n + \ldots + a_0x^n.$

**joll** · July 11th 2011, 12:57 AM

Ah..yes, I forgot that. Thank you.

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