# Determinant of M [closed]

How to find the determinant of the $n\times n$ matrix $M$, whose all the entries are zero except 1st row, 1st column and diagonal entries:

$$M= \begin{bmatrix} -x & a_2 & a_3 & \cdots & a_n\\ a_1+x & -x & 0 &\cdots & 0\\ a_1+x&0&-x&\cdots &0\\ \vdots &\vdots & &\ddots & \vdots \\ a_1+x&0&0&\cdots&-x\\ \end{bmatrix}.$$

## closed as off-topic by Fly by Night, Ken, hardmath, 6005, Davide GiraudoAug 4 '15 at 14:01

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Fly by Night, Ken, hardmath, 6005, Davide Giraudo
If this question can be reworded to fit the rules in the help center, please edit the question.

• How about expanding along the first row? – Omnomnomnom Aug 4 '15 at 12:19
• Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. – Fly by Night Aug 4 '15 at 12:45
• Your question was put on hold, the message above (and possibly comments) should give an explanation why. (In particular, this link might be useful.) You might try to edit your question to address these issues. Note that the next edit puts your post in the review queue, where users can vote to reopen this. (Therefore it would be good to avoid minor edits and improve your question as much as possible with the next edit.) – Martin Sleziak Aug 4 '15 at 14:03

If $\lambda$ is an eigenvalue of $M$, then $\mu = \lambda + x$ is an eigenvalue of $A = M + xI$. Equivalently, if $\mu$ is an eigenvalue of $A$, then $\mu - x$ is an eigenvalue of $M$. Note that $A$ is given by $$A = \pmatrix{ 0 & a_2 & a_3 & \cdots & a_n\\ a_1+x & 0 & 0 &\cdots & 0\\ a_1+x&0&0&\cdots &0\\ \vdots &\vdots & &\ddots & \vdots \\ a_1+x&0&0&\cdots&0\\ }$$ $A$ has a rank of $2$. So, all of its eigenvalues except for $2$ are equal to zero. The sum of the eigenvalues is the trace of $A$, which is zero. So, the non-zero eigenvalues of $A$ are given by $\mu,-\mu$ for some $\mu \in \Bbb C$.
So, the non-zero eigenvalues of $A^2$ are $\mu^2,\mu^2$, so that the trace of $A^2$ is $2 \mu^2$. We therefore calculate $$2\mu^2 = 2(a_1+x)(a_2 + a_3 + \cdots + a_n)$$ So, we have $$\mu = \sqrt{(a_1+x)(a_2 + a_3 + \cdots + a_n)}$$ Finally, the determinant of $M$ is the product of its eigenvalues. So, we have $$\det(M) = (\mu - x)(-\mu - x)(-x)^{n-2} = (x^2 - \mu^2)(-x)^{n-2} =\\ (-1)^n x^{n-2}(x^2 - (a_1+x)(a_2 + a_3 + \cdots + a_n))$$
Call the determinant $D_n(a_1,a_2,\ldots,a_n;x)$. Perform the following operations:
• row $n$ replaced by row $n$ minus row $n-1$; and then
• column $n-1$ replaced by column $n-1$ plus column $n$.
This gives an equal determinant, in which the last row is all zero except for $-x$ at the end. Expanding along the last row, $$D_n(a_1,a_2,\ldots,a_n;x)=-xD_{n-1}(a_1,a_2,\ldots,a_{n-1}+a_n;x)\ .$$ Repeating the procedure gives \eqalign{D_n(a_1,a_2,\ldots,a_n;x) &=(-x)^{n-2}D_2(a_1,a_2+\cdots+a_n;x)\cr &=(-x)^{n-2}(x^2-(a_1+x)(a_2+\cdots+a_n))\ .\cr}