3
$\begingroup$

So, I have a matrix
$$ A = \begin{pmatrix} 0 & 1 & 1 & ... & 1 \\ 1 & 0 & x & ... & x \\ 1 & x & 0 & ... & x \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & x & x & ... & 0 \end{pmatrix} $$ I need to evaluate it's determinant. At first I calculated $\det A$ for $n=2,3,4$. And I got a pattern $\det A=(-1)^{n-1}(n-1)x^{n-2}$ for every $n\ge2$. But I need to solve it differently. I added all rows to the first, then multiplied every row (except 1) by $(1+x(n-2))$ and subtracted first row multiplied by $x$. This is what I got: $$ \begin{pmatrix} n-1 & 1+x(n-2) & 1+x(n-2) & ... & 1+x(n-2) \\ 1-x & -x(1+x(n-2)) & 0 & ... & 0 \\ 1-x & 0 & -x(1+x(n-2)) & ... & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1-x & 0 & 0 & ... & -x(1+x(n-2)) \end{pmatrix} $$ Now I can multiply diagonal elements, but I don't know what can I do with the rest of it. Any hints will be helful.

$\endgroup$

1 Answer 1

3
$\begingroup$

First factor out $1+x(n-2)$ from all columns except the first to get $$\det A(x)=\begin{vmatrix} n-1 & 1 & 1 & \cdots & 1 \\ 1-x & -x & 0 & \cdots & 0 \\ 1-x & 0 & -x & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1-x & 0 & 0 & \cdots & -x \end{vmatrix}$$ and note that this cancels the multplication by $(1+x(n-2))^{n-1}$ from before. Now multiply the first column by $x$ and the rest of them by $(1-x)$ and then add them all to the first one: $$\det A(x)=\frac1{x(1-x)^{n-1}}\begin{vmatrix} n-1 & 1-x & 1-x & \cdots & 1-x \\ 0 & -x(1-x) & 0 & \cdots & 0 \\ 0 & 0 & -x(1-x) & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & -x(1-x) \end{vmatrix}$$

which yields

$$\det A = \frac{(n-1) (-x(1-x))^{n-1}}{x(1-x)^{n-1}} = (-1)^n(n-1)x^{n-2}.$$

Note that this calculation was assuming that $x \notin \left\{0,1,-\frac1{n-2}\right\}$ so these cases should be considered separately. Alternatively, $\det A(x)$ is a continuous function in $x$ equal to $(-1)^n(n-1)x^{n-2}$ on the dense set $\Bbb{R}\setminus \left\{0,1,-\frac1{n-2}\right\}$ so it must be $\det A(x) = (-1)^n(n-1)x^{n-2}$ everywhere.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .