How to solve the determinant?

Question

Ok so we know that it is an n order determinant but I do not know how to calculate it $\begin{vmatrix} 0 & a & a^2 & ... &a^{n-1} \\ \frac{1}{a} &0 & a & ... &a^{n-2} \\ \frac{1}{a^2}&\frac{1}{a} & 0 & ... &a^{n-3} \\ ...& ... & ... &... &... \\ \frac{1}{a^{n-1}} & \frac{1}{a^{n-2}} & \frac{1}{a^{n-3}} & ... & 0 \end{vmatrix}$

Answer 1

Scalar multiplication on second row with an 'a', third row with a squared, and so on until the last row with an 'a' to the power of n-1 on the last row. You will end up seeing something interesting. Do some row operation to clean out such as using the first row to subtract the last row. Using the property of determinant from 3 types of row operated matrix. You'll find the answer. I'm new to this site, so I didn't learn how to type in this format very well. sorry bro

Answer 2

Denote the matrix in question as $A$. The matrix $$A + I = \begin{bmatrix} 1 \\ \frac{1}{a} \\ \cdots \\ \frac{1}{a^{n-1}}\end{bmatrix} \cdot \begin{bmatrix} 1, a, \cdots, a^{n-1}\end{bmatrix}$$ is of rank $1$. Therefore, its eigenvalues are $0$ with multiplicity $n - 1$ and $n$ with multiplicity $1$. See for example this question.

Note that, if $\lambda$ is an eigenvalue for $A + I$, then $\lambda - 1$ is an eigenvalue for $A$. Therefore, the eigenvalues of $A$ are $-1$ with multiplicity $n - 1$ and $n - 1$ with multiplicity $1$. Thus $$ |A| = (-1)^{n-1}(n-1) $$ is the final answer.

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Another method that is via basic row operations is as follows. I show the details so that you can follow it.

• First, multiplying the second row of $A$ by $a$ and subtracting it from the first row, we get $$\begin{vmatrix} -1 & a & 0 & ... & 0 & 0 \\ \frac{1}{a} &0 & a & ... &a^{n-3} & a^{n-2} \\ \frac{1}{a^2}&\frac{1}{a} & 0 & ... & a^{n-4} & a^{n-3} \\ ...& ... & ... &... &... & ... \\ \frac{1}{a^{n-2}} & \frac{1}{a^{n-3}} & \frac{1}{a^{n-4}} & ... & 0 & a \\ \frac{1}{a^{n-1}} & \frac{1}{a^{n-2}} & \frac{1}{a^{n-3}} & ... & \frac{1}{a} & 0 \end{vmatrix}$$

• Second, multiplying the $3$rd row of $A$ by $a$ and subtracting it from the $2$nd row, we get $$\begin{vmatrix} -1 & a & 0 & ... & 0 & 0 \\ 0 & -1 & a & ... & 0 & 0 \\ \frac{1}{a^2}&\frac{1}{a} & 0 & ... & a^{n-4} & a^{n-3} \\ ...& ... & ... &... &... & ... \\ \frac{1}{a^{n-2}} & \frac{1}{a^{n-3}} & \frac{1}{a^{n-4}} & ... & 0 & a \\ \frac{1}{a^{n-1}} & \frac{1}{a^{n-2}} & \frac{1}{a^{n-3}} & ... & \frac{1}{a} & 0 \end{vmatrix}$$

• You can continue this routine by multiplying the $i+1$th row by $a$ and substracting it from the $i$th row, for $3 < i \leq n - 1$ and we finally obtain $$\begin{vmatrix} -1 & a & 0 & ... & 0 & 0 \\ 0 & -1 & a & ... & 0 & 0 \\ 0 & 0 & -1 & ... & 0 & 0 \\ ...& ... & ... &... &... & ... \\ 0 & 0 & 0 & ... & -1 & a \\ \frac{1}{a^{n-1}} & \frac{1}{a^{n-2}} & \frac{1}{a^{n-3}} & ... & \frac{1}{a} & 0 \end{vmatrix}$$

• We deal with the last row as follows: For $1 \leq i \leq n - 1$ in order, repeatedly multiply the $i$th row by $\frac{i}{a^{n-i}}$ and add it to the last row, leading to an upper triangle matrix $$\begin{vmatrix} -1 & a & 0 & ... & 0 & 0 \\ 0 & -1 & a & ... & 0 & 0 \\ 0 & 0 & -1 & ... & 0 & 0 \\ ...& ... & ... &... &... & ... \\ 0 & 0 & 0 & ... & -1 & a \\ 0 & 0 & 0 & ... & 0 & n-1 \end{vmatrix}$$ Therefore, the determinant is $(-1)^{n-1}(n-1)$, which is the product of the elements on the diagonal of the triangle matrix.