How to evaluate the determinant of this $n \times n$ matrix?

Find the following determinant: $$\begin{vmatrix} 0 &1&0&0 &\cdots& 0\\ 0 &0&1&0 &\cdots& 0\\ 0 &0&0&1 &\cdots& 0 \\ \vdots & \vdots &\cdots&\cdots&\cdots&\vdots\\0&0&0&0 &\cdots&1\\1&0&0&0 &\cdots&0\end{vmatrix}_{n \times n}$$

Attempt:
I switched $$1^{st}$$ row with the last one, then second with the last, and so on till the $$(n-1)^{th}$$ row with the $$n^{th}$$ (last) row and there I had all $$1$$'s in the diagonal so I came up with the answer $$1$$.

Am I correct? Can anyone solve this exercise please?

• Yes. Zero. ${}{}{}{}$ – copper.hat Apr 7 at 7:53
• how? can you explain please – Jr Someone Apr 7 at 7:54
• @copper.hat I don't think so. Note that the bottom left corner is a $1$. – YiFan Apr 7 at 7:55
• Have you tried any simple examples to see if there is a pattern? Have you tried anything at all? Do you expect people to click on your link to look at your problem and them explain it to you without any context or obvious effort? This is sheer & utter laziness. – copper.hat Apr 7 at 7:58
• Also, welcome to the website. Your original question did not meet the quality criteria here since you didn't provide any context: information about what you've tried to solve the question, where you found it, why is it relevant, etc.. Failing to provide context may result in downvotes/closure. Additionally, it is not considered good practice to post links to images of questions, since the links may expire with time. Refer here for a basic guide on Mathjax to be able to format your Math equations. – Shubham Johri Apr 7 at 8:27

Hint: row exchanges change the value of the determinant by a factor of $$-1$$, and the determinant of the identity is $$1$$. How do you make your matrix the identity by exchanging rows?
The corresponding matrix is the companion matrix for the polynomial $$f(x)=x^n-1$$ So the characteristic polynomial for this matrix is $$f(x)$$. So determinant is same as product of all $$n$$ th root of unity, which is $$(-1)^{n+1}$$
Expand along the first column. $$\Delta=(-1)^{n+1}|I_{n-1}|=(-1)^{n+1}$$, where $$I_{n-1}$$ is the identity matrix of order $$n-1$$.