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Here is the matrix, which I'm asked to give the determinant of. How do I read this?

$$R^{(1)}:= \begin{bmatrix} 1 \\ & \ddots \\ & & 0 & \cdots & 1 & &\\ & & \vdots & \ddots & \vdots \\ & & 1 & \cdots & 0\\ & & & & & \ddots \\ & & & & & & 1 \end{bmatrix}$$

How do I deal with the dots and find the determinant?

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  • $\begingroup$ @FraGrechi I'm not given the dimensions, but the problem does mention row operations on a quadratic matrix; however, I'm not sure how that fits with this question. $\endgroup$ – cdignam Mar 21 '17 at 1:39
  • $\begingroup$ I have never heard of the term "quadratic matrix", what does this refer to in the course/book you are currently taking/reading? Does this mean that you are dealing with an $n \times n$ square matrix? $\endgroup$ – FraGrechi Mar 21 '17 at 1:53
  • $\begingroup$ @FraGrechi Yeah, it means square matrix. $\endgroup$ – cdignam Mar 21 '17 at 1:54
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    $\begingroup$ Oh I am an idiot, of course it is a square matrix. To find the determinant simply swap those two rows, obtain $I$, and find that $\det R = -1$. $\endgroup$ – FraGrechi Mar 21 '17 at 2:02
  • $\begingroup$ @FraGrechi Oh, thanks. That's what I needed. $\endgroup$ – cdignam Mar 21 '17 at 2:06
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As FraGrechi said, we can transform this matrix into the identity matrix, whose determinant is $1$; however, since we swap two rows, we need to multiply this by $-1$, giving us our final determinate of $-1$.

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