## Related questions with answers

What can you say about the determinant of an $n \times n$ matrix with the following form?

$\left[\begin{array}{ccccc}{0} & {0} & {\cdots} & {0} & {1} \\ {0} & {0} & {\cdots} & {1} & {0} \\ {\vdots} & {\vdots} & {} & {\vdots} & {\vdots} \\ {0} & {1} & {\cdots} & {0} & {0} \\ {1} & {0} & {\cdots} & {0} & {0}\end{array}\right]$

Solution

VerifiedLet the given matrix be $P_n$ where $n$ is the size of $P$. Notice that $P_n$ is an exchange matrix which can be reduced to an identity matrix by interchanging the first and last rows.

```
When $n = 2$ or 3, det($P_n$) = $-1$. When $n = 4$ or 5, det($P_n$) = $+1$. When $n = 6$ or 7, det($P_n$) = $-1$. And so on and so forth with this pattern. Therefore, we can summarize this pattern as,
```

$\begin{align*} \text{det}\left( P_{2k} \right) &= \text{det}\left( P_{2k+1} \right) = -1 \quad \text{for} \; k=1,3,5,7,... \\ \text{det}\left( P_{2k} \right) &= \text{det}\left( P_{2k+1} \right) = +1 \quad \text{for} \; k=2,4,6,8,... \end{align*}$

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