# Determinant of $n\times n$ matrix

How can I calculate the determinant of the following $n\times n$ matrix, where $n$ is a multiple of $3$ ? $$\begin{pmatrix} 0 & 0 & 1 & & & & & & &\\ & & & 0 & 0 & 1 & & & &\\ & & & & & &\ddots\\ & & & & & & & 0 & 0 & 1\\ 0 & 1 & 0 & & & & & & &\\ & & & 0 & 1 & 0 & & & &\\ & & & & & &\ddots\\ & & & & & & & 0 & 1 & 0\\ 1 & 0 & 0 & & & & & & &\\ & & & 1 & 0 & 0 & & & &\\ & & & & & &\ddots\\ & & & & & & & 1 & 0 & 0 \end{pmatrix}$$

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If you permute the rows in the right way, you can obtain the identity matrix. So, the determinant is $1$ or $-1$ depending on the sign of the permutation. – wisefool Dec 23 '12 at 14:53
You could do the case where $n=3$ and $n=6$ and permute columns, then try for $n=3m$ ,$m\in \mathbb N$ in general. – jandrew Dec 23 '12 at 14:58
Both of the first two comments tell you how you can find it ... – Michael Joyce Dec 23 '12 at 15:13

You can use induction on $m$, using $n = 3m, \quad m\in \mathbb{N}$.

First, compute the determinants for base cases $m = 1$ where $n = 3\cdot 1 =3$, and, say, $m=2$, where $n = 3 \cdot 2 = 6$. (It would also be insightful to see what happens for a couple more "smallish" $m$.)

You'll find by using elementary row operations, you can obtain the identity matrix for any $n = 3m$. The only elementary row operation you'll need that changes the value of the original determinant is "exchange row i with row j", so you need only record the number of times "row exchangement" is required :

• When an odd number of exchanges is needed (e.g., n = 3), the determinant will be $-1$.
• When an even number of exchanges is required, then the determinant will be 1.

Through exploration, you'll note a pattern

After computing the determinants for $n = 3m$ for "small-ish" values of $m$, you should be able to establish a pattern and determine an expression in terms of $m$ that represents the required number of exchanges for any given $n = 3\cdot m$.

Use this expression to propose an inductive hypothesis, and using the inductive hypothesis, see if you can show that for $n = 3\cdot (m + 1)$, the inductive hypothesis is satisfied.

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For $n=6$, why I need only two exchanges? – Boh Dec 23 '12 at 15:49
Count the number of exchanges needed for $n = 6$ to determine if it needs an odd number of exchanges, or even number. – amWhy Dec 23 '12 at 15:59
The determinant for $n=3\cdot 3$ is $1$, and the determinant for $n=3\cdot 5$ is $-1$. It's not always the same for $n=3\cdot m$ with $m$ odd... – Boh Dec 23 '12 at 16:49
Good work...so see what pattern you come up with to determine your inductive hypothesis. – amWhy Dec 23 '12 at 16:52