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Fing $G_2$ and $G_3$ and then by row operations $G_4$. Can you predict $G_n$?

The matrices $G_2, G_3, G_4$ are below.

$$\begin{vmatrix}0 & 1\\ 1 & 0 \end{vmatrix}$$

$$\begin{vmatrix}0 & 1 &1 \\ 1 & 0 &1 \\ 1 &1 &0 \end{vmatrix}$$

$$\begin{vmatrix}0 & 1 &1 &1\\ 1 & 0 &1 &1 \\ 1 &1 &0 &1 \\ 1 &1 &1 &0 \end{vmatrix}$$

I understand how to calculate the determinants with pivots and cofactors.

The first $3$ make sense:

$$ G_2 = −1, G_3 = 2, G_4 = −3$$

However, I do not yet see how he got

$$G_n = (−1)^{n−1}(n − 1).$$

vitamin d
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Steak
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  • Probable duplicate of https://math.stackexchange.com/questions/2677780/find-the-general-formula-to-compute-deta-n-and-then-proof-by-induction-pro – lhf Mar 23 '21 at 20:58
  • See also https://en.wikipedia.org/wiki/Matrix_determinant_lemma – lhf Mar 23 '21 at 21:00

1 Answers1

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The formula for $G_n$ has two parts that accomplish two important things. First, disregarding the signs, you can see a pattern in which the determinant of the $n$th matrix has a magnitude of $n-1$. This is where $(n-1)$ comes from.

The second part involves the signs on the determinants. The determinants appear to be positive for odd $n$ and negative for even $n$. When $n$ is odd, $n-1$ is even, and thus $(-1)^{n-1}$ is positive. Likewise, when $n$ is even, $n-1$ is odd, and $(-1)^{n-1}$ is negative.

Kman3
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