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Algebra Level 5

Let $I =[|1,1567|] , (A_{i})_{1 \leq i \leq 2^{1567}-1}$ a non-empty part of $I$ numbered in any order and $B \in M_{2^{1567}-1}( \mathbb R)$ matrix defined by $b_{i,j}=1$ if $A_{i} \cap A_{j} \ne \emptyset$ and $b_{i,j}=0$ otherwise. Submit your answer as $\det(B)$ .

The answer is -1.

1 solution

Brian Moehring
Feb 26, 2017

Here's a proof sketch:

We generalize the problem first. Define $[n] := \{1,2,3,\ldots,n\}$ , let $(A_{i,n})$ denote the $2^n-1$ nonempty subsets of $[n]$ in some order, and define $B_n \in M_{2^n-1}(\mathbb{R})$ with the same definition as given in the problem. Then by elementary row operations, we may show that for all $n>1,$ $|B_n| = -|B_{n-1}|^2.$ Also, it's an immediate computation that $|B_1| = 1$ , so by induction on $n$ , we may see $|B_n| = -1$ for all $n\geq 2$ . In particular, by setting $n=1567,$ $|B| = |B_{1567}| = \boxed{-1}$

Nice solution! we can even start by showing that the determinant doesn't involve the order of how we choose $A_{i}$ that make it more easier when using induction.

Mehdi Elkouhlani - 4 years, 3 months ago

More of determinant

The answer is -1.

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