Let $a \in \mathbb{Z}, a\ge2$ and $A_{n \times n}$ be a $n$ by $n$ matrix with all entries equal to $\pm a$ or $0$. I want to show $\text{det}(A)$ is divisible by $a^n$. I am aware of the basic properties of matrices, however, can't quite put my finger on how to approach this in an elegant way.
Show $\text{det}(A)$ is divisible by $a^n$
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linear-algebra
matrices
determinant
2 Answers
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By the Leibniz formula for determinants
$$\det(A)=\sum\limits_{\sigma\in S_n}\text{sgn}(\sigma)\prod\limits_{i=1}^na_{\sigma(i),i}$$
Since for every pattern $\sigma$ one has $\prod\limits_{i=1}^n a_{\sigma(i),i}$ will equal either $\pm a^n$ (in the case none of the terms was zero) or equal to zero (in the case at least one of the terms was zero), and since $\text{sgn}(\sigma)$ is always equal to $\pm 1$, the summation is adding $n$ terms with each of which equal to a multiple of $a^n$.
Since adding multiples of $a^n$ to other multiples of $a^n$ results in multiples of $a^n$, we have the overall sum is a multiple of $a^n$.
(note: this is the more common definition for determinants of matrices and coincides with the definition of determinants for other operators which might not necessarily be expressible as matrices as opposed to using Laplace Expansion)
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Let's do it by induction on $n$. Suppose that it holds for $n$ and let's show it for $n+1$. Expanding through any row or column, we get a sum of the form $\sum_{i=1}^{n+1}\lambda_i \det B_i$, where $\lambda_i \in \{0,\pm a\}$ and $B_i$ is $n\times n$ and composed of $0$'s and $\pm a$'s. It follows that $\det A$ is divisible by $a^{n+1}$.