Divisibility of the Determinant of a Class of Matrices ... - MAA Sections

Theorem 7 Let A = [a ij ] ∈ M n (Z) and fix a nonzero k ∈ Z . Set b =
b1 · · · b n
T
where bi = a i1 (10 n−1 )+a i2 (10 n−2 )+...+a in (10 0 ). If b 1 , b 2 , .., b n
are all divisible by k, then det(A) is also divisible by k.
Proof. Let A = [a ij ] ∈ M n (Z) be given. First, notice that if A is singular,
then det (A) = 0 is divisible by k. Hence, we may assume that A is
nonsingular.
Suppose A is nonsingular and set z = 10 n−1 · · · 10 0 T
. Then we
have Az = b. We are given that for each i = 1, ..., n, we have b i = kr i for
some r i ∈ Z. Set r = T
r 1 · · · r n . Then b = kr.
Now, A −1 = 1 adj(A). Hence
det(A)
so that
z = A −1 b =
1
det(A) adj(A)b = 1
adj(A)kr (11)
det(A)
det(A)
z = adj(A)r
k
Since adj(A)r ∈ M n (Z), then det(A)
k
∈ Z. Thus, det(A) is divisible by k.
A natural question to ask is: for A ∈ M n (Z), when is A −1 ∈ M n (Z)?
Theorem 8 Let A ∈ M n (Z) be nonsingular. Then A −1 ∈ M n (Z) if and only
if det(A) = ±1.
Proof. Let A ∈ M n (Z) be given. Then det(A) ∈ Z because the determinant
is calculated by adding, subtracting and multiplying entries of a matrix.
If A −1 ∈ M n (Z), then det(A −1 ) ∈ Z. But det(A −1 ) = 1 . Therefore,
det(A)
1
det(A)
is also an integer. Hence det(A) = ±1.
Conversely, if det(A) = ±1, then A −1 = 1 adj(A) ∈ M det(A) n(Z) since
adj(A) ∈ M n (Z).
References
[1] Problem Department, Pi Mu Epsilon Journal, 12 No.8 (2008) 494-496.
[2] S. J. Leon. Linear Algebra with Applications Sixth Edition. Prentice Hall,
Upper Saddle River, NJ 2002.
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