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anonymous
 4 years ago
2. Suppose that A is an n×n matrix with integer entries, and that A is invertible. Since
A is invertible, we can compute the matrix A^−1. The computations seem much cleaner
(and friendlier) when A^−1 also has only integer entries. The purpose of this question is
to figure out when that can happen.
Prove: That (if A is an invertible n × n matrix, with only integer entries) A^−1 has
integer entries if and only if det(A) = ±1 (where “= ±1” means equals 1 or equals −1).
Reminder: Since this is an “if and only if”, don’t forget that means that you have two
directions to prove.
anonymous
 4 years ago
2. Suppose that A is an n×n matrix with integer entries, and that A is invertible. Since A is invertible, we can compute the matrix A^−1. The computations seem much cleaner (and friendlier) when A^−1 also has only integer entries. The purpose of this question is to figure out when that can happen. Prove: That (if A is an invertible n × n matrix, with only integer entries) A^−1 has integer entries if and only if det(A) = ±1 (where “= ±1” means equals 1 or equals −1). Reminder: Since this is an “if and only if”, don’t forget that means that you have two directions to prove.

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Hey james once you are done with this please help me....

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2One way is easy. If A has all integral entries, then det(A) is also an (nonzero) integer. If A also has all integral entries, then det(A^1) is also integral . Now, as \[ 1 = \det(I) = \det(AA^{1}) = \det(A).\det(A^{1}) \] the only way that can be the case is if \[ \det(A) = \det(A^{1}) = 1 \] or \[ \det(A) = \det(A^{1}) = 1 \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Hey turn towards Unanswered question james.... Adherent points

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2Now the other way. We know that A has all integral entries and we have that \( \det A = \pm 1 \). This must mean that \( \det A^{1} = \pm 1 \) respectively also. Now think about why this can imply the entries of A^1 are all integers; or why if they're not all integers this leads to a contradiction.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It implies that they're all integers right? Because ... if the inverse of detA is equal to pm1, then they must be integers so that they could undo themselves?
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