1. ## proof involving determinants

Show that if A is an n x n matrix of integers, then A^-1 is a matrix of integers if and only if det(A) = +/- 1. This is actually part b of a problem. Part a) asks to prove that (x_i)det(A) = det(A_ib) where A_ib is the matrix formed by replacing the ith column of A by the column vector b. x_i is a scalar and is the component of the vector x. So since that was the previous part i tried using that for part b). So if det(A) = +/- 1, then x_i = +/- det(A_ib). but i also know that x = (A^-1)b. so the ith entry of x is also the ith row of A^-1 which i denote as a_i dot product with the vector b. So i have +/- det(A_ib) = a_i * b. i don't know what to do from here. b may or may not be a vector of integers so the det(A_ib) may or may not be an integer. and even if it were i don't know how that would show that A^-1 must be a matrix of integers. am i going about this problem in the correct way? please help.

2. Originally Posted by oblixps
Show that if A is an n x n matrix of integers, then A^-1 is a matrix of integers if and only if det(A) = +/- 1. This is actually part b of a problem. Part a) asks to prove that (x_i)det(A) = det(A_ib) where A_ib is the matrix formed by replacing the ith column of A by the column vector b.

This is known as Cramer's Rule

x_i is a scalar and is the component of the vector x. So since that was the previous part i tried using that for part b). So if det(A) = +/- 1, then x_i = +/- det(A_ib). but i also know that x = (A^-1)b. so the ith entry of x is also the ith row of A^-1 which i denote as a_i dot product with the vector b. So i have +/- det(A_ib) = a_i * b. i don't know what to do from here. b may or may not be a vector of integers so the det(A_ib) may or may not be an integer. and even if it were i don't know how that would show that A^-1 must be a matrix of integers. am i going about this problem in the correct way? please help.

You should know by now, besides Cramer's Rule, also the basic formula $Adj(A)\cdot A=|A|\cdot I$ , where $Adj(A)=$ the

classical adjoint of the matrix $A$ (whose entries are $\pm$ the minors of A ), $|A|=$ the determinant of A,

and $I=$ the unit matrix.

From this it follows that if A is invertible, then $A^{-1}=\frac{1}{|A|}\,Adj(A)$ , and then it is straightforward that if A is an

integer matrix and $|A|=\pm 1$ , then also $A^{-1}$ is an integer matrix.

The other direction follows from Cramer's rule: in $A^{-1}\cdot |A| = Adj(A)$ , the entry ij of $Adj(A)$ ,which of course

in an integer, is given by $x_{ij}|A|$ , where $x_{ij}=$ the ij-entry of $A^{-1}$ . Continue from here.

Tonio

3. i understand the first part but i have trouble continuing from where you left off in the second part.

4. Originally Posted by oblixps
i understand the first part but i have trouble continuing from where you left off in the second part.

My message entered at 8:52 AM, and your last post entered at 9:08...16 minutes difference: where did the

"sit down, grab a pencil and paper, THINK REALLY HARD and begin doing maths" fit in this lapse of time?!?

Tonio

5. sorry i am still not figuring it out. i have that x_ij |A| = Adj(A)_ij where x_ij is the i,jth entry of A^-1 and Adj(A)_ij is the i,jth entry of Adj(A). Then i have formula for Cramer's Rule which is x_i det(A) = det(A_ib). I've tried manipulating the two equations i had together but i keep getting dead ends. I still have trouble seeing how if A^-1 and A are integer matrices then det(A) must be +/- 1 follows from Cramer's rule. Can you give me a hint or two? or maybe show the next step or two? thanks

6. Originally Posted by oblixps
sorry i am still not figuring it out. i have that x_ij |A| = Adj(A)_ij where x_ij is the i,jth entry of A^-1 and Adj(A)_ij is the i,jth entry of Adj(A). Then i have formula for Cramer's Rule which is x_i det(A) = det(A_ib). I've tried manipulating the two equations i had together but i keep getting dead ends. I still have trouble seeing how if A^-1 and A are integer matrices then det(A) must be +/- 1 follows from Cramer's rule. Can you give me a hint or two? or maybe show the next step or two? thanks

It's not that: by the product theorem, we have that $det(AB)=det(A)\cdot det(B)$ , so if $A,\,A^{-1}$ are both integer

matrices, then both determinants are integers, but

$1 = det(I) = det(AA^{-1})=det(A)\cdot det(A^{-1})=det(A)\cdot det(A)^{-1}$ .

Now, how many pairs of integers number do you know that their product equals 1...?

7. oh so both det(A) and det(A^-1) must be integers and the only two integers that multiply to 1 are 1 and 1. also, since det(A^-1) = det(A)^-1, the inverse of an integer must be an integer and the only integer with that property is 1 and -1. Thank you! i understand now. I was just going about the problem the wrong way since i thought i had to use Cramer's rule somewhere in the proof.

8. Originally Posted by oblixps
oh so both det(A) and det(A^-1) must be integers and the only two integers that multiply to 1 are 1 and 1. also, since det(A^-1) = det(A)^-1, the inverse of an integer must be an integer and the only integer with that property is 1 and -1. Thank you! i understand now. I was just going about the problem the wrong way since i thought i had to use Cramer's rule somewhere in the proof.

To be honest, so did I in the beginning, and perhaps this misled you. But it is way easier than that, as you saw above.

Tonio