Question

Let A and B be 2 x 2 matrices with integer entries such that A, A + B, A + 2B, A + 3B, and A + 4B are all invertible matrices whose inverses have integer entries. Show that A + 5B is invertible and that its inverse has integer entries.

Answer 1

This is a problem in which i have no idea where to start, just looking for some ideas or suggestions :)

Answer 2

What? Something you don't know. I'm surprised to see

Answer 3

If you don't know, I don't think anyone else knows

Answer 4

i would like alchemista to look at it, he/she is pro:) im not really looking for a solution, just ideas.

Answer 5

I don't think alchemista is around at the moment

Answer 6

i can tell :( im gonna mess with it for a bit.

Answer 7

joemath can u tell me abt invertible matrices... i forgot abt them

Answer 8

well, there are a couple of things, namely that the det of the matrix isnt 0, or that there exists another matrix such that when you multiply them you get the identity matrix.

Answer 9

what if adding the matrices helps us...

Answer 10

we can have a new matrice if we add a+3b and a+2b

Answer 11

and subtracting a

Answer 12

woul still give us an invertible matrice and with int in it

Answer 13

so you have that A+5B = (A+2B) + (A + 3B) - (A)

Answer 14

how do you know that? i dont think that is necessarily true.

Answer 15

a have some int so if we subtract a frm 2a it still would be int

Answer 16

thats right, we know that A + 5B is going to have integers in it, but we dont know that its inverse is

Answer 17

The inverse of a matrix will have integer entries if the determinant of the matrix is 1 or -1

Answer 18

is that an "if and only if" statement?

Answer 19

yes of course because

Answer 20

if the det is nonzero its invertible

Answer 21

err wait a second

Answer 22

determinants of matrice is non zero then if we add two non zero det matrice don't u think the result would be non zero only ? i'm not sure abt tht but i've an intuition fr it

Answer 23

the reverse implication is, if the inverse has integer entries is the determinant of the original -1 or 1

Answer 24

The intuition is based on elementary operations again

Answer 25

I dont know if it will help you with the problem.

Answer 26

anything is helpful :) although im gonna go to bed, thinking about it now is making me tired lol.

Answer 27

i'll look at it from that angle tomorrow when i have time, thank you both for your input :)

Answer 28

Also make sure you are thinking about this
\[\det \left[\begin{matrix}a & b \\ c & d\end{matrix}\right] = ad - bc\]

Answer 29

yeah, i have that written out for both A and B in an arbitrary way, but there are so many variables its a tad confusing.

Answer 30

anywhos, sleeeeeep, thanks again.

Answer 31

a and b must be nonzero ....

Answer 32

so 4b will only give 4det b

Answer 33

which will be nonzero

Answer 34

or even if b is zero then a can't be zero.... so a + 5b must be non zero matrice

Answer 35

alchemista in the question its already given that a+4b is non zero...so either a is non zero or b or maybe both..........then a + 5 b should be non zero too...

Answer 36

we aren't talking about weather the matrix has non zero entries.

Answer 37

invertibility is a stronger condition

Answer 38

there are nonzero 2x2 matrices that are not invertible

Answer 39

when i say non zero i mean non zero determinant

Answer 40

yeah but its given in the question that a is invertible

Answer 41

if we can sum a+3b then we can divide them too (A) + 3(B)

Answer 42

and it doesn't matter whether b is invertible or not .....becuz we know a is..

Answer 43

You will need to show the following
\[(a_1+ka_2)(d_1+kd_2)-(b_1+kb_2)(c_1+kc_2)\] is 1 or -1

Answer 44

for k = 5 given that it holds for 1,2 ,3, 4

Answer 45

You are also given that \[a_1d_1 - b_1c_1\] is 1 or -1

Answer 46

now we know a1d1 - b1c1 /= 0

Answer 47

from det of a

Answer 48

What I said before is an if and only if. Since the inverse has integer entries the det is 1/-1

Answer 49

1 or -1