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anonymous
 one year ago
Find all solutions of the congruences
\[3x+y+3z\equiv 1(mod 5) \]\[x+2y+z\equiv 2(mod 5)\]\[ 4x+3y+2z\equiv 2(mod 5)\]
anonymous
 one year ago
Find all solutions of the congruences \[3x+y+3z\equiv 1(mod 5) \]\[x+2y+z\equiv 2(mod 5)\]\[ 4x+3y+2z\equiv 2(mod 5)\]

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0are you using matrices?

freckles
 one year ago
Best ResponseYou've already chosen the best response.4I will come up with an example: \[x+y+z \equiv 1 (\mod 11) \\ 2x+3y+z \equiv 2 (\mod 11 ) \\ x+2y+3z \equiv 1 (\mod 11 ) \\ \text{ so we will attempt \to solve this using matrix } \\ \left[\begin{matrix}1 & 1 & 1 & 1 \\ 2 & 3 & 1 & 2 \\ 1 & 2 & 3 & 1\end{matrix}\right] \\ \text{ now we know } 2+9 \equiv 0 (\mod 11) \\ \text{ So I will go ahead and plus 9 to row 2} \\ \text{ and we also know } 1+10 \equiv 0 (\mod 11) \\ \text{ so I will plus 10 to row 3 } \\\] \[\left[\begin{matrix}1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & 2 & 0\end{matrix}\right] \] now multiply bottom row by 1 and add bottom two rows after that \[\left[\begin{matrix}1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 3 & 0\end{matrix}\right] \] so we have: \[3z \equiv 0 (\mod 11) \\ 3z \equiv 0 (\mod 11 ) \\ z \equiv 0 (\mod 11) \\ \text{ and then we have } \\ yz \equiv 0 (\mod 11) \\ y0 \equiv 0 (\mod 11) \\ y \equiv 0 (\mod 11) \\ \text{ so last equation gives } x \equiv 1 ( \mod 11) \text{ (first equation really )}\] I think I did this right.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0WOW! SO we can actually use linear algebra to solve questions like these? The part I knew was the mod

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i wonder if you can do this the regular way, keeping in mind that you are working mod 5 when you multiply and add not really sure, but it seems like it should work (probably harder)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What if we used determinant and the inverse. How would you go about it. @freckles

freckles
 one year ago
Best ResponseYou've already chosen the best response.4solution for the example is \[(1+11i,11t,11k) ; \text{ where } i,t,k \text{ are integers } \\ (1+11i)+11t+11k=1+11(i+t+k)=1+0=1 \text{ which is good } \\ \text{ for first equation } \\ 2(1+11i)+3(11t)+11k=2+11(2i+3t+k)=2+0=2 \text{ which is good for } \\ \text{ second equation } \\ 1+11i+2(11t)+3(11k)=1+11(i +2t+3k)=1+0=1 \text{ which is good for } \\ \text{ third equation } \\ \text{ this answer for my example seems good }\]

ybarrap
 one year ago
Best ResponseYou've already chosen the best response.1One way you can turn this into a system of equations is, for example, take the 1st equation $$ 2x+y+3z\equiv 1 (mod 5) $$ this is the same as $$ 5n+1=2x+y+3z $$ For all n Now just choose and n, say n=0 $$ 1=2x+y+3z $$ Similarly for the others Then you can use $$ x=A^{1}b $$ Or rref, as was done by @freckles

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0mod 5 is just [0,1,2,3,4] so less numbers to deal with

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you could check all possible \(5^3\) possibilities and see what works!

freckles
 one year ago
Best ResponseYou've already chosen the best response.4by the way I'm not sure if it was clear from my example but when I said add 9 to row 2 or add 10 to row 3 I can do that since my first row was all one's you know since: 9(1)=9 (so maybe I should have said multiply row 1 by 9 and add to row 2 but it was just shorter to say the other thing) 10(1)=10 (and blah blah for this other thing I did)
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