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anonymous

  • one year ago

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

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  1. anonymous
    • one year ago
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    are you using matrices?

  2. freckles
    • one year ago
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    I 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: \[-3z \equiv 0 (\mod 11) \\ 3z \equiv 0 (\mod 11 ) \\ z \equiv 0 (\mod 11) \\ \text{ and then we have } \\ y-z \equiv 0 (\mod 11) \\ y-0 \equiv 0 (\mod 11) \\ y \equiv 0 (\mod 11) \\ \text{ so last equation gives } x \equiv 1 ( \mod 11) \text{ (first equation really )}\] I think I did this right.

  3. UsukiDoll
    • one year ago
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    WOW! SO we can actually use linear algebra to solve questions like these? The part I knew was the mod

  4. anonymous
    • one year ago
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    i 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)

  5. anonymous
    • one year ago
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    What if we used determinant and the inverse. How would you go about it. @freckles

  6. freckles
    • one year ago
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    solution for the example is \[(1+11i,11t,11k) ; \text{ where } i,t,k \text{ are integers } \\ (1+11i)+11t+11k=1+11(i+t+k)=1+0=1 \text{ which is good } \\ \text{ for first equation } \\ 2(1+11i)+3(11t)+11k=2+11(2i+3t+k)=2+0=2 \text{ which is good for } \\ \text{ second equation } \\ 1+11i+2(11t)+3(11k)=1+11(i +2t+3k)=1+0=1 \text{ which is good for } \\ \text{ third equation } \\ \text{ this answer for my example seems good }\]

  7. ybarrap
    • one year ago
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    One 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

  8. UsukiDoll
    • one year ago
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    mod 5 is just [0,1,2,3,4] so less numbers to deal with

  9. anonymous
    • one year ago
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    you could check all possible \(5^3\) possibilities and see what works!

  10. freckles
    • one year ago
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    by 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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