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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

are you using matrices?
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.
WOW! SO we can actually use linear algebra to solve questions like these? The part I knew was the mod

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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)
What if we used determinant and the inverse. How would you go about it. @freckles
solution for the example is \[(1+11i,11t,11k) ; \text{ where } i,t,k \text{ are integers } \\ (1+11i)+11t+11k=1+11(i+t+k)=1+0=1 \text{ which is good } \\ \text{ for first equation } \\ 2(1+11i)+3(11t)+11k=2+11(2i+3t+k)=2+0=2 \text{ which is good for } \\ \text{ second equation } \\ 1+11i+2(11t)+3(11k)=1+11(i +2t+3k)=1+0=1 \text{ which is good for } \\ \text{ third equation } \\ \text{ this answer for my example seems good }\]
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
mod 5 is just [0,1,2,3,4] so less numbers to deal with
you could check all possible \(5^3\) possibilities and see what works!
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)

Not the answer you are looking for?

Search for more explanations.

Ask your own question