What is the step how can I solve following system of congruences (that is one system):
- $7x-8y≡5 \pmod {11}$
- $2x+5y≡9 \pmod {11}$
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$11$ is a prime, so integers modulo $11$ form a field. This means that you can use the same techniques as is taught over the reals in Linear Algebra. Form the matrix, and use elementary row operations. See my old answer for a walk-thru example of inverting a matrix modulo $29$. Of course, you can simply use the thinking behind Gaussian elimination. Solve $x$ in terms of $y$ from one equation, plug it into the other etc. – Jyrki Lahtonen Oct 14 '19 at 07:37
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But, please check out our guide for new askers. Your question falls a bit short of what is expected from a question here. – Jyrki Lahtonen Oct 14 '19 at 07:39
3 Answers
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Adding the first equation with twice the second one we obtain
$$7x-8y+2(2x+5y)≡5+2(9) \implies 11x+2y\equiv23 \implies2y\equiv 1 \mod 11$$
$$\implies 6\cdot 2y\equiv 6\cdot 1 \implies 12y\equiv6\implies y\equiv 6 \mod 11$$
Then from the second equation
$$2x+5y≡9 \implies 6\cdot 2x+6\cdot 5y \equiv 6\cdot 9 \implies12x+4\equiv 10 \implies x\equiv 6 \mod 11$$
user
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@Maris It's a trick to obtain $11x\equiv 0 \mod 11$ so we only remain with $2y\equiv 1 \mod 11$ – user Oct 14 '19 at 07:31
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@Maris That's not a new trick. It is the same thing you would with a system over the reals: use one equatioin to eliminate one variable from the other. – Jyrki Lahtonen Oct 14 '19 at 07:41
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You can solve with gaussian elimination of matrix over $\mathbb{F}_{11}$. Noting that $-8\equiv 3\pmod{11}$ we get the following matrix:
$$ \left[ \begin{array}{cc|c} 2 & 5 & 9 \\ 7 & 3 & 5 \end{array} \right] \underset{R_{2}:2R_{1}+R_{2}}{\longrightarrow} \left[ \begin{array}{cc|c} 2 & 5 & 9 \\ 0 & 2 & 1 \end{array} \right] \underset{R_{1}:3R_{2}+R_{1}}{\longrightarrow} \left[ \begin{array}{cc|c} 2 & 0 & 1 \\ 0 & 2 & 1 \end{array} \right] $$
Now we want to divid the second by $2$, for that we need to know what is the multiplicative inverse of $2$ in $\mathbb{F}_{11}$, that is were looking for a positive integer $n\in\{n\in\mathbb{N}\mid n<11\}$ such that $2n\equiv 1 \pmod{11}$, easily enough we know it's $6$ so the solution is: $$ \left[ \begin{array}{cc|c} 1 & 0 & 6 \\ 0 & 1 & 6 \end{array} \right] $$ Or just $x=6,\,y=6$
omer
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2@N.F.Taussig More conceptually it is $,R_2 \leftarrow R_2 - 7/2, R_1,,$ where $,-7/2\equiv 4/2\equiv 2,,$ i.e. eliminate $x$ by scaling $R_1$ so its $x$ coef becomes $-7$ which is the negative of that of $R_2$ then add it to $R_2$ to eliminate $x.,$ This way the row operation is not pulled out of hat - like magic. Rather, this view makes it clear that it is essentially the standard row reduction step in Gaussian elimination. – Bill Dubuque Oct 14 '19 at 13:14
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Hint:
$5\cdot(1)+8\cdot(2)$
$$(35+16)x\equiv5\cdot5+9\cdot8\pmod{11}$$
$$-4x\equiv3-5\pmod{11}\iff2x\equiv1\iff x\equiv2^{-1}\equiv6$$
lab bhattacharjee
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@Maris, I have eliminated $y$ See http://mathsfirst.massey.ac.nz/Algebra/SystemsofLinEq/EMeth.htm – lab bhattacharjee Oct 14 '19 at 07:10
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1For me there is no problem whith usual linear equation syatems. The biggest problem for me is to understand congruence... there cannot be decimal numbers etc. – Maris Oct 14 '19 at 07:23
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@Maris, In general we understand congruence in integers and preferably non-negative. See http://mathworld.wolfram.com/Congruence.html . But there is also something called http://mathworld.wolfram.com/FractionalCongruence.html. See also : http://mathworld.wolfram.com/CompleteResidueSystem.html – lab bhattacharjee Oct 14 '19 at 07:27
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