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Given

$$\begin{pmatrix}0&1&2\\ 4&3&a\\ 1&2&4\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix}=\begin{pmatrix}1\\ b\\ 4\end{pmatrix}$$

I need to find for what values $a,b \in \mathbb{F}_{11}$ there are solutions and what the general solution is (without determinan).

I created the following matrix

$$\begin{pmatrix}0&1&2&|1\\ 4&3&a&|b\\ 1&2&4&|4\end{pmatrix}$$

and after Gaussian elimination I got

$$\begin{pmatrix}4&3&a&|b\\ 0&5&10&|5\\ 0&0&a+5&|b\end{pmatrix}$$

How to find witch values of $a,b$ this system is trace?

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    $\begingroup$ Note: There aren't infinitely many solutions in any case, because in $\mathbb{F}_{11}$, there are only $11$ possible values for each parameter in a solution. $\endgroup$ – paw88789 Dec 5 '17 at 12:54
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Writing $Ax=b$ for your system of linear equations, the determinant of $A$ is given by $$ \det(A)=2(2a-27). $$ Over the field $\mathbb{F}_{11}$ the determinant is non-zero if and only if $a\neq 8$ (we could write $a\neq -3=8$ as well). Then the unique solution is given by $$ x=A^{-1}b=\frac{1}{a+3}\begin{pmatrix} 9(a + 7b + 7) \cr 3(a + 9b + 5) \cr b + 10 \end{pmatrix}. $$ In case that $a=8$, we get a $1$-parameter solution, because the rank of $A$ is then equal to $2$. More precisely, we obtain $b=1$ and $x=(x_1,x_2,x_3)$ is a the general solution with $x_1=8(x_3 + 8)$ and $x_2=5(x_3 + 5)$. The parameter $x_3$ is free, and can be any element of the finite field.

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  • $\begingroup$ good answer, but How i can do it using determinant ? $\endgroup$ – E.Roi Dec 5 '17 at 13:12
  • $\begingroup$ Do you mean without determinant? Without determinant, just compute $A^{-1}$ and your solution is $A^ {-1}b$. If you fail, $A$ was not invertible (hence the determinant was zero), and you can proceed as above to obtain $11$ solutions. $\endgroup$ – Dietrich Burde Dec 5 '17 at 13:13
  • $\begingroup$ oh my bad i meant without determinant $\endgroup$ – E.Roi Dec 5 '17 at 13:15

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