$\left( {\begin{array}{ccc|c} 3 & 1 & 3 & 3 \\ 4 & 4a & 2a & 3 \\ 2 & 2 & 2a & 1 \end{array} } \right) a \in \mathbb Z_5$ I would need advice on how to proceed with this example. I need to find and those for which the equation has no solution. I know that in z5 I am limited to the set {0-4} and I try to perform Gaussian elimination operations.
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Does GEM mean Gaussian elimination? That's how you do it. What exactly is the problem? – saulspatz Mar 23 '21 at 13:31
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@saulspatz, I would assume Gauss elementary/elimination matrices, or something to that extent. – Ennar Mar 23 '21 at 13:33
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2If this were over $\Bbb R$ instead of $\Bbb Z_5$, would you have been able to do it? – Arthur Mar 23 '21 at 13:37
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What exactly is your problem, martin12888? The procedure is the same, you are just working over different field. For example, you would like $1$ to appear at place $(1,1)$ in the matrix. You can achieve it by multiplying the first row by $2$. And so on. – Ennar Mar 23 '21 at 13:37
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Hint: The determinant of the matrix is $8 (3 a^2 - 5 a + 3)$, which is zero iff $a \equiv \pm 2 \bmod 5$. So, for $a \equiv 0,1,4 \bmod 5$, the system always has a solution.
lhf
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But a determinant of 0, doesn't mean that the system does not have a solution – samabu Mar 23 '21 at 13:59
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Yes, that's true. Just wanted to be sure, there is no missunderstanding – samabu Mar 23 '21 at 14:41
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Compute the Gaussian elimination. Just take $a$ as it is any number. If this is to difficult, you can use 4 equality systems and check if there is a solution or not (set a=0 and solve then a=1 and solve, ...)
samabu
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