Cramers Rule and unique solutions

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SUMMARY

The discussion focuses on determining the values of s and t for which the given system of equations has a unique solution using Cramer's Rule. The equations presented are: sx1 + 2x2 + x3 + 3x4 = 1, tx1 + x2 + 3tx3 + s^2x4 = 0, sx3 + 2tx4 = 2, and x3 + x4 = 1. It is established that specific values such as s = 2 and t = 1 lead to no solution, while other combinations like s = 4 and t = 2 also yield contradictions. The key takeaway is that the determinant of the coefficient matrix must be non-zero for a unique solution to exist.

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nick484
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Hey

Im having trouble of how to go about this. Afterwards we have to perform some Cramers rule operations however I am currently struggling. For:

Find all values of s and t for which the following system has a unique solution:
sx1 + 2x2 + x3 + 3x4 = 1
tx1 + x2 + 3tx3 + s^2x4 = 0
sx3 + 2tx4 = 2
x3 + x4 = 1

Do I just say that (s/2)=/1 and t=/1 because of the last two equations or is there more to it?

Thanks
 
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Well, s= 2, t= 1 will cause there to be no solution but those are not the only values. For example, if s= 4 and t= 2, the final two equations wil be 4x3+ 4x4= 2 which reduces to x3+ x4= 1/2, contradicting the last equation.

Since you talk about Cramer's rule, I assume you know that you can write the solutions to such a system as one determinant divided by the other- and you can do that as long as the denominator, the determinant of the coefficient matrix, is non-zero.
 
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Thanks
 

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