SOLUTION: Hi, I'm having trouble with this question. I don't know how I'm supposed to work it out, so any help would be great appreciated! Thank you! Determine the values for "a" for which

Algebra ->  Matrices-and-determiminant -> SOLUTION: Hi, I'm having trouble with this question. I don't know how I'm supposed to work it out, so any help would be great appreciated! Thank you! Determine the values for "a" for which      Log On


   



Question 1050415: Hi, I'm having trouble with this question. I don't know how I'm supposed to work it out, so any help would be great appreciated! Thank you!
Determine the values for "a" for which the following system of linear equations has no solutions, a unique solution, or infinitely many solutions. I need to solve using a matrix, and then row reducing.
3X1 + aX2 + 6X3 = 6
-X1 + 3X2 - X3 = 1
2X1 - 2X2 + 6X3 = 11

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Determine the values for "a" for which the following system of linear equations has no solutions, a unique solution, or infinitely many solutions. I need to solve using a matrix, and then row reducing.
3X1 + aX2 + 6X3 = 6
-X1 + 3X2 - X3 = 1
2X1 - 2X2 + 6X3 = 11
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You have to find the determinant of the "coefficient matrix"::
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D = (3*3*6 + (-1*2*6) + (2*-1*a) - [(6*3*2)+(-1*-2*3)+(6*-1*a))
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D = (54 - 12 - 2a - [36 + 6 - 6a])
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D = (42-2a)-(42-6a)
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D = 4a
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One unique solution if a is not zero
No solution when a = 0
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I'll leave the row reducing to you.
Cheers,
Stan H.
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