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Expert: - 9/11/2010


Question
Mr. Wilson,

I appreciate the time you put in to help with questions like mine. I have a question that I've been trying to figure out for a while now - I pretty sure I'm over-thinking the problem and making harder than it actually is.

The question is:
----
For what value(s) of K (if any) does the system of linear equations below have a unique solution? No solution? An infinite number of solutions?

x +  y - z = 1
2x - 5y + z = -1
7x - 7y - z = k

I think I can use Gauss-Jordan or Guassian Elimination, I'm just unsure how to handle the 'K'.

Thanks again for your time.

Frank N.

Answer
Solve the 1st equation for x.  This gives x = 1 - y + z.  Put the solution into the 2nd and 3rd equations.  Now we are reduced to two equations in y and z.  They should be -7y + 3z = -3 and
-14y + 6z = k - 7.

If we take -7y + 3z = -3 times -2 we get 14y - 6z = 6.
Adding this to the other equation gives 0 = k - 1.

Thia means for a solution to exist, we need k = 1.
It also means that when k = 1, there are infinite solutions since two of the equations are dependant on each other.

Note that another way to approach this would be to calculate the determinant of the matrix and find out it is 0.  I'm not sure if  you have heard about how to calculate determinants, though.

Another approach would be to try and reduce the matrix to upper triangular form and see that the last row is all 0s, but I'm not sure you've heard of upper triangular form or not either.  Upper triangular form of the multiplying matrix is where the rows are subtracted from each other to make it have 0's below the diagonal.

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Scott A Wilson

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