Thread: Nonzero determinants giving an inconsistent system with no solution?

Hello,

For a 3 by 2 matrix, which is not square, how does one determine values that make the system inconsistent, since by definition, we cannot find the determinant of such a non-square system?

If we augment the matrix by including the solution vectors, the system becomes a 3x3 matrix, at which point the determinant is calculable. I have found that when this determinant is 0, the system has infinite solutions. Ironically, when it is manipulated to have a nonzero determinant, it's row echelon form always contains the row [0,0,1] implying there is no solution, even though the determinant is nonzero. Does the determinant of the augmented system have any bearing on the 3x2 matrix ? What is this strange relationship I am noting?


The actual question and system are as follows:
Find the values of k for which the system is inconsistent




Det(A) = 70- 7k and therefore is 0 when k = 10. In this state, the reduced matrix contains a row of all 0's implying infinite solutions.
For k not 10, (determinant nonzero), the system has a reduces to contain the row 0,0,1 . This seems to go against rules that a non-zero determinant implies solutions.
Thanks for your help in clearing this up. I seem to be confusing myself, or missing some fundamental piece of data.

Try solving the first two equations by themselves for x and y. Then what could you do?

Using the solution from the first two eq's for x and y, in the third, k must solve to 10. So for k not 10, the system would seem to be inconsistent. This seems intuitive but I'm not sure if its complete.

I believe your answer is correct and complete. To convince yourself of that fact, you could show that 3x + 4y = 10 is a linear combination of the other two equations, whereas 3x + 4y = k, for k not 10, is not.

Ah excellent, I understand now, thanks.

You're very welcome!