Question

For what values of k is the linear system consistent?

Answer 1

A linear system is inconsistent is if it represents a contradiction, for instance the system

[030−2−101][00−103−21]

is inconsistent because the first line represents a linear equation 0x+0y=−100x+0y=−10 , i.e. 0=−100=−10 , which is a contradiction. Geometrically, when you solve a 2x2 linear system, you are finding the intersection between a pair of lines. If you reach a contradiction, like the system above, then your lines do not intersect, i.e. they must be parallel.

If you are being asked this question, you have probably already covered Gauss-Jordan elimination. Inconsistencies in linear systems can be readily identified if the system is brought to reduced row echelon form (can you see why?), so I would start with that. The steps are simple:

[12h−8−56][1h−52−86] Multiply the second row by 1/21/2 : [11h−4−53][1h−51−43] Subtract the second row from the first: [01h+4−4−83][0h+4−81−43] Without even proceeding further, it is obvious that one way for the system to be inconsistent is if the first line is 00|−800|−8 , since this would be equivalent to saying 0x+0y=−80x+0y=−8 , that is 0=−80=−8 , a contradiction. The first row would have this form only if h=−4h=−4 , so h=−4h=−4 makes the system inconsistent.

Now it is pretty clear at this point that no other value of hh would make the system inconsistent, and after you are comfortable with Gauss-Jordan elimination this fact would be apparent to you as well, though you should really try to understand why first. So let's say h≠−4h≠−4 . Then we can multiply the first row by 1h+41h+4 :

[011−4−8h+43][01−8h+41−43] And now add 4 times the first row to the second: ⎡⎣0110−8h+43−32h+4⎤⎦[01−8h+4103−32h+4] To really be precise, you can swap the two rows: ⎡⎣10013−32h+4−8h+4⎤⎦[103−32h+401−8h+4]

Thus for any value of hh other than −4−4 , we can solve the system - there is no way to make the system displayed above have a row which looks like 00|c00|c , for any non-zero number cc .