d =
| 1 2 -3 |
|3 -1 5 |
| 4 1 (a^2-14)|
d = -(a^2-14)-6(a^2-14)+40 - 9-12-5
d = -7(a^2-14)+14
d = -7a^2 + 8*14 = (-a^2 +16)7
if d is zero, all solutions have 0 in the denominator
so if a - 4 or -4, solution is undefined
Now if a is such that equation 3 is a multiple of one of the first two, there is no unique solution (lines are parallel)
now solutions are of form:
P/d
where P is the determinant of the matrix resulting from replacing row one with the right side for x, row 2 for the right for y, row 3 for the right for z
so for z for example
| +1 +2 +4 |
| +3 -1 +2 |
| +4 +1 (a+2)|
--------------------
(-a^2 +16)7
[-(a+2)+16+12+16-2 -6(a+2) ] /(-a^2 +16)7
[-7(a+2)+42]/(-a^2 +16)7
[-7a + 28 ] /(-a^2 +16)7
-7 [ a-4 ] / -7[a^2-16]
1/(a+4)
hmmm, z is undefined if a = -4
etc
For which values of 'a' will the following system have no solutions? Exactly one solution? Infinitely many solutions?
x + 2y - 3z = 4
3x - y +5z = 2
4x + y + (a^2-14)z = a+2
2 answers
I do not see how the lines can be parallel and the only real problem I see so far is singularity when a = +/-4