Linear Algebra
posted by Linus on .
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^214)z = a+2

d =
 1 2 3 
3 1 5 
 4 1 (a^214)
d = (a^214)6(a^214)+40  9125
d = 7(a^214)+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+162 6(a+2) ] /(a^2 +16)7
[7(a+2)+42]/(a^2 +16)7
[7a + 28 ] /(a^2 +16)7
7 [ a4 ] / 7[a^216]
1/(a+4)
hmmm, z is undefined if a = 4
etc 
I do not see how the lines can be parallel and the only real problem I see so far is singularity when a = +/4