Posted by **Linus** on Monday, January 17, 2011 at 5:41pm.

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

- Linear Algebra -
**Damon**, Monday, January 17, 2011 at 6:11pm
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

- Linear Algebra -
**Damon**, Monday, January 17, 2011 at 6:13pm
I do not see how the lines can be parallel and the only real problem I see so far is singularity when a = +/-4

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