Re: question on linear system

In article <1190659233.912103.259480@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Carl R. <solrac140@xxxxxxxxxxx> wrote:


Hello ,

Im given the system x + y - z =1
2x + 3y - az = 3
x + ay + 3z = 2

Im asked the following:
Find the value of "a" such that the system has NO solution.

So i compute the determinant which is equal to a^2 + a - 6 = (a-2) (a
+3)

You do? Here's what I come up with:

| 1 1 -1 |
| 2 3 -a | = |3 -a| - |2 -a| - |2 3|
| 1 a 3 | |a 3| |1 3| |1 a|

= 9+a^2 - (6+a) - (2a-3)
= a^2 -3a + 6

which has negative discriminant and no real roots.

Of course, you may have mis-copied the system above.

Personally, I would work by doing row reduction directly instead of
determinants in any case. From the original system we have


(1 1 -1 | 1) (1 1 -1 | 1)
(2 3 -a)| 3) ---> (0 1 2-a | 1) (subtract twice the 1st row from 2nd)
(1 a 3)| 2) (0 a-1 4 | 1) (subtract 1st row from 3rd)

(1 1 -1 | 1 )
---> (0 1 2-a | 1 )
(0 0 a^2-3a+6| 2-a)

So the system has no solution if and only if a^2-3a+6 = 0 and 2-a
<>0. As it happens, there are no solutions to a^2 - 3a + 6 = 0 (in the
real numbers), so the system always has a solution.


--
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"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
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Arturo Magidin
magidin-at-member-ams-org

.