Hello: I am having a hard time with this problem:
For which values of "a" does the following system have zero solutions? One
solution? Infinitely many solutions?
x1 + x2 + x3 = 4
x3 = 2
(a^2 - 4)x3 = a - 2
I wasnt sure if I should substitute x3 into the third equation, then solve for
a. I re-read the chapter section and there are not any examples.
x1 + x2 + x3 = 4
x3 = 2
(aČ-4)x3 = a-2
Since x3 = 2, if we substitute 2 for x3 in the third equation we get
(aČ-4)(2) = a-2
2aČ - 8 = a - 2
2aČ - a - 6 = 0
(2a + 3)(a - 2) = 0
2a + 3 = 0. | a - 2 = 0
| a = 2
2a = -3 |
a = -3/2 |
In order to have a solution at all, a has to be one of these.
So if a is neither of these, there are no solutions.
If a = -3/2 the bottome equation comes out x3 = 2, identical with the
second equation.
And if we substitute 2 for x3 in the first equation, we get
x1 + x2 + 2 = 4 or
x1 + x2 = 2 or
x1 = 2 - x2.
So there are infinitely many solutions in this case, because there are
infinitely many choices for x2 and each one gives a different value for x1.
If a = 2 , the bottome equation becomes 0x3 = 0, so any value of x3 would
satisfy that, but the middle equation tells us that x3 = 2. So that still
gives us an infinite number of solutions.
So a = -3/2 or a = 2 gives infinitely many solutions. Any other value of a
gives no solutions. There are no values of a which give just one solution.
Edwin
