So I've got the following problem:
Find the values of $a$ for which the following system has:
A. A unique solution
B. An infinite number of solutions
C. No solutions
$$x+y-z=4$$ $$x+2y+z=7$$ $$3x+6y+(a^2 - 5a+9)z = a+18$$
I solved A by using Cramer's rule, expressing $x$, $y$ and $z$ in terms of $a$ and then replacing inside the first equation to find that $a=21$. However, I can't find a way to solve the other two. I thought about making the third equation a multiple of one of the other two but no value of $a$ satisfies that.
Sorry for formatting/english. Any help is appreciated.
This system of equations "row-reduces" to
$$ \left(\begin{array}{cc}1&1&-1&4 \\0&1&2&3\\0&0&a^2-5a+6&a-3\\\end{array}\right) \ \ . $$
So we can answer the latter parts of the question first. In order to have an infinite number of solutions (system is "dependent"), we would need to have $ \ (a-2) \ (a-3) \ = 0 \ \ $ and $ \ a-3 \ = \ 0 \ $ simultaneously, which occurs for $ \ a = 3 \ $. In order to have no solutions (system is "inconsistent"), we require that $ \ (a-2) \ (a-3) \ = 0 \ $ , with $ \ a \ $ equal to anything other than 3 ; so $ \ a = 2 \ $ satisfies this condition.
Any other value of $ \ a \ $ leads to a unique solution of the system.
EDIT -- Sorry, just corrected my factoring goof...
Response to comments:
I am getting for the numerators from Cramer's Rule
$$ x : \ \ a^2 \ - \ 2a \ - 3 \ = \ (a+1)(a-3) \ \ , $$
$$ y : \ \ 3a^2 \ - \ 17a \ + 24 \ = \ (a-\frac{8}{3})(a-3) \ \ , $$
$$ z : \ \ a-3 \ \ , $$
so all the numerators and the denominator are zero for $ \ a = 3 \ $ , which is the "dependent" case, and the denominator is also zero (with all numerators non-zero) for $ \ a = 2 \ , $ indicating "inconsistency". [Thanks to WolframAlpha for assistance with the tedious determinant computations. All the agony involved in that technique is why a lot of people generally use some "elimination" method, such as row-reduction or Gauss-Jordan...]