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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.

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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...]

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  • $\begingroup$ Thanks! Does this mean that not only 21 but any other $a$ (other than 2 and 3) leads to an unique solution? If so, why does Cramer's rule lead me to $a=21$? Did i do something wrong? $\endgroup$ – Juan José Castro Mar 25 '14 at 21:13
  • $\begingroup$ If you are looking at the determinant of the coefficient matrix, you should find it to be $ \ (a^2 - 5a + 6) \ $ , so there are just two values of $ \ a \ $ which cause that matrix to be singular. All other values make the determinant non-singular, which means the system will have a unique solution. $\endgroup$ – colormegone Mar 25 '14 at 21:17
  • $\begingroup$ Wow, I suck pretty badly at this. Thanks, that makes sense, I hadn't thought of that (clearly). I still don't understand why would Cramer's rule lead me to $a=21$, though. I guess I must've done something wrong. Thanks again! $\endgroup$ – Juan José Castro Mar 25 '14 at 21:21
  • $\begingroup$ When you say "Cramer's Rule", are you talking about the version for finding determinants of 3 x 3 matrices, or the version for solving a system using adjoints? In either event, you want to be particularly careful about arithmetic when working in linear algebra. $\endgroup$ – colormegone Mar 25 '14 at 21:21
  • $\begingroup$ It would be helpful if you posted the work you have been doing; it makes it a lot easier for someone on M.SE to "troubleshoot" difficulties. $\endgroup$ – colormegone Mar 25 '14 at 21:24

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