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Given the system : \begin{array}{ccccrcc} x & + & 2y & + & z & = & 3 \\ x & + & 3y & - & z & = & 1 \\ x & + & 2y & + & (a^2-8)z & = & a \end{array}

Find values of $a$ such that the system has a unique solution, infinitely many solutions, or no solutions.


I begin by placing the system into an augmented matrix.

$\displaystyle \left[ \begin{array}{rrr|r} 1 & 2 & 1 & 3 \\ 1 & 3 & -1 & 1 \\ 1 & 2 & a^2-8 & a \\ \end{array} \right] $

I then perform operations on the matrix.

$\displaystyle \left[ \begin{array}{rrr|r} 1 & 2 & 1 & 3 \\ 1 & 3 & -1 & 1 \\ 1 & 2 & a^2-8 & a \\ \end{array} \right] $ $: (R_3-R_1)\rightarrow$ $\displaystyle \left[ \begin{array}{rrr|r} 1 & 2 & 1 & 3 \\ 1 & 3 & -1 & 1 \\ 0 & 0 & a^2-9 & a-3 \\ \end{array} \right] $

$\displaystyle \left[ \begin{array}{rrr|r} 1 & 2 & 1 & 3 \\ 1 & 3 & -1 & 1 \\ 0 & 0 & a^2-9 & a-3 \\ \end{array} \right] $ $: (R_2-R_1; R_3/(a-3))\rightarrow$ $\displaystyle \left[ \begin{array}{rrr|r} 1 & 2 & 1 & 3 \\ 0 & 1 & -2 & -2 \\ 0 & 0 & a+3 & 1 \\ \end{array} \right] $

$\displaystyle \left[ \begin{array}{rrr|r} 1 & 2 & 1 & 3 \\ 0 & 1 & -2 & -2 \\ 0 & 0 & a+3 & 1 \\ \end{array} \right] $ $: (R_1-2R_2)\rightarrow$ $\displaystyle \left[ \begin{array}{rrr|r} 1 & 0 & 5 & 3 \\ 0 & 1 & -2 & -2 \\ 0 & 0 & a+3 & 1 \\ \end{array} \right] $


Truth be told, at this point, I'm not too clear on how to progress (or if my approach is even ideal for this issue). I desire to find intervals across all of $a\in\mathbb{R}$ that explain where $a$ causes the system to have a unique, infinite, or inconsistent solution set.

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Just from the first reduction you can find: $$ (a^2-9)z=a-3 $$ Now you simply have to discuss such equation ( the others does not contain the parameter). And you see that this equation has a solution only if $a^2-9 \ne0$ i.e $a \ne \pm 3$.

For $a=3$ the equation becomes: $ 0=0$ that is an identity and this means that the system has infinitely many solutions.

For $a=-3$ the equation becomes $ =-6$ and this means that there are no solutions for the equation and so for the system.

For all other values $a \ne \pm 3$ you find $ z= 1/(a+3)$ and back-substituting in the other equations you find the solutions $x, y$ of the system that, obviously, depend from the parameter.


Substituting $z=\dfrac{1}{a+3}$ in the first two equations we have: $$ \begin{cases} x+2y=3-\dfrac{1}{a+3}\\ x+3y=1+\dfrac{1}{a+3} \end{cases} $$ now subtracting the two equations: $$ y=-2+\dfrac{2}{a+3}=\dfrac{-2(a+2)}{a+3} $$ and substituting this value in the first equation we have: $$ x-\dfrac{4(a+2)}{a+3}=3-\dfrac{1}{a+3} $$ that gives $$ x=3-\dfrac{1}{a+3}+\dfrac{4a+8}{a+3}=\dfrac{7a+16}{a+3} $$

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  • $\begingroup$ If I set $a=-2$, and inspect my last matrix, then can I not add $2R_3$ to $R_2$ and subtract $5R_3$ from $R_1$ to find $x=-2,y=0, z=1$? This is then a value of $a$ for which the system has a unique solution. The answer you've given doesn't talk about that value. Have I made an error here? $\endgroup$ – alxmke Sep 7 '15 at 17:40
  • $\begingroup$ My answer talk of all possible values of $a$. $a=-2$ is one of the values $ a\ne \pm 3$ and your result has just $z=1/(a+3)=1(-2+3)=1$. $\endgroup$ – Emilio Novati Sep 7 '15 at 19:36
  • $\begingroup$ Is there any way to determine for which values of $a \neq \pm 3$ the solution is either unique, infinite, or inconsistent, so that I may give intervals for which they are respectively true? $\endgroup$ – alxmke Sep 7 '15 at 22:45
  • $\begingroup$ For all value of $a \ne \pm 3$ the solution is unique . $\endgroup$ – Emilio Novati Sep 8 '15 at 6:10
  • $\begingroup$ Forgive my ignorance, but how have you determined this? I don't understand by what analysis one might come to this conclusion. $\endgroup$ – alxmke Sep 8 '15 at 7:01
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What strikes me immediately is that if $a^2-8 = 1$, the first and last equations have the same LHS. Since choosing $a=3$ makes these equations identical, this becomes only two equations in three unknowns, so there are an infinite number of solutions.

On the other hand, if you choose $a=-3$, then the first and third equations are contradictory, so there are no solutions in this case.

I will leave the discussion of other values of $a$ to others.

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  • $\begingroup$ Those are the only two values I really came to as well. That leaves a lot of other $a$'s though. On an unrelated note, my friend lives in Santa Monica. $\endgroup$ – alxmke Sep 7 '15 at 17:12

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