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As a disclosure, this question is more for me to confirm that I did my work correctly. More specifically, the "solution" provided to me claims there are two values of $a$ and $b$ that yield infinite solutions, but I found only one.

That said, this is the problem:

For what values of $a$ and $b$ does this system have infinitely many solutions \begin{equation} A = \begin{pmatrix} a & 0 & b & |2 \\ a & 2 & a & |b \\ b & 2 & a & |a \\ \end{pmatrix} \end{equation}

I reduced the matrix to

\begin{equation} A = \begin{pmatrix} a & 0 & b & |2 \\ 0 & 2 & a-b & |b-2 \\ 0 & 0 & b-\frac{b^{2}}{a} & |2-b+a-2\frac{b^{2}}{a} \\ \end{pmatrix} \end{equation}

For this system to have infinite solutions, then one row must be $0$. In order for that to happen, I determined that $b=0$, $a=-2$.

However, the solution I have claims $b=0$, $a=-2$, or $a = b \neq 0$. I don't see how the latter is possible based on my reduction. Either I did something wrong, or the solution is wrong. Any insight would be appreciated.

Mlagma
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  • That is also one way to check, but that could also imply there are no solutions. I don't believe it guarantees that there are infinite solutions. – Mlagma Jan 31 '15 at 23:19

1 Answers1

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You can check easily if the solution is wrong. (It is not.) Put $a=b$. Then you have:

\begin{equation} A = \begin{pmatrix} a & 0 & a & |2 \\ a & 2 & a & |a \\ a & 2 & a & |a \\ \end{pmatrix} \end{equation} Reducing to (III - I) and (II - I) \begin{equation} A = \begin{pmatrix} a & 0 & a & |2 \\ 0 & 2 & 0 & |a-2 \\ 0 & 0 & 0 & |0 \\ \end{pmatrix} \end{equation} For $a\neq 0$ this has infinitely many solutions. (For $a=0$ it has none.) Thus there is a problem in your computation.

quid
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    Yep, you're right. I have a $b^{2}$ where it should be $b$ in $a_{34}$. Thanks. Must have done one too many problems today. – Mlagma Jan 31 '15 at 23:23
  • I am glad it was useful and you found the place were the problem arises. – quid Jan 31 '15 at 23:24