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Suppose I have a system of equations as such, with $\alpha$ and $\beta$ being constants:

$$ x_1+5x_2+x_3=1\\ x_1+6x_2-x_3=1\\ 2x_1+\alpha x_2-6x_3=\beta $$

Suppose I want to calculate for which values of $\alpha$ and $\beta$ my system is consistent. I can do that by making a coëfficiënt matrix and using Gaussian elimination to derive the solution space, it being:

$$ x_1=1-\frac{11\beta -22}{2\alpha-28}\\ x_2=\frac{\beta -2}{\alpha -14}\\ x_3=\frac{\beta -2}{2\alpha -28} $$

Then I can see that for the value $\alpha=14$ the denominator = 0, which is not possible. So, I can conclude that the system is only consistent for $\alpha\neq14$. This is clear to me. However, my answer sheet says that $\alpha=14$ and $\beta=2$ is also an answer. Plugging these values into $x_1$, $x_2$ and $x_3$ gives these answers:

$$ x_1=1-\frac{0}{0}\\ x_2=\frac{0}{0}\\ x_3=\frac{0}{0} $$

I thought $\frac{0}{0}$ was supposed to be undefined? How can it be then that using these values for $\alpha$ and $\beta$, we get a consistent system?

To try to figure it out, I checked the answer of $A\overrightarrow{x}=\overrightarrow{b}$, which gives me this (I think):

$$ \begin{bmatrix}1 & 5 & 1\\1 & 6 & -1\\2 & \alpha & -6\end{bmatrix} \begin{bmatrix}1-\frac{0}{0}\\\frac{0}{0}\\\frac{0}{0}\end{bmatrix} = \begin{bmatrix}1-\frac{0}{0}+5(1-\frac{0}{0})+1-\frac{0}{0}\\\frac{0}{0}+6\frac{0}{0}-\frac{0}{0}\\2\frac{0}{0}+\alpha\frac{0}{0}-6\frac{0}{0}\end{bmatrix} $$

Which in turn implies that the equation below must be true, since the answer sheet says that the system is consistent:

$$ \begin{bmatrix}1-\frac{0}{0}+5(1-\frac{0}{0})+1-\frac{0}{0}\\\frac{0}{0}+6\frac{0}{0}-\frac{0}{0}\\2\frac{0}{0}+\alpha\frac{0}{0}-6\frac{0}{0}\end{bmatrix} = \begin{bmatrix}1\\1\\\beta\end{bmatrix} $$

If I substitute $\frac{0}{0}$ for something else (lets say $\frac{0}{0}=p$), I get this:

$$ \begin{bmatrix}1-p+5(1-p)+1-p\\p+6p-p\\2p+\alpha p-6p\end{bmatrix} = \begin{bmatrix}7-7p\\6p\\\alpha p-4p\end{bmatrix} = \begin{bmatrix}1\\1\\\beta\end{bmatrix} $$

I get this equation, which when looking at the first 2 components of the vectors already doesn't make any sense, since for $7-7p=1$ to make sense, $p$ would have to be $\frac{6}{7}$. But for $6p=1$, $p$ would have to be $\frac{1}{6}$.

Elijah
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    When you do Gaussian elimination, you already assume that $2\alpha\neq 28$. So you need to check the case $\alpha=14$ manually using the original equations. – Maximilian Janisch Nov 05 '21 at 15:05
  • @MaximilianJanisch so you're saying that when I conclude that $\alpha \neq 14$, I need to plug $\alpha$ back in the original equation and determine a value for $\beta$? – Elijah Nov 05 '21 at 15:10
  • Yeah exactly: When you assume $\alpha\neq 14$, then you need to check $\alpha=14$ manually by inserting that into the original equations, and trying to see how many solutions this system of equations has for each $\beta$. – Maximilian Janisch Nov 05 '21 at 15:11
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    This part: "Then I can see that for the value α=14 the denominator = 0, which is not possible. So, I can conclude that the system is only consistent for α≠14." is false. All you have actually done is expressed the original problem in a form where, provided $\alpha\neq 14$, it's solutions will match that of the original problem. That is to say that the approach you took works for all cases except when $\alpha =14$. – Justin Benfield Nov 05 '21 at 15:13
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    @MaximilianJanisch omg, when I do that I do eventually get $b=2$! Ludicrous, thx man – Elijah Nov 05 '21 at 15:13
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    @elijah I'm happy to help :) – Maximilian Janisch Nov 05 '21 at 15:13
  • @JustinBenfield ah, so since it doesn't work for $\alpha = 14$, then like Maximilian said I have to manually check for when $\alpha$ is indeed $14$. Makes sense... – Elijah Nov 05 '21 at 15:16
  • Check if the conditions of the Rouche-Capelli hold. It is possible that the system not be compatible with these values of $\alpha$ and $\beta$. – Vasile Nov 05 '21 at 15:54

1 Answers1

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I made a summary of the comments by @MaximilianJanish and @JustinBenfield in the comments under the answer.

After I calculate the solution set, I state that:

Then I can see that for the value α=14 the denominator = 0, which is not possible. So, I can conclude that the system is only consistent for α≠14.

This is actually not the full picture. When I calculate the solution set, I write it in a way where the solution only makes sense if $\alpha \neq14$, so I cannot conclude that there can be no answer for when $\alpha=14$. To check this case, I can plug $\alpha=14$ into the original set of equations, which when solved, gives me a value for $\beta$, namely $\beta=2$.

The final conclusion therefore is that the system is consistent for $\alpha \neq14$ and $\alpha=14, \beta=2$.

Elijah
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  • Note that the $\alpha=14$, $\beta=2$ case is a bit special. The equations are consistent, but not independent. There are infinitely many solutions for the $x_i$. On the other hand, when $\alpha\neq14$ the equations are independent and the $x_i$ have exactly only one solution, namely the one you calculated. – Jaap Scherphuis Nov 05 '21 at 15:40