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I have the following equation system matrix:

$$\left[\begin{array}{ccc|c} c & 1 & 1 & 1 \\ 1 & c & 1 & 1 \\ 1 & 1 & c & 1 \end{array}\right]$$

From this one I'm supposed to be able to define the constant $c$, for creating an equation with either no solutions, one solution or infinite solutions.

I have successfully gotten the row reduce matrix:

$$\left[\begin{array}{ccc|c} 1 & 0 & 0 & \frac{1}{c+2} \\ 0 & 1 & 0 & \frac{1}{c+2} \\ 0 & 0 & 1 & \frac{1}{c+2} \end{array}\right]$$

But I'm not sure where to go from this. Is it possible just from the this form to see which definitions of $c$ gives the different solutions?

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That form should suggest something funny will happen (that may be useful to you) if $c=-2$, but I don't think you can prove what you need directly from that form. – dfeuer Sep 16 '13 at 15:27
What you did does not look right. It seems clear that for $c=1$ one does not get a unique solution. Row reduction without constraining $c$ seems a difficult route here. – Marc van Leeuwen Sep 16 '13 at 15:28
Note: I did not check any of the work, so my comment may not make sense if the work is wrong. – dfeuer Sep 16 '13 at 15:29
If you have studied determinants then $D=c^3-3c+2=(c-1)^2(c+2)$ indicates that for c other than 1 and -2 you can always solve such an equation regardless of any constant right hand side. The you have to check $c=1$ and $c=-2$ to see if you have none or infinitely many answers. – Maesumi Sep 16 '13 at 15:36
up vote 3 down vote accepted

Notice that

$$\det \begin{bmatrix} c & 1 & 1 \\ 1 & c & 1 \\ 1 & 1 & c \end{bmatrix} = c^3 - 3c + 2$$

has two roots: $-2$ and $1$. So, for all other values of $c$, the system is regular and, hence, has a single solution.

It should be easy to check what happens for $c = -2$ and $c = 1$.

To check the above: determinant, roots.

Since this one is not as obvious as the other one, here is a hint for $c = -2$: summarize all three equations. What do you get?

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If you have the deterimant of $A$, where the system is $Ax=(1,1,1)^t$, then you have made one step forward, too. We have $\det(A)=(c+2)(c-1)^2$, so that for all $c\neq 1,-2$ there is a unique solution $x=A^{-1}(1,1,1)^t$. Then check the cases $c=1$ and $c=-2$.

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Row reduction looks OK (algebraically), and suggests $c=-2$ to give no solution. But note that something is assumed during the automatic process of row reduction (such as nonzero "pivots"), since if $c=1$ in the original system it's the same equation three times, so infinitely many solutions. Probably any $c$ other than $1,-2$ will give the unique solution as in the row reduced version. But I'd suggest not relying on the row reduced algebraic form but instead just pick three values of $c$ say $1,-2,10$ and going back to the original system with those numbers put in to get the three desired examples.

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I would recommend the others' answers over mine, as they have wider applicability, but here is a way to see things more concretely. Let

$$ J=\begin{bmatrix}1&1&1\\ 1&1&1\\ 1&1&1\end{bmatrix}, \ \mathbf{v}=\begin{bmatrix}x\\ y\\ z\end{bmatrix}, \ \mathbf{e}=\begin{bmatrix}1\\ 1\\ 1\end{bmatrix}. $$ Then your system of equations can be rewritten as $[(c-1)I+J]\mathbf{v}=\mathbf{e}$. Since $J\mathbf{v}=(x+y+z)\mathbf{e}$, the equation can be further rewritten as $$(c-1)\mathbf{v}=(1-x-y-z)\mathbf{e}.\tag{1}$$ So, we immediately see how many solutions there are if $c=1$.

What if $c\ne1$? Divide both sides of $(1)$ by $c-1$, we see that the solution $\mathbf{v}$, if exists, must be a scalar multiple of $\mathbf{e}$. Hence $x=y=z$. Therefore, $(1)$ reduces to a single one-variable equation $(c-1)x=1-3x$, or equivalently, $(c+2)x=1$. The number of solutions again depends on the value of $c$.

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