# Discussing this system of equations in terms of a parameter, $m$

I need to tell if the following system of equations has either:

• A unique solution, or
• Infinite solutions, or
• No solutions at all.

(for all possible $R$eal values of the given parameter, $m$)

To solve this question I decided to use the well-known Rouché-Capelli theorem. The system of equations under study is: $$\begin{cases} x + my + z = 0 \\ mx – y + 3z = 0\\ x - 3y + mz =0\end{cases}$$

Let's study the rank of the augmented matrix, which I will call $A^*$: $$A^*=\left(\begin{array}{ccc|c} 1 & m & 1 & 0\\ m & -1 & 3 & 0\\ 1 & -3 & m & 0\end{array}\right)\sim \ldots \sim \left(\begin{array}{ccc|c} 1 & m & 1 & 0\\ 0 & (1+m^2)(3+m) & m^2-9 & 0\\ 0 & 0 & 10-m-m^3 & 0\end{array}\right)$$

I obtained the last equivalent matrix after several operations I performed on the original augmented matrix. I double-checked that it is correct, which I'm 100% confident it is.

The thing is that, for $m=2$, both the augmented matrix ($A^*$) and the coefficient matrix ($A$) have rank equal to 2, so there would be infinite solutions in this case.

## Now my question

If you give the previous matrix a second look, you will notice that, for $m=-3$, the rank of both $A^*$ and $A$ would be 2 again: $$A^*(m=-3)=\left(\begin{array}{ccc|c} 1 & -3 & 1 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 40 & 0\end{array}\right)\sim \left(\begin{array}{ccc|c} 1 & -3 & 1 & 0\\ 0 & 0 & 40 & 0\\ 0 & 0 & 0 & 0\end{array}\right)$$

And the solutions would be given by: $$\left(\begin{array} \\x \\ y \\ z\end{array}\right)=\lambda\,\left(\begin{array} \\3\\ 1\\ 0\end{array}\right)$$

However this is incorrect. Following my book's solution, $m=-3$ is no special case. But why? We can clearly see that the rank of the system would still be 2, and in theory there would be infinite solutions (rather than just the trivial solution, $(x,y,z)=(0,0,0)$)

Where did I go wrong? Why do I get $m=-3$ as a special case?

## EDIT

Let me add the operations I performed on $A^*$ to put it in its Row echelon form:

• What are exactly the operations you performed ? just taking the determinant of $A$ ($10-m-m^3$) leads to exactly the same result as your textbook so the problem is most certainly here. – Delta-u Apr 16 '18 at 12:50
• @Moo It was a typo, nothing changes. Thanks for pointing it out, I corrected it! – Jose Lopez Garcia Apr 16 '18 at 13:04
• Hi @Delta-u, let me add an image with the operations I performed, it's pretty straightforward. However I don't know why it didn't work. I know that the determinant of $A$ gives $m=2$ as the only solution, but I just want to know why my solution didn't work as it should. – Jose Lopez Garcia Apr 16 '18 at 13:05
• @JoseLopezGarcia: For that matrix, the CP is $$-m^3-m+10$$ The roots of this are $$m = 2, -1~ \pm ~2i$$ What am I missing since $3$ is not a root of the determinant? – Moo Apr 16 '18 at 13:13
• @Moo I followed another procedure. Instead of calculating the determinant of $A^*$, I reduced $A^*$ to its row-echelon form so I can study its rank. From there I will be able to use the Rouche-Capelli theorem (Please check the end of the updated answer). Following this procedure, which should be correct, should yield all special values of $m$ that I would need to have into account – Jose Lopez Garcia Apr 16 '18 at 13:16

In your computation of the rank of $A^*$ to go from the third matrix to the fourth you multiply the second row by $(3+m)$. But this is a rank preserving operation only if $3+m \neq 0$.
Hint: eliminating the variables $y,z$ i got $$x(m^3+m-10)=0$$ Can you finish?