# Let be $\lambda$ and $\mu$ complex numbers. Find the general solution $(x,y,z)\in \mathbb{C}^{3}$ to the system of equations:

Let be $$\lambda$$ and $$\mu$$ complex numbers. Find the general solution $$(x,y,z)\in \mathbb{C}^{3}$$ to the system of equations:

\begin{align*} \lambda x+y+z&=1\\ x+\lambda y+z&=\mu\\ x+y+\lambda z &=\mu^{2} \end{align*}

I found that the determinant of the coefficient matrix is distinct to zero, this is: \begin{align} \begin{vmatrix} \lambda & 1 & 1\\ 1& \lambda & 1\\ 1 & 1 & \lambda \end{vmatrix}=\lambda^3+2-3\lambda=\lambda(\lambda^2-3)+2\neq0 \ \iff\lambda\neq0 \end{align}

So, there should be an unique solution if $$\lambda\neq0$$. By otherside \begin{align*} x=\frac{1-y-z}{\lambda} \end{align*} I solved it using $$z$$ as a paramter, and I got that the general solution is:

$$\begin{pmatrix} \mu^2-\lambda z-\left [ \mu-z-\frac{1-z}{\lambda} \right ]\frac{\lambda}{\lambda^2-1}\\ \left [ \mu-z-\frac{1-z}{\lambda} \right ]\frac{\lambda}{\lambda^2-1}\\ z \end{pmatrix}$$

My question is, how can I analyze the system of equations to know how to solve it? I mean, with this problem I decide to use the equations and I "luckely" solved it, but I would like to know how to proceed in every case. Hope you can help me to understand how to decide how to solve a system of equations, or at least in this case how could I knew how to solve it?. I would really appreciate your help!

There are some problems here. You have computed, correctly, $$\begin{vmatrix} \lambda & 1 & 1\\ 1& \lambda & 1\\ 1 & 1 & \lambda \end{vmatrix}=\lambda^3-3\lambda + 2.$$ However, this is not $$0$$ if and only if $$\lambda = 0$$. Indeed, substituting $$\lambda = 0$$ into this produces a determinant of $$2$$.

Further, if the determinant is non-zero, as you say, you'll get a unique solution. This means you can't use $$z$$ (or any of the variables) as a free parameter. The solution is unique, hence there must be one and only one value of $$z$$ that produces a solution.

Let's look at the polynomial $$\lambda^3-3\lambda + 2$$. We can solve this via the usual methods: guess a root, use the factor theorem, then solve the remaining quadratic. If we look at the matrix itself, it jumps out at me that $$\lambda = 1$$ must be a root, as the corresponding matrix would have all the same rows/columns, making the rows/columns linearly dependent. So, I expect $$\lambda = 1$$ to be a root (it is, check it!), and $$\lambda - 1$$ to be a factor. Using polynomial division, $$\lambda^3 - 3\lambda + 2 = (\lambda - 1)(\lambda^2 + \lambda - 2).$$ Solving/factorising the quadratic by whatever means you prefer, you should then get $$\lambda^3 - 3\lambda + 2 = (\lambda - 1)^2(\lambda - 2).$$

While this information will be important, it will not, by itself, help us solve the system in place. Cramer's Rule could be used when $$\lambda \neq -2, 1$$, but just because the system doesn't have a unique solution doesn't mean it can't be solved.

Instead, just set up an augmented matrix, as usual:

$$\left[\begin{array}{ccc|c} \lambda & 1 & 1 & 1 \\ 1 & \lambda & 1 & \mu \\ 1 & 1 & \lambda & \mu^2 \end{array}\right].$$

We can now perform all the usual elementary row operations. Remember, $$\lambda$$ is just a scalar. Though, we must be careful when dividing a row by a scalar expression in terms of $$\lambda$$, in case that expression happens to be $$0$$. First, let's swap row $$1$$ and $$3$$, to get $$1$$ in the top left:

$$\left[\begin{array}{ccc|c} 1 & 1 & \lambda & \mu^2 \\ 1 & \lambda & 1 & \mu \\ \lambda & 1 & 1 & 1 \end{array}\right].$$

Subtract row $$1$$ from $$2$$, and $$\lambda$$ lots of row $$1$$ from row $$3$$:

$$\left[\begin{array}{ccc|c} 1 & 1 & \lambda & \mu^2 \\ 0 & \lambda - 1 & 1 - \lambda & \mu - \mu^2 \\ 0 & 1 - \lambda & 1 - \lambda^2 & 1 - \lambda\mu^2 \end{array}\right]. \tag{1}$$

The next step I'd want to do is divide a row by $$\lambda - 1$$ or $$1 - \lambda$$. But, what if this scalar is $$0$$, i.e. if $$\lambda = 1$$? Then the matrix turns into:

$$\left[\begin{array}{ccc|c} 1 & 1 & 1 & \mu^2 \\ 0 & 0 & 0 & \mu - \mu^2 \\ 0 & 0 & 0 & 1 - \mu^2 \end{array}\right],$$

which can only have a solution if $$1 - \mu^2 = 0$$ and $$\mu - \mu^2 = 0$$. The first equation is satisfied if and only if $$\mu = \pm 1$$, while the second is satisfied if and only if $$\mu = 0$$ or $$1$$. Thus, $$\mu = 1$$ is the only possible value that produces solution. In this case, we get one non-zero row: $$x + y + z = 1,$$ which we solve by making $$y$$ and $$z$$ both free parameters.

Moving back to $$(1)$$, we can now divide rows $$2$$ and $$3$$ by $$1 - \lambda$$, giving us,

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

Add row $$2$$ to row $$3$$, then multiply row $$2$$ by $$-1$$ to get

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

Again, we find ourselves wanting to divide by the scalar $$\lambda + 2$$, but we don't know if it's $$0$$. Try substituting $$\lambda = -2$$ into the matrix, and find the two values of $$\mu$$ that produce solutions. This time, when $$\mu$$ is one of those two values, there will only be a one-parameter solution.

Otherwise, $$\lambda + 2 \neq 0$$, and we get

$$\left[\begin{array}{ccc|c} 1 & 1 & \lambda & \mu^2 \\ 0 & 1 & -1 & \frac{\mu^2 - \mu}{1 - \lambda} \\ 0 & 0 & 1 & \frac{1 - \lambda\mu^2 + \mu - \mu^2}{(1 - \lambda)(\lambda + 2)} \end{array}\right].$$

This gives us immediately our unique value of $$z$$ (assuming $$\lambda \neq 1, -2$$), and the other values can be found through back-substitution.

It's a long question for sure, but it can be done with all the elementary techniques.