Solutions of a system of linear equations with a parameter

Question

I've come across a linear algebra problem that I'm not sure how to solve. It's a generic problem - I have to find the solutions of a system of equations dependent on a parameter. So, my first though was to check for which values of the parameter the matrix on the left is not invertible (1 and -2). I think that for these values there are no solutions.

Is that enough, or should I check the rank of the augmented matrix and use the Rouché–Capelli theorem (for which values of the parameter the planes meet at a point, a line or a plane)?

Thanks,

$$\begin{pmatrix} \lambda & 1 & 1\\ 1 & \lambda & 1\\ 1 & 1 & \lambda \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix}= \begin{pmatrix} 1 \\ \lambda \\ \lambda^2 \end{pmatrix} $$

Answer 1

You have checked that for $\lambda \ne 1,-2$ the system has a solution and it is unique. It is enough to apply Cramer's rule to get the solution.

Now you have to consider the remaining cases and you need to make use of the Rouch\'e-Capelli theorem.

For $\lambda=-2$ you have that

$$rank\left( \begin{matrix}-2 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -2 \end{matrix}\right)=2\neq 3 =rank \left( \begin{matrix}-2 & 1 & 1 & 1 \\ 1 & -2 & 1 & -2\\ 1 & 1 & -2 & 4 \end{matrix}\right)$$ and so the system has no solution. But for $\lambda=1$ you have that

$$rank\left( \begin{matrix}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{matrix}\right)=1 =rank \left( \begin{matrix}1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1\\ 1 & 1 & 1 & 1 \end{matrix}\right),$$ which tells you that the system has infinitely many solutions. Indeed, its solutions form a plane through the origin. Note that a plane has dimension two which is equal to the number of variables ($3$) minus the common rank of the matrices ($1$).

Answer 2

$$\left(\begin{array}{ccc|c} \lambda & 1 & 1 & 1 \\ 1 & \lambda & 1 & \lambda \\ 1 & 1 & \lambda & \lambda^2 \end{array}\right)\sim \left(\begin{array}{ccc|c} \lambda+2 & \lambda+2 & \lambda+2 & 1+\lambda+\lambda^2 \\ 1 & \lambda & 1 & \lambda \\ 1 & 1 & \lambda & \lambda^2 \end{array}\right)\overset{(1)}\sim \left(\begin{array}{ccc|c} 1 & 1 & 1 & \frac{1+\lambda+\lambda^2}{\lambda+2} \\ 1 & \lambda & 1 & \lambda \\ 1 & 1 & \lambda & \lambda^2 \end{array}\right)\sim \left(\begin{array}{ccc|c} 1 & 1 & 1 & \frac{1+\lambda+\lambda^2}{\lambda+2} \\ 0 & \lambda-1 & 0 & \frac{\lambda-1}{\lambda+2} \\ 0 & 0 & \lambda-1 & \frac{\lambda^3+\lambda^2-\lambda-1}{\lambda+2} \end{array}\right)\overset{(2)}\sim \left(\begin{array}{ccc|c} 1 & 1 & 1 & \frac{1+\lambda+\lambda^2}{\lambda+2} \\ 0 & 1 & 0 & \frac{1}{\lambda+2} \\ 0 & 0 & 1 & \frac{(\lambda+1)^2}{\lambda+2} \end{array}\right)\sim \left(\begin{array}{ccc|c} 1 & 0 & 0 & \frac{-\lambda-1}{\lambda+2} \\ 0 & 1 & 0 & \frac{1}{\lambda+2} \\ 0 & 0 & 1 & \frac{(\lambda+1)^2}{\lambda+2} \end{array}\right)\sim $$

The step (1) is only valid for $\lambda\ne -2$.
The step (2) is only valid for $\lambda\ne 1$.

For values $\lambda\ne1,-2$ we get that there is only one solution $x_1=\frac{-\lambda-1}{\lambda+2}$, $x_2=\frac{1}{\lambda+2}$, $x_3=\frac{(\lambda+1)^2}{\lambda+2}$.

We can do the sanity check.
$\lambda x_1 + x_2 + x_3 = \frac{\lambda(-\lambda-1)+1+(\lambda+1)^2}{\lambda+2} = \frac{-\lambda^2-\lambda+1+\lambda^2+2\lambda+1}{\lambda+2} = \frac{\lambda+2}{\lambda+2} = 1$ $x_1 + \lambda x_2 + x_3 = \frac{(-\lambda-1)+\lambda+(\lambda+1)^2}{\lambda+2} = \frac{\lambda^2+2\lambda}{\lambda+2} = \lambda$ $x_1 + x_2 + \lambda x_3 = \frac{(-\lambda-1)+1+\lambda(\lambda+1)^2}{\lambda+2} = \frac{-\lambda+\lambda^3+2\lambda^2+\lambda}{\lambda+2} = \frac{\lambda^3+2\lambda^2}{\lambda+2}=\lambda^2$

Notice that the check works for $\lambda=1$. So we know that there is at least one solution for $\lambda=1$.

For $\lambda=1$

We get the system $$\left(\begin{array}{ccc|c} 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \end{array}\right)$$ which has solutions $x_1=1-s-t$, $x_2=s$, $x_3=t$ (where $s$, $t$ are arbitrary).

For $\lambda=-2$

$\left(\begin{array}{ccc|c} -2 & 1 & 1 & 1 \\ 1 & -2 & 1 & -2 \\ 1 & 1 & -2 & 4 \end{array}\right)\sim \left(\begin{array}{ccc|c} -2 & 1 & 1 & 1 \\ 1 & -2 & 1 & -2 \\ 0 & 0 & 0 & 3 \end{array}\right)$.

So in this case there is no solution.