Find all values of a for which the following system has a solution, no solution and infinitely many solution.

Question

Find all values of a for which the following system $$\begin{cases}x+2y+z=a^2 \\x+y+3z=a \\3x+4y+8z=8\end{cases}$$

has a solution, no solution and infinitely many solution.

I found the reduced row echelon form of this system which is:

$\begin{pmatrix} 1 & 0 & 0 & 4a^2+12a-40\\ 0 & 1 & 0 & -a^2-5a+16\\ 0 & 0 & 1 & -a^2-2a+8 \end{pmatrix}$

Does that mean this system has only one solution and there is no value for a which makes the system infinitely many solutions and no solution?

Answer 1

Yes since the RREF is the following

$$\begin{pmatrix} 1 & 2 & 1 &a^2\\ 1 & 1 & 3 & a\\ 3 & 4 & 8 & 8 \end{pmatrix} \to \begin{pmatrix} 1 & 2 & 1 &a^2\\ 0 & 1 & -2 & a^2-a\\ 0 & 0 & 1 & 8-2a-a^2 \end{pmatrix}$$

by Rouché–Capelli theorem, we always have exacty one soution for the system.

Answer 2

The system of equation if written as $AX=B$, then $\det|A|=-1 \ne 0$. so this system will have a unique solution for any value of $a$.