I have the following linear system:
$$x + 2y - 3z = 4$$ $$3x - y + 5z = 2$$ $$4x + y + (s^2 - 14)z = s+2$$
Im trying to solve for $s$ to figure out how many solutions it has (if any).
I know how to implement Gauss-Jordan Elimination on matrices without variables but any help on how to go about solving this?
First we eliminate the first terms taking 3 of the first row from the second and 4 of the first row from the third, to get $$x+2y-3z=4$$ $$-7y+14z=-10$$ $$-7y+(s^{2}-2)z=s-14$$
Then we simply take 1 of the new second row from the new third row, and our transformed system is just $$x+2y-3z=4$$ $$-7y+14z=-10$$ $$(s^{2}-16)z=s-4$$
If $s=+4$, the last equation becomes $0=0$ and your set is underdetermined, with solution $$x=z+\frac{20}{7}, \quad y=2z+\frac{10}{7}$$ for any $z$. The solutions lie on a line.
If $s=-4$, the last equation becomes $0=-8$, and your set is overdetermined, with no solution.
If $s$ is any other value, then simply $z=\frac{1}{s+4}$ and the other variables can be found by back-substitution. This solution is unique.