I've run into somewhat of a problem during my Linear Algebra homework and I can't make heads or tails of it for some odd reason. I'm hopeful that one of you could help me out. It's worded as so :
Consider the System of Linear Equations
|x + y - z = 2|
|x + 2y + z = 3|
|x + y + (k^2 - 5)z = k|
where k is an abitrary constant. For which value(s) of k does this system have
a unique Solution, infinite many solutions and no solutions
I've already looked through the whole chapter, and did a lot of searching but I can't seem to figure out how to prove it. I can tell that if K = 2 then the first and third equations will be the same, but I don't know how I could word that, and I dont know if the second one has to be the same in order to qualify as infinitely many solutions. I feel like I'm missing something. Any help would be appreciated!
Indeed, if $k = 2$, the first and third equations will be the same, so the system of equations will have infinitely many solutions.
To exhaust all such values that lead to infinite solutions, I suggest you construct the augmented coefficient matrix for you system of equations, $$\begin{pmatrix} 1 & 1 & -1 &|& 2 \\ 1 & 2 & 1 &|&3\\ 1 & 1 & k^2 - 5 &|& k\end{pmatrix}$$and then put it into row-echelon form (e.g. add $-R_1$ to $R_2$ and to $R_3$). That will help simplify matters.
If there are any values of $k$ at which one or more rows become all zeros, then there will be infinitely many solutions for each such values $k$.
Again, looking at your the augmented coefficient matrix, after row reducing it: If there are values of $k$ at which any row has all zeros except for a non-zero entry in the right-most column (the augmented portion), then the system of equations is inconsistent, meaning NO solution exists at that value of $k$.
Note that $k = -2$ is such a value.
Finally, each and every value of $k$ not leading to infinitely many solutions and not leading to no solution will give rise to a unique solution.