Is this the right solution for an unsolvable system?

Question

I have to find the values of $k$ such that this system $\left[ \begin{array}{ccc|cc} 1&0&-3&-3\\ 2&k&-1&-2\\ 1&2&k&1 \end{array} \right]$ is:

a) unsolvable b) solvable with infinite solutions c) one solution.

After some steps we can see in this matrix $\left[ \begin{array}{ccc|cc} 1&0&-3&-3\\ 0&k&5&4\\ 0&2&k+3&4 \end{array} \right]$ for $k=2$ the second and the thrid row are the same, so if the substract them on row becomes all zeros. (And the system is unsolvable)

The next step: $\left[ \begin{array}{ccc|cc} 1&0&-3&-3\\ 0&k&5&4\\ 0&0&k^2-7&4k-8 \end{array} \right]$ now im not that sure but I think, the solutions to $k^2-7=0$ should be the values of $k$ such that there are infinite solutions.

Then every possbile value k besides 2 and the solutions for $k^2-7=0$ would return a solvable system with one solution?

Answer 1

Note that the system has no/infinite or unique solutions if the determinant of the system is zero or nonzero. The determinant is: $$k^2+3k-10.$$ Thus: $$k^2+3k-10=0 \Rightarrow k_1=-5 \ (\text{no}); k_2=2 \ (\text{infinite}).$$ For $k\ne -5;2$ there is a unique solution.

Answer 2

from the first equation we get $$x_1=-3+3x_3$$ plugging this in the second equation and solving for $x_3$ we have $$x_3=\frac{4}{5}-\frac{k}{5}x_2$$ and for the third equation we get $$x_2\left(2-\frac{k}{3}(3+k)\right)=4-\frac{4}{5}(3+k)$$ can you finish? from the last equation we get $$x_2(6-3k-k^2)=\frac {3}{5}(8-4k)$$

Answer 3

After the first step you should consider $k=0$, in which case there is a unique solution, just plug in $k=0$.

If in your second step, if $k=\pm\sqrt7$, the system has no solution because for these values $4k-8\neq0$.