Answer: k = 7
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Explanation:
There are probably a number of ways to approach this problem, but row reduction was the only method I could think of at the moment.
If you were to follow the steps shown in the attached image, then you'll be on the process of applying row reduction. The last step in that diagram isn't in full REF (row echelon form), but that doesn't technically matter.
What does matter is the 2k-14 entry in the bottom row. If that entry was 0, while the entry just to the right of it was nonzero, then this would lead the entire system of equations to be inconsistent. This is because the bottom equation would be in the form 0x+0y+0z = m, where m is some nonzero constant. As you can see, that equation would simplify to 0 = m; however, m is nonzero, so we have a contradiction.
If 2k - 14 were 0, then
2k - 14 = 0
2k = 14
k = 14/2
k = 7
This is the only k value in which the system is inconsistent. In other words, the system wouldn't have any solutions with this k value. You can verify this through completing the row reduction with k = 7 (it should be far easier now that we can nail down a fixed k value) and find that you'll get a contradiction. You could also use substitution to find an inconsistency would arise when k = 7.
Side notes:
- You don't need to do full RREF, though you can if you want. REF should be sufficient.
- I drew each matrix as a grid of boxes to help separate the terms. This is also done to space out each step. Usually those grid lines aren't present.
