Find the value of $k$ such that the following system of equations does not have a unique solution: $$kx+y+2z = 4$$ $$-y+4z = 5$$ $$3x+4y+2z = 1$$
I am allowed to use a calculator.
The answer given is:
Which I'm guessing has to do with matrices. Can anyone explain this?
Edit:
The answer given above is for a different question. My mistake. Here's the answer:
Since $$\det\left(\begin{vmatrix} k &1 & 2\\ 0 & -1 & 4\\ 3 & 4 & 2\end{vmatrix}\right)=(-2k +12+0) -(-6+16k+0)=18(1-k)$$ then the system has a unique solution iff $k\not=1$. Moreover $$\det\left(\begin{vmatrix} 1 &2 & 4\\ -1 & 4 & 5\\ 4 & 2 & 1\end{vmatrix}\right)=(4+40-8)-(64+10-2)-36\not = 0$$ therefore for $k=1$ the system has no solutions.
P.S. Remember that the system $Ax=b$ has at least a solution iff: $rank(A)=rank(A∣b)$. In order to compute the determinant of a $3\times 3$ matrix I used Sarrus Rule.