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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: enter image description here

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:enter image description here

StopReadingThisUsername
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  • Since the matrix' determinant is $-18k+18$, the answer is $k=1$. – Michael Hoppe Aug 12 '16 at 08:18
  • For $k=5$, there is a unique solution: $x=1/2, y = -2/3, z = 13/12$. For $k=1$ the system has no solutions. Something is wrong here. Are you sure that the answer you showed us is the answer to that exercise? – Robert Z Aug 12 '16 at 08:28
  • @RobertZ Good catch. It is my mistake. The edit shows the correct answer. How do you get the determinant to equal $-18k+18$? – StopReadingThisUsername Aug 12 '16 at 08:34

1 Answers1

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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.

Robert Z
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