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Okay, so this is probably an elementary level question, but I am going to ask anyways since I cannot figure out what to do next.

Given the system of equations:

x + 2y - 5z = 1

x + y + 4z = 1

4x + 10y + kz = 5

Find a k such that there is no solution to the system. How I started on the problem: From looking at the first two equation I figured out that y=9z. I am not sure what to do with this information. My reasoning could be wrong but I feel that there would be many k such that there would be no solution and only one that there would be a solution. Am I wrong in my reasoning?

I also know that if I could get a row in the matrix that was untrue there would be no solution.

3 Answers 3

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You are right when you are talking about a matrix, it could provide an elegant solution.

Try to compute the derminant

$$\begin{vmatrix} 1 & 2 & 5 \\ 1 & 1 & 4 \\ 4 & 10 &k \end{vmatrix}$$

(why this one?) and then find a value of $k$ to make it equals to $0$.

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    The $k$ determined this way promises that the system does not have a *unique* solution. You need one more step in order to distinguish between no solutions at all vs. inifinitely many solutions. – 2017-01-24
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    @dxiv Absolutely, but the way the question is formulated, it is sufficient to find the value of $k$ requested. If you want to prove that there is not infinitely many solutions, you can prove that this matrix have a rank $\geqslant 2$. – 2017-01-24
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    The question asks to "*Find a value such that linear system has* ***no solution***". If you replace the $=5$ in the last equation with $=4$ then the system will have infinitely many solutions for the $k$ you determined, but your answer does not even look at the free terms on the RHS. – 2017-01-24
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    @dxiv You are right! Though I was trusting the question to be correct :) (maybe not the best way to go I agree) – 2017-01-24
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I figured out that $y=9z\,$.

That's a good observation. Take it one step farther, and substitute $y=9z$ back into the first equation to get $x=1-2y+5z=1-18z+5z=1-13z\,$.

Now substitute $x,y$ into the last equation:

$$ 4 \cdot (1-13z) + 10 \cdot 9z + k z = 5 \quad \iff \quad (38 +k)\,z = 1 $$

Can you see a value of $k$ for which this last equation has no solution?

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Possible methods:

  1. Find $k$ so that the determinant of the coefficient matrix is $0$.
  2. Row reduce and see what happens to the last row.