# Thread: Linear equations with a parameter

1. ## Linear equations with a parameter

How do I find for which K parameter we have zero solutions, single solution and infinite number of solutions? I can't seem to do it with the Gaussian Elimination.

2. Hello, GIPC!

Here's a start . . .

$\begin{array}{ccccccc}
x & + & (k^2-6)y &+& (4k+4)z &=& 5k+3 \\
\text{-}x &+& (2k^2-6)y &+& (4k+4)z &=& 6k+3 \\
& +&(3k^2-12)y &+& (6k+6)z &=& 9k+6 \end{array}$

Find for which parameter $k$ we have zero solutions, single solution and infinite number of solutions.
Consider the determinant of the system: . $D \;=\;\begin{vmatrix}1 & k^2-6 & 4(k+1) \\ \text{-}1 & 2(k^2-3) & 4(k+1) \\ 0 & 3(k^2-4) & 6(k+1) \end{vmatrix}$

Factor $2(k+1)$ from column 3: . $D \;=\;2(k+1)\begin{vmatrix}1 & k^2-6 & 2 \\ \text{-}1 & 2(k^2-3) & 2 \\ 0 & 3(k^2-4) & 3 \end{vmatrix}$

Evaluate: . $D \;=\;6(k+1)\bigg[\left[2(k^2-3) - 2(k^2-4)\right] + \left[(k^2-6) - 2(k^2-4)\right] + 0\,\bigg]$

And we have: . $D \;=\;6(k+1)(-2k^2+8) \;=\;-12(k+1)(k^2-4) \;=\;-12(k+1)(k-2)(k+2)$

. . Hence: . $D \:=\: 0\,\text{ for }\,k \:=\:\text{-}1,\: 2,\:\text{-}2$

Can you take it from here?

3. I have another question, if the system changes to:

x + (k^2 -6)y + (4k+4)z = 5k+3
(3k^2 -3)y +(6k+6)z = 9k+6

for which k do i have a single solution, no solution and infinite solotions? How do I solve it?

4. bump
anyone?

this problem is giving me fits.

5. Are you not trying anything yourself? For example, it shouldn't take long to see that you can reduce the second equation to $(k^2- 1)y+ 2(k+1)z= 3k+ 2$. And, as long as $k\ne \pm 1$, we must have $y= \frac{3k+ 2- 2(k+1)z}{k^2- 1}$ for any value of z. And then $x= 5k+3- (4k+4)z- (k^2- 6)y$.