# Linear equations with a parameter

• March 27th 2010, 04:19 AM
GIPC
Linear equations with a parameter
http://img34.imageshack.us/img34/3480/basicq.jpg

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.
• March 27th 2010, 05:32 AM
Soroban
Hello, GIPC!

Here's a start . . .

Quote:

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

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?

• March 27th 2010, 08:08 AM
GIPC
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?
• March 28th 2010, 04:45 AM
GIPC
bump
anyone? (Worried)

this problem is giving me fits.
• March 28th 2010, 05:44 AM
HallsofIvy
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 $x= 5k+3- (4k+4)z- (k^2- 6)y$.