1. Linear Algebra help

hey, i need some hints on how do this problem. i really have no idea how to do it... i don't need an answer, i just need to understand how to do the problem...

it is:

x + y + kx = 1
x + ky + z = 1
kx + y + z = -2

no solution, unique solution, or infinitely many solutions

2. The symmetry of the equations suggests to look at the particular case $k=1$ first. What happens in this case?

Then, suppose $k\neq 1$. You can try to solve the system just as you would have with known numbers: for instance, using the second and the last equation, you can get an equation without $z$. Repeating this with another subset of two equations, you find a system of two equations in $x$ and $y$ (where the coefficients depend on $k$). Do these equations have one solution? none? infinitely many?

I hope this helps. If you know matrices and determinants, there's a quicker solution by looking at the system only for the values of $k$ which make the determinant of the system equal to 0 (you know $1$ is one of them), since you know that if it is non-zero then there's a unique solution.
Laurent.

3. Originally Posted by aeubz
hey, i need some hints on how do this problem. i really have no idea how to do it... i don't need an answer, i just need to understand how to do the problem...

it is:

x + y + kx = 1
x + ky + z = 1
kx + y + z = -2

no solution, unique solution, or infinitely many solutions
First of all, I'm assuming you made a misprint, and that by:

$\left\{\begin{array}{c}x + y + kx = 1 \\ x + ky + z = 1 \\ kx + y + z = -2 \end{array}\right\}$

...you actually meant:

$\left\{\begin{array}{c}x + y + kz = 1 \\ x + ky + z = 1 \\ kx + y + z = -2 \end{array}\right\}$

I actually don't know how to do this using matrices, but you can do it using high school algebra. According to the first equation:

$x=1-y-kz$

Plug that into the second equation, then find y in terms of z:

$1-y-kz+ky+z=1$

$y=z$

Now plug your x and y equations into the third equation to find z in terms of k:

$k(1-z-kz)+z+z=-2$

$z=\frac{k-2}{(k+2)(k-1)}$

Which of course means:

$y=\frac{k-2}{(k+2)(k-1)}$

Which in turn means:

$x=1-\frac{k-2}{(k+2)(k-1)}-k\frac{k-2}{(k+2)(k-1)}$

or, simplified:

$x=\frac{2k}{(k+2)(k-1)}$