If the system of equations: x-ky-z=0 , kx-y-z=0 , x+y+z=0, has a non zero solution, then the possible values of k are ?
Have you tried anything yourself? In particular, have you tried to solve for x, y, and z to see if you get a "zero solution" (x= y= z= 0) and what might prevent you?
From the first equation, x- ky- z= 0, z= ky- x. Putting that into the third equation, x+ y+ z= 0, x+ y+ ky- x= y+ ky= (1+ k)y= 0. Now, if the product of two numbers is 0, then at least one of them must be 0. So either y= 0 or ....
Sigh. From the first equation, x- ky- z= 0, z= x -ky. Putting that into the second equation, kx- y- z= kx- y- (x- ky)= (k- 1)x+ (k- 1)y= (k-1)(x+ y)= 0. If k is NOT 1, we can divide both sides by k- 1 to get x+ y= 0. In that case, the last equation becomes x+ y+ z= z= 0. If z= 0, t we have x= ky, y= kx so that x= k(kx)= k^2x and (1- k^2)x= 0. If 1- k^2 is not 0, then x must be 0 which means, since x+ y= 0, y= 0 and, since x+ y+ z= 0, z= 0. Was that what you meant by "a long process of substitution"?
If k= 1 the equations are x- y- z= 0, x- y- z= 0, and x+ y+ z= 0. The first two equations give x= y+ z so that last equation becomes 2x= 0. x= 0, y= -z and z anything will satisfy that.
If k= -1, the equations are x+y- z= 0, -x-y- z= 0, and x+ y+ z= 0. The last two equations are the same. Together with z= x+y. we have 2(x+ y)= 0. So x anything, y= -x, and z= 0 will satisfy that.
Here is what I would do (naively). Form the last equation, we have:
$z = -(x+y)$ (so far we haven't assumed anything about $k$).
We can thus eliminate $z$ from the first two equations, replacing it everywhere we see it with $-(x+y)$.
That gives us 2 equations in 2 unknowns, a definite improvement:
$x - ky - (-(x+y)) = 0$
$kx - y - (-(x+y)) = 0$, or more simply:
$2x + (1-k)y = 0$
$(1+k)x = 0$
Since we do not know if $1-k,1+k$ are non-zero, we should avoid dividing by them unless we have to. Multiplying is OK, though.
Let's look at the second equation: we have two choices- either $1+k = 0$ (and we don't know what $x$ is), or we have $x = 0$ (which tells us nothing about $k$). Or both (which tells us everything about both).
Let's assume, for the moment $k \neq -1$. Then $x$ HAS to be 0. SO our first equation becomes:
$(1 - k)y = 0$. Here, we have a similar dilemma, one of the two factors (or both) must be 0. If $k$ isn't 1, either, then $y$ has to be 0. And if $x = y = 0$, then $z = -(x + y) = -(0 + 0) = -0 = 0$, as well.
So, so far, we have discovered:
If $k \not\in \{-1,1\}$, we have the UNIQUE solution: $(x,y,z) = (0,0,0)$.
So now we have to ask: what if $k = -1$, or $k = 1$? Let's back up a bit, and see what happens in our "two equations" in each case:
$k = -1$ makes these equations:
$2x + 2y = 0$
$0 = 0$
The second equation tells us absolutely nothing, but the first says $y = -x$. Here, $x$ could be anything. If we say, for example, $x = a$, then $y = -a$, and thus $z = -(x + y) = -(a + (-a)) = -0 = 0$.
So we have $(x,y,z) = (a,-a,0)$ for any choice of a real number $a$. Since we have infinitely many real numbers, we have infinitely many solutions. Note that specifying $y$ would do as well, since we would then have $x = -y$.
Now we look at $k = 1$, which makes our two equations:
$2x = 0$
$2x = 0$
This tells us $x = 0$, but says nothing about $y$. So $y$ could be any real number, say $b$. This means that $z = -(x + y) = -(0 + b) = -b$, and we have the infinitely many solutions $(x,y,z) = (0,b,-b)$.
We can combine the two conditions: $k \neq 1$ and $k \neq -1$ into one condition: $k^2 - 1 \neq 0$ (since $k^2 -1 = (k + 1)(k - 1)$, if neither factor is 0, neither is their product).
So our "final answer" is:
The system of equations:
$x - ky - z = 0$
$kx - y - z = 0$
$x + y + z = 0$
has a UNIQUE solution (0,0,0), if and only if $k^2 - 1 \neq 0$, and infinitely many otherwise.
If we are looking for a non-zero solution, it follows that $k^2 - 1 = 0$ (so that either $k = 1$ , or $k = -1$), and we must choose a non-zero value for $y$.