1. ## System of equations

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 ?

2. ## Re: System of equations

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 ....

3. ## Re: System of equations

How about the determinant of the coefficient matrix must be zero?

$k^2-1=0$

4. ## Re: System of equations

Originally Posted by Idea
How about the determinant of the coefficient matrix must be zero?

$k^2-1=0$
Matrix is probably too advanced for this question, with the assumption that the person asking this is in high school... (unless they have already learned it. Curriculum does differ from place to place.)

5. ## Re: System of equations

Originally Posted by Idea
How about the determinant of the coefficient matrix must be zero?

$k^2-1=0$
I would like to know more about how you derive k2-1=0. Please explain.

6. ## Re: System of equations

First off, you take out the co-efficients and put it into a matrix.
x-ky-z=0: [1, -k, -1 | 0]
Likewise, [k, -1, -1| 0] and
[1, 1, 1|0]

7. ## Re: System of equations

For determinant calculation, I will advise you to look it up since it's really hard to explain it here.

8. ## Re: System of equations

Have you tried solving the equations as I suggested before? What do you get as the solution in terms of k?

9. ## Re: System of equations

Originally Posted by HallsofIvy
Have you tried solving the equations as I suggested before? What do you get as the solution in terms of k?
Yes, I have found the value of k = +1 and -1. But that was with a long process of substitution and it was not exactly your method. I was actually looking for a simple and easy solution. Also I didn't understand x=y=z=0 or zero solution. How did you derive that ?

10. ## Re: System of equations

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.

11. ## Re: System of equations

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).

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 $y$.