1. ## Simultaneous equations question

Hi there, I'm trying to work out this question, but not having much luck:

Solve this system of equations for all possible values of $\displaystyle a$ and $\displaystyle b$, where $\displaystyle a$ and $\displaystyle b$ are real constants.

$\displaystyle ax+y+z=1$

$\displaystyle x+by+z=1$

$\displaystyle ax+by+z=1$

The question says to be careful with special values.

2. Originally Posted by Stroodle
Hi there, I'm trying to work out this question, but not having much luck:

Solve this system of equations for all possible values of $\displaystyle a$ and $\displaystyle b$, where $\displaystyle a$ and $\displaystyle b$ are real constants.

$\displaystyle ax+y+z=1$

$\displaystyle x+by+z=1$

$\displaystyle ax+by+z=1$

The question says to be careful with special values.

Multiply equation 2 by a, to get

$\displaystyle ax + y + z = 1$

$\displaystyle ax + aby + az = a$

$\displaystyle ax + by + z = 1$.

Now apply $\displaystyle R_2 - R_1 \to R_2$ and $\displaystyle R_3 - R_1 \to R_3$

$\displaystyle ax + y + z = 1$

$\displaystyle (ab - 1)y + (a - 1)z = a - 1$

$\displaystyle (b - 1)y = 0$.

Now you should be able to solve for $\displaystyle y$, back substitute to find $\displaystyle z$ and $\displaystyle x$.

3. Hmm, I still don't get it. The only possible values of a and b that I get are both 1; Giving infinitely many solutions, but I'm sure there must be more to it than this..

4. I have no idea what you mean by "The only possible values of a and b that I get are both 1". a and b can have any values at all. For example if a= b= 0, the equations reduce to y+ z= 1, x+ z= 1, and z= 1 which has solution x= y= 0, z= 1.

The "special values" referred to are a= b= 1 for which the equations do not have a single value.

5. Oh yeah, Thanks.

I still can't find any other values except x=0, y=0 and z=1 though... New to these simultaneous equations...

6. Hello Stroodle

This question is all about making sure you never divide by something that might be zero. So, continuing from here:
Originally Posted by Prove It
$\displaystyle ax + y + z = 1$

$\displaystyle (ab - 1)y + (a - 1)z = a - 1$

$\displaystyle (b - 1)y = 0$
If $\displaystyle b \ne 1$, we can divide both sides of
$\displaystyle (b-1)y=0$ ...(4)
by $\displaystyle (b-1)$ to get:
$\displaystyle y = 0$
Substitute into (2):
$\displaystyle (a-1)z = a-1$ ...(5)
So if $\displaystyle a\ne 1$ we can now divide both sides by $\displaystyle (a-1)$ to get:
$\displaystyle z=1$
Substitute these values of $\displaystyle y$ and $\displaystyle z$ into (1) to get:
$\displaystyle ax+0+1=1$

$\displaystyle \Rightarrow ax = 0$
...(6)
Again, we must ensure that we don't divide by zero. So, if $\displaystyle a \ne 0$, we can divide both sides by $\displaystyle a$ and get:
$\displaystyle x=0$
Therefore, provided $\displaystyle b \ne 1$ and $\displaystyle a \ne 1$ and $\displaystyle a \ne 0$, the solution is:
$\displaystyle (0,0,1)$
Now we need to re-trace our steps, and see what happens if one or more of these conditions is not satisfied.

Starting with the last condition first, if $\displaystyle a = 0$, then equation
(6) is satisfied by any value of $\displaystyle x$, say $\displaystyle x = \lambda$. So, if $\displaystyle b \ne 1$ and $\displaystyle a = 0$ we get the solution:
$\displaystyle (\lambda, 0,1)$, for any value of $\displaystyle \lambda$.
Next, if $\displaystyle b \ne 1$ and $\displaystyle a = 1$, then (5) is satisfied by any value of $\displaystyle z$, say $\displaystyle z = \mu$. Substituting this value into (1) (with $\displaystyle a = 1$ and $\displaystyle y = 0$) we get:
$\displaystyle x+\mu = 1$

$\displaystyle \Rightarrow x = 1-\mu$
So the solution for $\displaystyle b \ne 1$ and $\displaystyle a = 1$ is:
$\displaystyle (1-\mu,0,\mu)$, for any value of $\displaystyle \mu$.
Now you need to go back and see what happens if $\displaystyle b = 1$, where equation (4) is then satisfied by any value of $\displaystyle y$.