# Simultaneous equations question

• Mar 8th 2010, 03:41 AM
Stroodle
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 $a$ and $b$, where $a$ and $b$ are real constants.

$ax+y+z=1$

$x+by+z=1$

$ax+by+z=1$

The question says to be careful with special values.

• Mar 8th 2010, 04:42 AM
Prove It
Quote:

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 $a$ and $b$, where $a$ and $b$ are real constants.

$ax+y+z=1$

$x+by+z=1$

$ax+by+z=1$

The question says to be careful with special values.

Multiply equation 2 by a, to get

$ax + y + z = 1$

$ax + aby + az = a$

$ax + by + z = 1$.

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

$ax + y + z = 1$

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

$(b - 1)y = 0$.

Now you should be able to solve for $y$, back substitute to find $z$ and $x$.
• Mar 8th 2010, 08:53 PM
Stroodle
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..
• Mar 9th 2010, 02:07 AM
HallsofIvy
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.
• Mar 9th 2010, 03:17 AM
Stroodle
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...
• Mar 9th 2010, 06:05 AM
Hello Stroodle

This question is all about making sure you never divide by something that might be zero. So, continuing from here:
Quote:

Originally Posted by Prove It
$ax + y + z = 1$

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

$(b - 1)y = 0$

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

$\Rightarrow ax = 0$
...(6)
Again, we must ensure that we don't divide by zero. So, if $a \ne 0$, we can divide both sides by $a$ and get:
$x=0$
Therefore, provided $b \ne 1$ and $a \ne 1$ and $a \ne 0$, the solution is:
$(0,0,1)$
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 $a = 0$, then equation
(6) is satisfied by any value of $x$, say $x = \lambda$. So, if $b \ne 1$ and $a = 0$ we get the solution:
$(\lambda, 0,1)$, for any value of $\lambda$.
Next, if $b \ne 1$ and $a = 1$, then (5) is satisfied by any value of $z$, say $z = \mu$. Substituting this value into (1) (with $a = 1$ and $y = 0$) we get:
$x+\mu = 1$

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