Math Help - Simultaneous Equation

1. Simultaneous Equation

For what values of $\lambda$ and $\mu$, the simultaneous equations

$x+y+z=6$
$x+2y+3z=10$
$x+2y+\lambda z=\mu$

i) no solution
ii) a unique solution
iii) an infinite number of solution

2. Originally Posted by varunnayudu
For what values of $\lambda$ and $\mu$, the simultaneous equations

$x+y+z=6$
$x+2y+3z=10$
$x+2y+\lambda z=\mu$

i) no solution
ii) a unique solution
iii) an infinite number of solution
One way is to set up the augmented matrix and then do elementary row operations to get the left hand partition in echelon form. The last row then becomes:

$[~ 0 ~ 0 ~ \lambda - 3 ~ | ~ \mu - 10 ~]$

(You will need to check this result very carefully because my algebra and arithmetic is well known to be terrible).

Examine it carefully. The interpretation of the following cases is left for you:

Case 1: $\lambda - 3 \neq 0$.

Case 2: $\lambda - 3 = 0, ~ \mu - 10 \neq 0$.

Case 3: $\lambda - 3 = \mu - 10 = 0$.

3. Please can u be more explanatory if u can

I couldnt fully understand ur answer .Please if u can provide me with steps (atleast 1st few steps) on hw to evaluate the problem .

4. Originally Posted by varunnayudu
I couldnt fully understand ur answer .Please if u can provide me with steps (atleast 1st few steps) on hw to evaluate the problem .
Where do you get lost? Do you know what an augmented matrix is? Are you familiar with elementary row operations? Do you know what an echelon form is?

5. No i am not that famialiar with that

Sir i donot know can u guide me ..........

6. Originally Posted by varunnayudu
Sir i donot know can u guide me ..........
A much easier way: Subtract equation (2) from equation (3). You get $(\lambda - 3) z = \mu - 10$.

Now consider the three cases I listed.

7. case 1 listed

as per ur reply i have carried out some steps, please tell me if they are correct or no.......

case 1: if $\lambda -3 = 0$ then,

$3z = \mu - 10 - \lambda z$

$\Rightarrow z = \frac{\mu - 10 - \lambda z}{3}$
$\Rightarrow 3x + 6y + \lambda \mu - \lambda^2 z - 10 \lambda=3 \mu$

8. You seem to be losing your connection with what it is you are trying to do.

Try this version:

Use the first equation to eliminate 'x' in the other two.

I get:

$y + 2z = 4$

$y - z + \lambda z = \mu - 6$

Now use the first to eliminate y from the second.

I get:

$z = \frac{\mu - 10}{\lambda - 3}$

Now, reread the problem statement and see if you can make any inferences from this information.

9. Originally Posted by varunnayudu
as per ur reply i have carried out some steps, please tell me if they are correct or no.......

case 1: if $\lambda -3 = 0$ then,

$3z = \mu - 10 - \lambda z$

$\Rightarrow z = \frac{\mu - 10 - \lambda z}{3}$
$\Rightarrow 3x + 6y + \lambda \mu - \lambda^2 z - 10 \lambda=3 \mu$
Originally Posted by Mr Fantastic
[snip]
Case 2: $\lambda = 3$ and $\mu \neq 10$. Then 0 = some non-zero number.

Q: Does this make sense? A: No.

Therefore the system is inconsistent, that is, there's no solution.

You now need to give some thought as to what Case 1 and Case 3 mean.

10. Originally Posted by varunnayudu
For what values of $\lambda$ and $\mu$, the simultaneous equations

$x+y+z=6$
$x+2y+3z=10$
$x+2y+\lambda z=\mu$

i) no solution
ii) a unique solution
iii) an infinite number of solution
Try to solve: Seeing "x" begining each of the equations, subtract the first from the second to get y+ 2z= 4 and the first from the third to get $y+(\lambda- 3)z= \mu- 10$. Now subtract the first of those two from the second to cancel the "y"s: $(\lambda- 5)z= \mu- 14$.

If [tex]\lambda- 5[/itex] is NOT 0, we can divide both sides by that to get a specific value for z and then it would be easy to solve for x and y. As long as $\lamda- 5$ is not 0, we have a unique solution.

If $\lambda- 5=0$, in other words if $\lambda= 5$ we cannot divide to find z but if it also happens that $\mu= 14$ that equations just says "0z= 0" which is true for all z. If [tex]\lamba= 5[/itex] and $\m= 14$, there are an infinite number of solutions.

If $\lambda- 5= 0$ and $\mu$ is NOT 14, then 0z would have to be equal to a non-zero number and that is impossible. If $\lamba= 5$ and $\mu$ is not 14, there is no solution.

11. Originally Posted by HallsofIvy
Try to solve: Seeing "x" begining each of the equations, subtract the first from the second to get y+ 2z= 4 and the first from the third to get

$y+(\lambda- 3)z= \mu- 10$. Mr F says: Careless here. It's ${\color{red}y + (\lambda - 1)z = \mu - 6}$

Now subtract the first of those two from the second to cancel the "y"s: $(\lambda- 5)z= \mu- 14$. Mr F says: Wrong. A ripple effect from the above mistake. The correct result (found in a simpler way) is given in pst #6.

If [tex]\lambda- 5[/itex] is NOT 0, we can divide both sides by that to get a specific value for z and then it would be easy to solve for x and y.

Mr F says: Mixing your latex tags. Tsk tsk.

As long as $\lamda - 5$ is not 0, we have a unique solution. Mr F says: Darn missing b ..... "As long as ${\color{red}\lambda - 5}$"

If $\lambda- 5=0$, in other words if $\lambda= 5$ we cannot divide to find z but if it also happens that $\mu= 14$ that equations just says "0z= 0" which is true for all z.

If [tex]\lamba= 5[/itex] and $\m= 14$, there are an infinite number of solutions.

Mr F says: Darn missing d ..... mixing latex tags ..... darn \m
"If ${\color{red}\lambda= 5}$ and ${\color{red}m= 14}$, there are an infinite number of solutions."

If $\lambda- 5= 0$ and $\mu$ is NOT 14, then 0z would have to be equal to a non-zero number and that is impossible.

If $\lamba= 5$ and $\mu$ is not 14, there is no solution.

Mr F says: Darn missing d again ......
"If ${\color{red}\lambda= 5}$ and ${\color{red}\mu}$ is not 14, there is no solution."
To the OP: Reply # 6 gives the correct relationship.

To HallsofIvy: Your Christmas present is in the mail - How to Spell the Letters of the Greek Alphabet, With Particular Emphasis on the Eleventh Letter