Simultaneous Equation

**varunnayudu** · December 22nd 2008, 11:40 PM

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

**mr fantastic** · December 23rd 2008, 01:03 AM

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

**varunnayudu** · December 23rd 2008, 03:02 AM

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

**mr fantastic** · December 23rd 2008, 03:24 AM

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?

**varunnayudu** · December 23rd 2008, 03:39 AM

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

**mr fantastic** · December 23rd 2008, 03:55 AM

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.

**varunnayudu** · December 25th 2008, 08:12 PM

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,

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

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

**TKHunny** · December 25th 2008, 08:26 PM

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.

**mr fantastic** · December 25th 2008, 09:45 PM

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.

**HallsofIvy** · December 26th 2008, 05:27 PM

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.

**mr fantastic** · December 26th 2008, 07:11 PM

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

Thread: Simultaneous Equation

LinkBack

Thread Tools

Display

Simultaneous Equation

Please can u be more explanatory if u can

No i am not that famialiar with that

case 1 listed

Similar Math Help Forum Discussions

simultaneous equation help

Simultaneous equation

Simultaneous equation help

Simultaneous Equation [C1]

Simultaneous equation help

Search Tags