For what values of and , the simultaneous equations

i) no solution

ii) a unique solution

iii) an infinite number of solution

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- December 22nd 2008, 11:40 PMvarunnayuduSimultaneous Equation
For what values of and , the simultaneous equations

i) no solution

ii) a unique solution

iii) an infinite number of solution - December 23rd 2008, 01:03 AMmr fantastic
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:

(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: .

Case 2: .

Case 3: . - December 23rd 2008, 03:02 AMvarunnayuduPlease 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 .

- December 23rd 2008, 03:24 AMmr fantastic
- December 23rd 2008, 03:39 AMvarunnayuduNo i am not that famialiar with that
Sir i donot know can u guide me ..........

- December 23rd 2008, 03:55 AMmr fantastic
- December 25th 2008, 08:12 PMvarunnayuducase 1 listed
as per ur reply i have carried out some steps, please tell me if they are correct or no.......

case 1: if then,

- December 25th 2008, 08:26 PMTKHunny
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:

Now use the first to eliminate y from the second.

I get:

Now, reread the problem statement and see if you can make any inferences from this information. - December 25th 2008, 09:45 PMmr fantastic
- December 26th 2008, 05:27 PMHallsofIvy
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 . Now subtract the first of those two from the second to cancel the "y"s: .

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 is not 0, we have a unique solution.

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

If and is NOT 14, then 0z would have to be equal to a non-zero number and that is impossible. If and is not 14, there is no solution. - December 26th 2008, 07:11 PMmr fantastic