1. Originally Posted by b00yeah05
yes i realise that they are identical..but what i meant was why did you use -2 and 0. isnt it possible for you to use any value..for eg....5 or 10 since k= all real numbers except 2.
... i use -2 and 0 because that's what it is. -2 and 0 are the numbers in the position of k in the original matrix...

2. argh...im so stupid. i realised that you were looking at the linear equations.....im so sorry...they are seperate question. my question asks....for the original matrix in my 1st post....

a) Find the values of k for which A is invertible ---> for which the answer is all real numbers except 2

b) solve the linear systems. you may find your answer in(a) helpful..----> which means that i can substitute the values from the linear system as they are real numbers but not the value "2"..and then find the inverse of teh matrix and solve the linear equation or just doing it simultaneously.

i didnt mean to confuse you or myself...sorry

3. Originally Posted by b00yeah05
argh...im so stupid. i realised that you were looking at the linear equations.....im so sorry...they are seperate question. my question asks....for the original matrix in my 1st post....

a) Find the values of k for which A is invertible ---> for which the answer is all real numbers except 2

b) solve the linear systems. you may find your answer in(a) helpful..----> which means that i can substitute the values from the linear system as they are real numbers but not the value "2"..and then find the inverse of teh matrix and solve the linear equation or just doing it simultaneously.

i didnt mean to confuse you or myself...sorry
yes. so you can do the problems now, right? what are your solutions?

4. yes, i can do the problems now. thanks heaps for your help.

my solutions for the linear systems are:
i) x = 10
y = 5
z = 4

and i am still working on the second linear system...will post the answer up when done...

thanks again

5. Originally Posted by b00yeah05
yes i worked out the det(A) when k=2 it is 0. which means that at k=2, the matrix has no inverse.
[snip]
No you did not!

If you did you would know that there is another value of k that makes the determinant equal to zero.

Now you've got more trouble - your pants are on fire.

Originally Posted by Jhevon
the matrix is invertible for ANY value of k that is not 2. any real number, of course.

look at the matrices i gave you and match them up to the matrix you posted in your first post. do you realize the matrices are identical except for the k's? that's all i did. i realize i can get one of my matrices by replacing k with -2 in the original, and i can get the other by replacing k with 0 in the original.
Grud on a beanstalk! Has anyone bothered to work out the determinant??

Following the tip I gave to make life easier, the determinant is (k - 2)(-2 - k).
Last time I checked, (k - 2)(-2 - k) = 0 had two distinct and real solutions .....

Originally Posted by Jhevon
A is not invertible in the first question, since k = 2. so don't even bother with that one (no solution)
[snip]
No. You do have to bother! When det(A) = 0, there are two possibilities:

1. No solution.

2. Infinite number of solutions exist.

You have to go to a small trouble and check which of these two possibilities is present. If it's infinite number of solution, you've got to find'em!

6. Originally Posted by mr fantastic
... Grud on a beanstalk! ...
i'll remember that one

7. Originally Posted by b00yeah05
[snip]
(i) x - y - 2z = 7
2x - 2y - 2z = 2
x - y + 2z = 13

and

(ii) x - y = 5
2x = 3
x - y +2z = 1

i have to find the solutions to that.

Originally Posted by b00yeah05
yes, i can do the problems now. thanks heaps for your help.

my solutions for the linear systems are:
i) x = 10
y = 5
z = 4

and i am still working on the second linear system...will post the answer up when done...

thanks again
I am stunned that you get a unique solution for the first system, since the matrix A corresponds to k = -2 and, in case the penny hasn't dropped yet, k = -2 makes the determinant equal to zero. Which means that there is NOT a unique solution. Either no solution or infinite solutions and you have to check which.

Again the cutting of corners .... Did you bother to check that x = 10, y = 5 and z = 4 satisfy

x - y - 2z = 7

2x - 2y - 2z = 2

x - y + 2z = 13

To seal the deal, I did. They do not (unsurprisingly).

8. Originally Posted by b00yeah05
yes, i can do the problems now. thanks heaps for your help.

my solutions for the linear systems are:
i) x = 10
y = 5
z = 4

and i am still working on the second linear system...will post the answer up when done...

thanks again
in light of mr fantastic's post in post #20. the matrix is not invertible for k = -2 either. how did you get that solution for your system?

9. Originally Posted by Jhevon
i'll remember that one
Blame Judge Dredd. I get all my profanities from him.

10. Originally Posted by mr fantastic
I am stunned that you get a unique solution for the first system, since the matrix A corresponds to k = -2 and, in case the penny hasn't dropped yet, k = -2 makes the determinant equal to zero. Which means that there is NOT a unique solution. Either no solution or infinite solutions and you have to check which.

Again the cutting of corners .... Did you bother to check that x = 10, y = 5 and z = 4 satisfy

x - y - 2z = 7

2x - 2y - 2z = 2

x - y + 2z = 13

To seal the deal, I did. They do not (unsurprisingly).
For what it's worth, the system boils down to:

z = -6

x - y - z = 1

z = 4

The conclusion is obvious.

I hope you've learned more from this thread than just what the solution to you assignment question is ......

11. yeah..i have learned quite a lot. i went back and did the cofactor to find the detrerminant and i got -2..so im just doing the question again. and practising more questions like this.................................. thanks for your guys help....

12. Originally Posted by b00yeah05
yeah..i have learned quite a lot. i went back and did the cofactor to find the detrerminant and i got -2..so im just doing the question again. and practising more questions like this.................................. thanks for your guys help....
Then all's well that ends well.

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