1. ## matrix inversion

Find the values of k for which A is invertible.

How would i go on about solving this question??

thanks

2. Originally Posted by b00yeah05

Find the values of k for which A is invertible.

How would i go on about solving this question??

thanks
a matrix is invertible if and only if it's determinant is non-zero. so the easiest way, perhaps, is to find the expression that gives the determinant of the matrix and then pick appropriate values for k

3. Originally Posted by b00yeah05

Find the values of k for which A is invertible.

How would i go on about solving this question??

thanks
k= all real numbers EXCEPT for the one(s) that make the determinant equal to zero.

To get the determinant, make life easy for yourself by first doing the row op R1 --> R1 - R3. Then expand along the top row.

By inspection you should see that k = 2 will make the determinant equal to zero since you'll have a row of identical entries.

4. well...yes i see that if its a 2 the det = 0 which means that there is no inverse.
so does that mean....k = all real numbers except 2? is that the answer???

5. Originally Posted by b00yeah05
well...yes i see that if its a 2 the det = 0 which means that there is no inverse.
so does that mean....k = all real numbers except 2? is that the answer???
i didn't check, but yes, if k = 2 makes the determinant zero, that is the answer

6. if i was to go on to find the inverse of the matrix..would i just substitute any value into k apart from 2? and find the inverse?

7. Originally Posted by b00yeah05
if i was to go on to find the inverse of the matrix..would i just substitute any value into k apart from 2? and find the inverse?
it's not that kind of a question. you are to find k for which the matrix has an inverse, if you were to find the inverse as it is now, you'd have k's in it. don't worry about that, just know that you can find the inverse as long as k is anything but 2

8. yes, thanks i understand that but the question leading on to that is solving linear systems. of :

(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. if i find the inverse of the matrix above cant i just use matrices to solve these linear euqation as it is x = A-1*b

9. Originally Posted by b00yeah05
yes, thanks i understand that but the question leading on to that is solving linear systems. of :

(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. if i find the inverse of the matrix above cant i just use matrices to solve these linear euqation as it is x = A-1*b
A is not invertible in the first question, since k = 2. so don't even bother with that one (no solution)

for the second, k = 0, so just use the method you mentioned (if you must, there are actually many ways to solve a system), to find the solution

10. sorry to be a nusense but why is k=0 for the second one?

11. Originally Posted by b00yeah05
sorry to be a nusense but why is k=0 for the second one?
do you know how do find the matrix A? you get it from the coefficients of the variables in order

for the first question: $\displaystyle A = \begin{bmatrix} 1 & -1 & -2 \\ 2 & -2 & -2 \\ 1 & -1 & 2 \end{bmatrix} \implies k = -2$

ah, my bad, k = -2 not 2

for the second: $\displaystyle A = \begin{bmatrix} 1 & -1 & 0 \\ 2 & 0 & 0 \\ 1 & -1 & 2 \end{bmatrix} \implies k = 0$

12. Originally Posted by mr fantastic
k= all real numbers EXCEPT for the one(s) that make the determinant equal to zero.

To get the determinant, make life easy for yourself by first doing the row op R1 --> R1 - R3. Then expand along the top row.

By inspection you should see that k = 2 will make the determinant equal to zero since you'll have a row of identical entries.
I didn't mean that "By inspection ...." bit to imply k = 2 was the only solution that made the determinant equal to zero. I put it in to point out that at least one of the values k cannot be is obvious.

Originally Posted by b00yeah05
well...yes i see that if its a 2 the det = 0 which means that there is no inverse.
so does that mean....k = all real numbers except 2? is that the answer???
Did you work out the determinant and confirm that k = 2 is the only solution that solves det(A) = 0?? If you had done so instead of cutting corners (and I even suggested how to make life easier - to the point of almost trivial - in getting the determinant) you would have found that there is another value of k.

I don't want to sound rude - please take the following on board as a wake up call:

The questions you've been asking suggest to me that you haven't made much of an effort to get on top of this material. And to not even bother working out the determinant to get the other value of k - that suggests something to. This service is not here to do your assignment work - it's expected you'll do work too. Especially when you're given clear guidance on what to do.

13. yes i worked out the det(A) when k=2 it is 0. which means that at k=2, the matrix has no inverse. but i do not know how to find the value of k for which the matrix is invertible...i sort of understand what jhevon is saying but i dont know how to do it with the co-efficients

14. 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. but i do not know how to find the value of k for which the matrix is invertible...
the matrix is invertible for ANY value of k that is not 2. any real number, of course.

i sort of understand what jhevon is saying but i dont know how to do it with the co-efficients
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.

15. 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.

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