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.
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
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: $\displaystyle A = \begin{bmatrix} 1 & -1 & 0 \\ 2 & 0 & 0 \\ 1 & -1 & 2 \end{bmatrix} \implies k = 0$
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.
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.
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
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.i sort of understand what jhevon is saying but i dont know how to do it with the co-efficients