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
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.
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 .....
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!
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).