Thread: Problematic Inverses

1. Problematic Inverses

Hey Math Help Forum Community,

I need your help. I cannot solve these problem involving matrices and was hoping you could help me.

In the first problem I get a polynomial to the third degree with k which has to equal to zero. But I only get one root, not 3. Can anyone get the roots for me?

I can't do the second one at all, but I suspect -2 makes the rank = 2.

Thank you in advance!

2. Well, I can't see what you are to "choose" but I presume it is between "invertible" and "not invertible" (or "singular" and "not singular" which would just reverse the previous two). A matrix is invertible (not singular) if and only if its determinant is not 0. It is not invertible (singular) if and only if its determinant is 0. Find the determinant of the first matrix and set it equal to 0. That will be a cubic equation which may have 3 roots. Those three numbers will be values of k for which the matrix is not invertible (is singular). For any other values of k, the matrix is invertible (not singular).
If you are going ask someone to find the roots for you, you could at least supply the cubic polynomial! When you say you got only one root, do you mean it was a triple root? That is, that the polynomial reduced to $(k- a)^3= 0$ for your solution, a? I don't get that. It is fairly easy to row-reduce to a triangular matrix and then the determinant is just the product of the numbers on the main diagonal- and here, those are just linear terms in k!

For the other problem, again, row-reduce to diagonal form. A value of a that makes the third row all zeros but not the other 2 gives the matrix rank 1. A value of a, if any, that makes both the second and third row all zeros gives the matrix rank 1.