If $A$ is a $5\times 5$ matrix with trace 15 and if 2 and 3 are eigenvalues of A, each with algebraic multiplicity 2, then the determinant of A is equal to ?
I am preparing for an examination and this question popped up.
I have never solved the determinant of a matrix given its trace, eigenvalues and algebraic multiplicity and I have looked everywhere to find a relation.
What I do know is
Trace of a matrix = sum of the eigen values
Determinant of a matrix = product of eigen values
How are these concepts connected and how do I solve this question?
EDIT
Okay so I think I figured out the answer.
Algebraic multiplicity basically means the number of times the eigenvalues are repeated so in this situation the 4 of the 5 eigen values are:
$$2,2,3,3$$
Trace = sum(eigenvalues)
Therefore $$ 4+6+\lambda = 15 ; \ \lambda = 5$$
Therefore the determinant is simply the product of these eigenvalues
$$ |A| = 180$$
Thanks to everyone in the comments that helped me out. I can't believe I was toiling over such a simple question :)
I shall now go hid in my shell.