Values to whom the matrix is diagonalizable and invertible

Question

I've got this matrix A=$$\begin{bmatrix} 2&0&3\\0&L&0\\4&0&0\end{bmatrix}$$ And I must to find the values for what this matrix is diagonalizable and they are 5 and 1 with geometrical and algebrica molteplicity equal 2. And for all of them this matrix is diagonalizable. But I can't understand the second part of this exercise. I must to find the values for what this matrix is invertible and 5|A^-1|>=tr(A)-6. So I find the invertible matrix because the determinant it's different from 0. And this matrix is $$\begin{bmatrix} 8&0&4\\0&5&0\\-3&0&8\end{bmatrix}$$ (I used the value L=1 in this case. And |A^-1|=0*0-5*4=-20 so 5|A^-1|=-100 and tr(A)-6=2+1+4=7-6=1 and here I think to do something wrong, and I can't find the correct solution for the second part of this exercise. I try to search on web.

Answer 1

Your matrix has characteristic polynomial $P(\lambda) =\lambda^3 - (2+L) \lambda^2 - (12 + 2 L) \lambda + 12 L$. It is diagonalizable if this has distinct roots. The discriminant of the characteristic polynomial is $52\, \left( {L}^{2}-2\,L-12 \right) ^{2}$, so unless $L$ is one of the roots of that ($1 \pm \sqrt{13}$), the matrix is diagonalizable. But it turns out that for those values of $L$ it is also diagonalizable.