Find determinant of $B$ when $B=2A+A^{-1}-I$

Question

Let $A$ be a $4\times 4$ real matrix with eigenvalues $1,\ -1,\ 2,\ -2$. If $$B=2A+A^{-1}-I$$ then determinant of $B$ is?

We are provided with eigenvalues then from Cayley Hamilton theorem, $A$ satisfies its characteristics equation, $$(x-1)(x+1)(x-2)(x+2)=0$$hence $$A^4-5A^2+4I=0$$

Now how do I get that $B$ as $2A+A^{-1}-I$, I don't know please help.

Answer 1

It may be possible to solve this without picking a basis, but I can't think of such a solution.

Basically, because $B$ is a polynomial in $A$, it is simultaneously diagonalizable with $A$. So we can simply write $A = \operatorname{diag}(-2,-1,1,2)$ and then $B = \operatorname{diag}(-11/2,-4,2,7/2)$. Then $\det(B) = 154$.