Eigenvalues of a Matrix Function

Since $3$ is not an eigenvalue of $A$, the matrix $A-3I_2$ is invertible and $B$ is well defined. The eigenvalues $5$ and $7$ are distinct, so $A$ is diagonalisable: $A=PDP^{-1}$ with $D=\operatorname{diag}(5,7)$.

Then

Inverting gives

Cubing gives

So $B$ has eigenvalues $\frac18$ and $\frac1{64}$. Their sum is

Key idea. This is spectral mapping: if $Av=\lambda v$, then $(A-3I)^{-3}v=(\lambda-3)^{-3}v$. So the eigenvalues of $B$ are $f(\lambda)$ with $f(x)=(x-3)^{-3}$. The argument needs no diagonalisability.