Matrix $A$ is non-singular matrix and $(A-3 I)(A-5 I)=0$, then $\frac{15}{8} A^{-1}=$ _________

Given a non-singular matrix $A$ satisfying $(A-3I)(A-5I)=0$, expand the product:

$(A-3I)(A-5I)=A^2-8A+15I=0$.

Since $A$ is invertible, multiply both sides by $A^{-1}$:

$A^{-1}(A^2-8A+15I)=A-8I+15A^{-1}=0$.

Rearranging gives $15A^{-1}=8I-A$.

Dividing both sides by 8 yields:

$\frac{15}{8}A^{-1}=I-\frac{1}{8}A$.

This matches option C.