If A is a 2 × 2 matrix such that $A^2-4A+3I=0$ then the inverse of A + 3I is

$\begin{array}{l}{A}^2-4A+3I=0\Rightarrow A(A+3I)-7A+3I=0\\ \Rightarrow A(A+3I)-7(A+3I)=-24I\Rightarrow (A+3I)(A-7I)=-24I\\ (A+3I)\left(\frac{7}{24}I-\frac{A}{24}\right)=I\Rightarrow (A+3I{)}^{-1}=\frac{7}{24}I-\frac{A}{24}\end{array}$