Let λ be an eigenvalue of A and X be the eigenvector of A.
Then according to the definition of eigenvalues and eigenvector we get
AX = λ X ........................(1)
Now, multiplying both sides of equation 1 with A we get
A2X = A λ X = λ AX = λ λ X = λ2X ............... (2)
So, the matrix A2 has the same eigenvector as in A but the eigenvalue is square of the eigenvalue of A.
Now, multiplying both sides of equation 1 with -3 we get
-3 AX = -3 λX ..........................(3)
Now, adding equation (2) and (3) we get
(A2 - 3A)X = ( λ2 - 3 λ )X ........................(4)
Now, we add 4$I$ X to the both sides of equation 4
(A2 - 3A) X + 4$I$ X = ( λ2 - 3 λ )X + 4$I$X
=> (A2 - 3A + 4$I$)X = ( λ2 - 3 λ +4 )X [ since $I$X = X]
So, the eigenvalue of the matrix (A2 - 3A +4$I$) is ( λ2 - 3 λ + 4 ) and the eigenvector remains same as A.
Now, the value of λ is -2 and 1
So, putting those values in the expression of the eigenvalue we get the eigenvalues of the given matrix which are 2 and 14
So, Option A is the correct answer.