Consider the 2×2 matrix A=$\begin{bmatrix} a &b \\ c& d \end{bmatrix}$
$\lambda^2 -trace(A) \lambda+det(A)=0$
$\lambda^2 -(a+d )\lambda+(ad-bc)=0$
Put $\lambda$=1
$1-a-d+ad-(1-d)(1-a)=0$
$\text{as we know that a+c=1 and b+d=1}$
$1-a-d+ad-1+d+a-ad =0$
$0=0$ other option $\lambda=2,3,4$doesn't satisfy
$\mathbf{(D )ans}$