If $A=\left[\begin{array}{cc}1& 2\\ 2& 3\end{array}\right]$ and $A^2-\lambda A-I_2=0$, then $\lambda$ is equal to

we are given

$A=\left[\begin{array}{cc}1& 2\\ 2& 3\end{array}\right]$

And  $A^2-\lambda A-I_2=0\to \left(1\right)$

so $A^2=A.A=\left[\begin{array}{cc}1& 2\\ 2& 3\end{array}\right] \left[\begin{array}{cc}1& 2\\ 2& 3\end{array}\right] =\left[\begin{array}{cc}5& 6\\ 6& 13\end{array}\right]$

From eq(1)

$$\begin{gathered} A^2-\lambda A+I_2=\left[\begin{array}{cc}5& 8\\ 8& 13\end{array}\right]-\lambda \left[\begin{array}{cc}1& 2\\ 2& 3\end{array}\right]-\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=0 \\ \Rightarrow \left[\begin{array}{cc}5-\lambda -1& 8-2\lambda -0\\ 8-2\lambda -0& 13-3\lambda -1\end{array}\right] =\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right] \\ \Rightarrow \left[\begin{array}{cc}4-\lambda & 8-2\lambda \\ 8-2\lambda & 12-3\lambda \end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right] \\ \Rightarrow 4-\lambda =0 \\ \lambda =4 \end{gathered}$$

Hence option (d) is correct