Consider the following in respect of two non-singular matrices A...

Non-singular matrix is a matrix whose determinate Value is non-zero. Let A=\left[$\begin{array}{ll}1 & 1 \\ 1 & 2\end{array}$\right] and B=\left[$\begin{array}{ll}2 & 1 \\ 1 & 2\end{array}$\right] Here, $A$ and $B$ are non-singular matrix Now from Statement 1: $$\begin{aligned} & A+B=\left[\begin{array}{ll} 1 & 1 \\ 1 & 2 \end{array}\right]+\left[\begin{array}{ll} 2 & 1 \\ 1 & 2 \end{array}\right]=\left[\begin{array}{ll} 3 & 2 \\ 2 & 4 \end{array}\right] \\ & \operatorname{det}(A+B)=12-4=8 \\ & \text{ Now det }(A)=\left|\begin{array}{ll} 1 & 1 \\ 1 & 2 \end{array}\right|=1 \\ & \text{ and } \operatorname{det}(B)=\left|\begin{array}{ll} 2 & 1 \\ 1 & 2 \end{array}\right|=3 \end{aligned}$$ Now, $\operatorname{det}(A+B)=4 \neq \operatorname{det}(A+B)\therefore \quad$ Statement 1 is wrong. Now from Statement 2: $$\begin{aligned} & (A+B)^{-1}=\frac{1}{|A+B|} \operatorname{adj}(A+B) \\ & =\frac{1}{8}\left[\begin{array}{cc} 4 & -2 \\ -2 & 3 \end{array}\right] \\ & \text{ and } \mathrm{A}^{-1}=\frac{1}{|\mathrm{ A}|} \operatorname{adj}(\mathrm{A})=\frac{1}{1}\left[\begin{array}{cc} 2 & -1 \\ -1 & 1 \end{array}\right] \\ & \text{ and } B^{-1}=\frac{1}{|B|} \operatorname{adj}(B)=\frac{1}{3}\left[\begin{array}{cc} 2 & -1 \\ -1 & 2 \end{array}\right] \\ & \text{ Now, } A^{-1}+B^{-1}=\left[\begin{array}{cc} 2 & -1 \\ -1 & 1 \end{array}\right]+\left[\begin{array}{cc} \frac{2}{3} & \frac{-1}{3} \\ \frac{-1}{3} & \frac{2}{3} \end{array}\right] \end{aligned}$$=\left[$$\begin{array}{cc} \frac{8}{3} & \frac{-4}{3} \\ \frac{-4}{3} & \frac{5}{3} \end{array}$$\right] \neq(A+B)^{-1}\therefore \quad$Statement 2 is wrong$\therefore \quad$ Option (D) is correct.