If A=( array ccc 1 & 0 & 0 0 & θ & θ 0 & θ & - θ array ) , then...

Concept: If a matrix A=\left($\begin{array}{lll}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}$\right) then A^{\top}=\left($\begin{array}{lll}a_{11} & a_{21} & a_{31} \\ a_{12} & a_{22} & a_{32} \\ a_{13} & a_{32} & a_{33}\end{array}$\right) Explanation: If A=\left($\begin{array}{ccc}1 & 0 & 0 \\ 0 & \cos \theta & \sin \theta \\ 0 & \sin \theta & -\cos \theta\end{array}$\right), the cofactor matrix of A is given by, \operatorname{cofactor}(A)=\left($\begin{array}{ccc} -1 & 0 & 0 \\ 0 & -\cos \theta & -\sin \theta \\ 0 & -\sin \theta & \cos \theta \end{array}$\right) So, adj A=\left($\begin{array}{ccc}-1 & 0 & 0 \\ 0 & -\cos \theta & -\sin \theta \\ 0 & -\sin \theta & \cos \theta\end{array}$\right) (Transpose of cofactor matrix) $\Rightarrow \operatorname{adj} A=-A \_(i)\Rightarrow \operatorname{adj} A+A=0$ is a null matrix So, statement 1 is correct. Now, expanding A along Row 1 , $$\begin{aligned} & |A|=1\left(-\cos ^2 \theta-\sin ^2 \theta\right)=-1 \\ & \Rightarrow A^{-1}=\frac{1}{\mid A} \operatorname{adj} A=-\operatorname{adj} A \\ & \Rightarrow A^{-1}+\operatorname{adj} A=0 \ldots \text{ (ii) } \end{aligned}$$ So, statement 2 is correct. Putting the value of (i) in (ii), $$\begin{aligned} & \Rightarrow A^{-1}-A=0 \\ & \text{ or } A-A^{-1}=0 \end{aligned}$$ So, statement 3 is correct.