Q.If $A = \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}$ then $A^{-1} =$ (A) $\begin{bmatrix} 1/\cos\alpha & -1/\sin\alpha \\ 1/\sin\alpha & 1/\cos\alpha \end{bmatrix}$ (B) $\begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix}$ (C) $\begin{bmatrix} -\cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix}$ (D) $\begin{bmatrix} -\cos\alpha & \sin\alpha \\ \sin\alpha & -\cos\alpha \end{bmatrix}$

When a square matrix $A$ is non-singular ($|A| eq0$), its inverse can be computed directly by the formula $A^{-1} = \dfrac{1}{|A|}(\text{adj}\,A)$, which follows from the identity $A(\text{adj}\,A)=|A|\,I$. The working method has three clean stages: (1) evaluate $|A|$ and confirm it is nonzero (otherwise stop — the inverse does not exist); (2) compute the cofactor $A_{ij}$ of every entry of $A$ and transpose the resulting cofactor matrix to get $\text{adj}\,A$; (3) divide every entry of $\text{adj}\,A$ by the scalar $|A|$. For a $2\times2$ matrix A=\begin{bmatrix}a&b\c&d\end{bmatrix} this collapses to the shortcut $A^{-1}=\dfrac{1}{ad-bc}\begin{bmatrix}d&-b\-c&a\end{bmatrix}$ — swap the leading-diagonal entries, negate the other two, divide by the determinant. For a $3\times3$ (or larger) matrix all nine cofactors must be found individually before assembli …