Q.Find the inverse of $A = \begin{bmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{bmatrix}$ by elementary column transformations.

Step 1: For the column-transformation route, start instead from $A^{-1}A=I$ and apply column operations to the postfactor $A$ and to the right-hand $I$, keeping the prefactor slot ($A^{-1}$) fixed until the end.

Step 2: As in part (i), the third row and column of $A$ are already exactly those of $I_3$, so no column operation touching column 3 is needed -- column 3 of the final $A^{-1}$ stays $(0,0,1)^T$.

Step 3: On the top-left $2\times2$ block $\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}$, use $C_1\to C_1+\tan\theta\,C_2$-style column combinations (or equivalently the standard sequence of the two allowed column operations) to reduce it to $I_2$; carrying the identical operations through the right-hand identity block produces $\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix}$ on that side -- the transpose-like swap-and-negate pattern that the $2\times2$ shortcut inverse always gives for a determinant-$1$ blo …

Set up $A^{-1}A=I$ and clear the block-diagonal rotation part using COLUMN transforma …

• Mixing row and column transformations together inside one derivation (they must stay separate systems)
• Applying a column operation to the prefactor slot instead of to $A$ and the right-hand $I$ …