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 row transformations.

Step 1: $A=\begin{bmatrix}\cos\theta&-\sin\theta&0\\\sin\theta&\cos\theta&0\\0&0&1\end{bmatrix}$ has a block structure: a $2\times2$ rotation block in the top-left, and a lone $1$ in the bottom-right corner, with zeros elsewhere in that row/column.

Step 2: Since the third row and column are already exactly the third row/column of $I_3$, no row transformation involving row 3 is needed at all -- row 3 of $A^{-1}$ stays $(0,0,1)$ throughout.

Step 3: For the top-left $2\times2$ block $\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}$, its determinant is $\cos^2\theta+\sin^2\theta=1\neq0$, so it is invertible; using the shortcut $2\times2$ inverse (swap the diagonal entries, negate the off-diagonal entries, divide by the determinant $1$): inverse of the block is $\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix}$ -- a sequence of row operations achieving this same result can equally be written out explicitly (clear the (2,1) entry using a multiple of row 1, scale, then clear (1,2)), landing on the identical block. …

Set up $AA^{-1}=I$ and clear the block-diagonal $2\times2$ rotation part using row transformations; the extra …

• Trying to perform row operations that mix row 3 into rows 1/2 unnecessarily
• Sign slip on $-\sin\theta$ when swapping the diagonal of the $2\times2$ block …