Answer: \(P=\begin{bmatrix}1&0&0\\-1&1&0\\-1&1&1\end{bmatrix}\), \(Q=\begin{bmatrix}1&-1&-1\\0&1&-1\\0&0&1\end{bmatrix}\), \(PAQ=\begin{bmatrix}I_2&0\\0&0\end{bmatrix}\), rank \(=2\).
Write \(A=I_3\,A\,I_3\). Apply every row operation to \(A\) and the pre-factor, and every column operation to \(A\) and the post-factor.
Row operations:
\(R_2\to R_2-R_1\): \(\begin{bmatrix}1&1&2\\0&1&1\\0&-1&-1\end{bmatrix}=\begin{bmatrix}1&0&0\\-1&1&0\\0&0&1\end{bmatrix}A\,I_3\)
\(R_3\to R_3+R_2\): \(\begin{bmatrix}1&1&2\\0&1&1\\0&0&0\end{bmatrix}=\begin{bmatrix}1&0&0\\-1&1&0\\-1&1&1\end{bmatrix}A\,I_3\)
Column operations:
\(C_2\to C_2-C_1,\ C_3\to C_3-2C_1\): \(\begin{bmatrix}1&0&0\\0&1&1\\0&0&0\end{bmatrix}=P\,A\begin{bmatrix}1&-1&-2\\0&1&0\\0&0&1\end{bmatrix}\)
\(C_3\to C_3-C_2\): \(\begin{bmatrix}1&0&0\\0&1&0\\0&0&0\end{bmatrix}=P\,A\begin{bmatrix}1&-1&-1\\0&1&-1\\0&0&1\end{bmatrix}\)
So \(PAQ=\begin{bmatrix}I_2&0\\0&0\end{bmatrix}\) with
\[P=\begin{bmatrix}1&0&0\\-1&1&0\\-1&1&1\end{bmatrix},\quad Q=\begin{bmatrix}1&-1&-1\\0&1&-1\\0&0&1\end{bmatrix}.\]
\(|P|=1\neq0\), \(|Q|=1\neq0\), so both are non-singular. Rank of A = 2 (also \(|A|=0\), since \(R_3=R_1-R_2\)).
