Find nonsingular matrices P & Q such that PAQ is in normal form and hence find rank of the matrix \(A\) for the following matrix \(A=\begin{bmatrix}3&-3&4\\2&-3&4\\0&-1&1\end{bmatrix}\).

Write \(A=I_3\,A\,I_3\). Row operations act on A and the pre-factor; column operations act on A and the post-factor.

\(R_1\to R_1-R_2\): \(\begin{bmatrix}1&0&0\\2&-3&4\\0&-1&1\end{bmatrix}=\begin{bmatrix}1&-1&0\\0&1&0\\0&0&1\end{bmatrix}A\,I_3\)

\(R_2\to R_2-2R_1\): \(\begin{bmatrix}1&0&0\\0&-3&4\\0&-1&1\end{bmatrix}=\begin{bmatrix}1&-1&0\\-2&3&0\\0&0&1\end{bmatrix}A\,I_3\)

\(R_2\leftrightarrow R_3\): \(\begin{bmatrix}1&0&0\\0&-1&1\\0&-3&4\end{bmatrix}=\begin{bmatrix}1&-1&0\\0&0&1\\-2&3&0\end{bmatrix}A\,I_3\)

\(R_3\to R_3-3R_2\): \(\begin{bmatrix}1&0&0\\0&-1&1\\0&0&1\end{bmatrix}=\begin{bmatrix}1&-1&0\\0&0&1\\-2&3&-3\end{bmatrix}A\,I_3\)

\(C_3\to C_3+C_2\): \(\begin{bmatrix}1&0&0\\0&-1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}1&-1&0\\0&0&1\\-2&3&-3\end{bmatrix}A\begin{bmatrix}1&0&0\\0&1&1\\0&0&1\end{bmatrix}\)

\(R_2\to(-1)R_2\): \(I_3=\begin{bmatrix}1&-1&0\\0&0&-1\\-2&3&-3\end{bmatrix}A\begin{bmatrix}1&0&0\\0&1&1\\0&0&1\end{bmatrix}\)

\(P=\begin{bmatrix}1&-1&0\\0&0&-1\\-2&3&-3\end{bmatrix},\ Q=\begin{bmatrix}1&0&0\\0&1&1\\0&0&1\end{bmatrix}\), \(PAQ=I_3\) (verified by multiplication).

Normal form is \(I_3\), so rank of A = 3 (also \(|A|=3(1)+3(2)+4(-2)=1\ne0\)).