Question-38

permutation matrix

Which permutation matrix \(\displaystyle P\) makes \(\displaystyle PA\) upper triangular where:

\[ \begin{equation*} A=\begin{bmatrix} 0 & 0 & 6\\ 1 & 2 & 3\\ 0 & 4 & 5 \end{bmatrix} \end{equation*} \]

\(\displaystyle \begin{bmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ 1 & 0 & 0 \end{bmatrix}\)
\(\displaystyle \begin{bmatrix} 0 & 1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 1 \end{bmatrix}\)
\(\displaystyle \begin{bmatrix} 1 & 0 & 0\\ 0 & 0 & 1\\ 0 & 1 & 0 \end{bmatrix}\)
\(\displaystyle \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}\)

Answer

\(\displaystyle \begin{bmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ 1 & 0 & 0 \end{bmatrix}\)
\(\displaystyle \begin{bmatrix} 0 & 1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 1 \end{bmatrix}\)
\(\displaystyle \begin{bmatrix} 1 & 0 & 0\\ 0 & 0 & 1\\ 0 & 1 & 0 \end{bmatrix}\)
\(\displaystyle \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}\)

Solution

Recall the following facts:

A permutation matrix is obtained by permuting the rows of the identity matrix.
Pre-multiplying \(\displaystyle A\) by a permutation matrix is equivalent to permuting the rows of \(\displaystyle A\).

To get \(\displaystyle A\) into an upper triangular matrix, we need:

\[ \begin{equation*} \begin{bmatrix} - & R_{2} & -\\ - & R_{3} & -\\ - & R_{1} & - \end{bmatrix} \Longrightarrow P=\begin{bmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ 1 & 0 & 0 \end{bmatrix} \end{equation*} \]