Binary, Permutation, and Communication Matrices (VCE SSCE General Mathematics): Revision Notes

Worked Example: Applying a Permutation Matrix

Consider the column matrix $X = \begin{bmatrix} T \\ A \\ R \end{bmatrix}$ and the permutation matrix $P = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix}$.

Part (a)(i): Show that pre-multiplying $X$ by $P$ changes it to $Y = \begin{bmatrix} R \\ A \\ T \end{bmatrix}$

Calculate $PX$:

$PX = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix} \begin{bmatrix} T \\ A \\ R \end{bmatrix}$

For the first entry: $0 \times T + 0 \times A + 1 \times R = R$

For the second entry: $0 \times T + 1 \times A + 0 \times R = A$

For the third entry: $1 \times T + 0 \times A + 0 \times R = T$

Therefore: $PX = \begin{bmatrix} R \\ A \\ T \end{bmatrix}$

The permutation matrix has rearranged the elements from $T, A, R$ to $R, A, T$.

Part (a)(ii): Show that pre-multiplying $X$ by $P^2$ leaves $X$ unchanged

Calculate $P^2X$:

$P^2X = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix}^2 \begin{bmatrix} T \\ A \\ R \end{bmatrix} = \begin{bmatrix} T \\ A \\ R \end{bmatrix}$

The matrix $X$ remains unchanged.

Part (b): What can we deduce about $P^2$?

Since $P^2X = X$ for any matrix $X$, we can deduce that $P^2$ must be the identity matrix:

$P^2 = I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

Worked Example: Finding a Permutation Matrix

Find a permutation matrix that transforms $\begin{bmatrix} A \\ B \\ C \\ D \end{bmatrix}$ into $\begin{bmatrix} D \\ C \\ B \\ A \end{bmatrix}$.

Solution strategy:

• $D$ (in position 4) needs to move to position 1 → place a $1$ in row 1, column 4
• $C$ (in position 3) needs to move to position 2 → place a $1$ in row 2, column 3
• $B$ (in position 2) needs to move to position 3 → place a $1$ in row 3, column 2
• $A$ (in position 1) needs to move to position 4 → place a $1$ in row 4, column 1

The permutation matrix is:

$\begin{bmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} A \\ B \\ C \\ D \end{bmatrix} = \begin{bmatrix} D \\ C \\ B \\ A \end{bmatrix}$

This matrix reverses the order of the elements - it's sometimes called a reversal matrix.

Worked Example: Using the Inverse Property

A $4 \times 4$ permutation matrix $P = \begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{bmatrix}$ is applied to a $4 \times 1$ column matrix $A$, giving the result $\begin{bmatrix} X \\ Y \\ Z \\ W \end{bmatrix}$. Determine matrix $A$.

Solution:

We know that $PA = \begin{bmatrix} X \\ Y \\ Z \\ W \end{bmatrix}$.

To find $A$, we pre-multiply both sides by $P^{-1}$:

$A = P^{-1} \begin{bmatrix} X \\ Y \\ Z \\ W \end{bmatrix}$

Since $P^{-1} = P^T$, we need to find the transpose of $P$:

$P^T = \begin{bmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{bmatrix}$

Therefore:

$A = P^T \begin{bmatrix} X \\ Y \\ Z \\ W \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{bmatrix} \begin{bmatrix} X \\ Y \\ Z \\ W \end{bmatrix} = \begin{bmatrix} W \\ Z \\ X \\ Y \end{bmatrix}$

You can verify this using a calculator for larger or more complex matrices.

Worked Example: Constructing a Communication Matrix

Scenario:

Four students - Eva, Wong, Yumi and Kim - are staying in a backpacker's hostel. They face communication challenges because they speak different languages:

• Eva speaks English only
• Yumi speaks Japanese only
• Kim speaks Korean only
• Wong speaks English, Japanese and Korean

Network diagram:

We can represent this situation visually with a network diagram where arrows show who can communicate directly with whom.

Building the matrix:

To construct the communication matrix, follow these steps:

Step 1: Create a $4 \times 4$ matrix (since there are four people). Label both rows and columns with the initials: E, W, Y, K.

Step 2: Label the rows as "Speaker" and columns as "Receiver".

Step 3: Fill in the matrix entries using these rules:

• Entry = $1$ if the two people can communicate directly (they share a common language)
• Entry = $0$ if they cannot communicate directly (no common language)

The completed communication matrix:

Let's interpret some entries:

• Row E, column W = $1$ because Eva and Wong both speak English
• Row E, column Y = $0$ because Eva (English only) and Yumi (Japanese only) have no common language
• Row W, column Y = $1$ because Wong and Yumi both speak Japanese
• Notice the diagonal elements are $0$ - a person doesn't "communicate with themselves" in this context