VICGeneral MathematicsSyllabus dot point

How does a permutation matrix rearrange the rows of a state matrix, and how are binary matrices used to encode and reorder information?

Recognise and use a permutation matrix as a binary matrix with exactly one 1 in each row and column, apply it to reorder the entries of a state matrix, and identify the matrix that reverses or repeats the reordering

A focused answer to the VCE General Mathematics Unit 4 Matrices key-knowledge point on permutation matrices. The defining one-per-row-and-column structure, using a permutation matrix to reorder a state matrix, the inverse that reverses it, and powers that repeat the reordering.

Generated by Claude Opus 4.76 min answerUpdated 2026-05-29

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
How a permutation matrix reorders
Powers repeat the shuffle
Binary matrices more generally
Why this matters for the exams

What this dot point is asking

VCAA wants you to recognise and use a permutation matrix: a binary matrix with exactly one $1$ in each row and each column. Multiplying it by a state matrix shuffles the entries into a new order without changing their values. You apply a permutation to reorder a list, find the matrix that reverses the reordering, and use powers to repeat it. This is a clean application of matrix multiplication used for rotations, rosters and rankings.

How a permutation matrix reorders

Each row of the permutation matrix picks out exactly one entry of the state matrix to place in that position. Consider

P = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix}, \qquad S = \begin{bmatrix} a \\ b \\ c \end{bmatrix}.

The first row of $P$ has its $1$ in column 2, so the first entry of the result is $b$ . The second row's $1$ is in column 3, giving $c$ . The third row's $1$ is in column 1, giving $a$ .

Worked example

Applying a permutation and reversing it

With $P$ and $S$ above:

P S = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix}\begin{bmatrix} a \\ b \\ c \end{bmatrix} = \begin{bmatrix} b \\ c \\ a \end{bmatrix}.

The list $a, b, c$ has been rotated to $b, c, a$ .

To reverse this, use the inverse, which for a permutation matrix is simply its transpose $P^{-1} = P^{\mathsf{T}}$ :

P^{\mathsf{T}} = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}, \qquad P^{\mathsf{T}}\begin{bmatrix} b \\ c \\ a \end{bmatrix} = \begin{bmatrix} a \\ b \\ c \end{bmatrix}.

The original order is restored.

Powers repeat the shuffle

Applying the same permutation again multiplies by $P$ once more. Because a permutation just cycles the entries, repeating it enough times returns to the start: here $P^3 = I$ , so after three applications the list $a, b, c$ is back in its original order.

Binary matrices more generally

Permutation matrices are a special case of binary matrices (entries only $0$ or $1$ ), which also appear as communication, dominance and adjacency matrices. What sets a permutation matrix apart is the strict one-per-row-and-column structure that makes it reorder rather than count.

Why this matters for the exams

Permutation matrix questions test whether you understand multiplication as a reordering operation and whether you can reverse it with the transpose. Markers reward the correct reordered state, the transpose as the inverse, and recognising the cycle length that restores the original. The topic reinforces matrix multiplication and the special structure of binary matrices used across the module.

Exam-style practice questions

Practice questions written in the style of VCAA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2025 VCAA1 marksConsider the matrix G with rows [0, 1, 0], [1, 0, 1] and [0, 0, 0]. Which one of the following correctly describes matrix G? A. a binary matrix B. a permutation matrix C. an identity matrix D. a diagonal matrix

Show worked answer →

Every entry of G is either 0 or 1, which is exactly the definition of a binary matrix, so option A is correct.

It is not a permutation matrix: a permutation matrix needs exactly one 1 in each row and each column, but the third row of G is all zeros, so option B is wrong.

It is not the identity matrix (the leading diagonal is not all 1s) and not a diagonal matrix (it has non-zero entries off the leading diagonal), so the answer is A.