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

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

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 $a$.

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 the transpose is the inverse

For an ordinary matrix the inverse requires a determinant calculation, but a permutation matrix is special: its inverse is simply its transpose. The reason is that the columns of a permutation matrix are the standard basis vectors in some order, so multiplying $P$ by $P^{\mathsf{T}}$ pairs each row of $P$ with the matching column, producing the identity. Practically, this means you never run a full inverse routine for a permutation matrix; you just swap its rows and columns. Stating in a written answer that the inverse equals the transpose, and that this undoes the reordering, is exactly the reasoning markers look for.

Cycle length and powers

Repeatedly applying the same permutation eventually returns the list to its original order, and the number of applications needed is the permutation's cycle length. For the three-cycle $a \to b \to c \to a$ this length is $3$, so $P^3 = I$; for a permutation built from a $2$-cycle and a separate $3$-cycle the list returns only after the lowest common multiple, here $6$, applications. To find when an order recurs, track where each element goes and how the disjoint cycles fit together. Questions often ask after how many rounds a roster or rotation returns to its start, which is precisely this cycle-length calculation.

Left versus right multiplication

Whether a permutation reorders the rows or the columns of a matrix depends on which side you multiply. To reorder the entries of a column state matrix $S$, the permutation goes on the left, $PS$, so each row of $P$ selects an entry of $S$. Multiplying on the right, $SP$, instead reorders columns and is non-conformable for a column vector. Because VCE most often reorders a column state matrix, the standard form is $PS$. Getting the side right, and checking the matrices are conformable, avoids the most common setup error in these questions.

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.