LINEAR-ALGEBRA · Unit 4 · Video 4 · Interactive Practice

Permutation Matrices — Row Exchanges, Groups, and P⁻¹ = Pᵀ

IKey Formulas

Formula	Name	Description
$P^{-1} = P^T$	Inverse rule	The inverse of any permutation matrix is its transpose
$PP^T = I$	Orthogonality	Product with transpose gives the identity
$n!$	Count of $n \times n$ permutation matrices	e.g., $3! = 6$ , $4! = 24$ , $5! = 120$
$PA = LU$	LU with pivoting	Permutation fixes zero pivots in elimination

IIVisualization 1 — Row Exchanges Build P

Each swap left-multiplies by a transposition; stack a few and watch $P$ reorder the rows of $A$ .

IIIVisualization 2 — Products Stay in the Group

The six $3 \times 3$ permutations are closed — any product $P_1 P_2$ is again one of the six.

First P₁ Second P₂

IVVisualization 3 — The Symmetric Group S₃

Every product in the table is one of the six: closure, the identity, and each element's inverse at a glance.

💡 These six matrices are the symmetric group $S_3$ ; an $n \times n$ board has $n!$ of them ( $4! = 24$ , $5! = 120$ ).

VVisualization 4 — Pᵀ Undoes P

Apply $P$ , then $P^T$ , and the rows snap back to $1,2,3$ — so $P^T P = I$ .

Permutation P Step start

💡 A permutation matrix has orthonormal rows, so it is orthogonal — the same reason $Q^{-1} = Q^T$ holds for every orthogonal $Q$ .

VIQuiz Questions

Question 1

Consider the permutation matrix $P_{13}$ that swaps rows 1 and 3:

$P_{13} = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix}$

If $A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}$ , what is the first row of $P_{13} \cdot A$ ?

✅ Correct! The 1 in column 3 of row 1 selects row 3 of A.

❌ Not quite. Look at where the 1 sits in the first row of P₁₃ — its column tells you which row of A to pick.

Show solution

Solution:

Row 1 of $P_{13}$ is $[0, 0, 1]$ . The dot product with each column of $A$ :

$[0, 0, 1] \cdot A = 0 \cdot \text{(row 1)} + 0 \cdot \text{(row 2)} + 1 \cdot \text{(row 3)} = [g, h, i]$

The single 1 in column 3 selects row 3 of $A$ .

The answer is [g, h, i].

Question 2

What is the product $P_{12} \cdot P_{12}$ ?

$P_{12} = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

✅ Correct! Swapping the same two rows twice undoes the swap — you get the identity.

❌ Not quite. Think about what happens physically: swap rows 1 and 2, then swap them again.

Show solution

Solution:

$P_{12}$ swaps rows 1 and 2. Applying it twice:

First: swap rows 1 ↔ 2
Second: swap rows 1 ↔ 2 again (undoing the first swap!)

$P_{12}^2 = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I$

Swapping the same two rows twice returns everything to its original position. Every row-swap permutation is its own inverse: $P_{12}^{-1} = P_{12}$ .

Question 3

True or False: Every permutation matrix satisfies $P^2 = I$ (i.e., every permutation matrix is its own inverse).

✅ Correct! Row-swap permutations are self-inverse, but cyclic permutations are not. The universal rule is P⁻¹ = Pᵀ.

❌ Not quite. Consider the cyclic permutation (1→2→3→1) — does applying it twice return to the identity?

Show solution

Solution: False

This is true for row-swap (transposition) permutations like $P_{12}$ , $P_{13}$ , $P_{23}$ , but not for cyclic permutations.

Counterexample: The cyclic permutation $P = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}$

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

$P^2$ is the reverse cycle. You need $P^3 = I$ for this permutation.

The correct general rule is: $P^{-1} = P^T$ (the transpose), which works for every permutation matrix.

Question 4

How many distinct $4 \times 4$ permutation matrices exist?

✅ Correct! There are 4! = 24 ways to arrange 4 rows, giving 24 permutation matrices.

❌ Not quite. The count equals n! — the number of ways to arrange n rows. For n = 4, compute 4 × 3 × 2 × 1.

Show solution

Solution: 24

An $n \times n$ permutation matrix is the identity with its rows rearranged. The number of ways to arrange $n$ rows is $n!$ (n factorial):

$4! = 4 \times 3 \times 2 \times 1 = 24$

Size	Count
$2 \times 2$	$2! = 2$
$3 \times 3$	$3! = 6$
$4 \times 4$	$4! = \mathbf{24}$
$5 \times 5$	$5! = 120$

These $n!$ matrices form the symmetric group $S_n$ .

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