LINEAR-ALGEBRA · Unit 2 · Video 3 · Interactive Practice

Permutation Matrices and Inverses

IKey Formulas

Formula	Name	Description
$PA$ swaps rows; $AP$ swaps columns	Left / Right rule	Multiplication side determines what moves
$P^{-1} = P^{T}$	Permutation inverse	Transpose undoes any permutation
Flip the sign of the multiplier	Elimination inverse	Subtracting $\ell$ is undone by adding $\ell$
$P^2 = I$ (for single row swaps)	Self-inverse	Swapping the same pair twice restores original

IIVisualization 1 — Permutation as Row Selector

A permutation matrix rearranges the identity's rows; the product $PA$ applies that same reordering to $A$ .

Permutation

IIIVisualization 2 — Same P, Rows vs Columns

The side you multiply on decides the axis: $PA$ reorders the rows of $A$ , while $AP$ reorders its columns.

Permutation

IVVisualization 3 — Inverse as Undo

$E_{21}$ subtracts $\ell$ times row 1 from row 2; $E_{21}^{-1}$ flips that sign to add it back.

Multiplier ℓ 3

💡 This sign-flip shortcut works only for elementary elimination matrices with a single off-diagonal entry — general matrices need full inversion.

VQuiz Questions

Question 1

Let $P = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$ and $A = \begin{bmatrix} 3 & 4 \\ 7 & 8 \end{bmatrix}$ .

What is the product $PA$ ?

✅ Correct! Left-multiplying by $P$ swaps the rows of $A$ .

❌ Not quite. Remember: left-multiplying swaps rows, not columns. Each row of $P$ selects a row from $A$ .

Show solution

Solution:

Left-multiplying by $P$ swaps the rows of $A$ :

$PA = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 3 & 4 \\ 7 & 8 \end{bmatrix} = \begin{bmatrix} 7 & 8 \\ 3 & 4 \end{bmatrix}$

Row 1 of $P$ is $[0,\; 1]$ , which selects row 2 of $A$ : $[7,\; 8]$ .
Row 2 of $P$ is $[1,\; 0]$ , which selects row 1 of $A$ : $[3,\; 4]$ .

Question 2

True or False: For any permutation matrix $P$ , the inverse equals the transpose: $P^{-1} = P^{T}$ .

✅ Correct! The transpose of any permutation matrix is always its inverse.

❌ Not quite. Permutation matrices have a special property: their transpose is always their inverse.

Show solution

Solution: True.

Permutation matrices are orthogonal — their rows (and columns) are orthonormal vectors. This means $P^{T} P = I$ , so $P^{T}$ is the inverse of $P$ .

Special case: For a matrix that swaps exactly two rows, $P^{T} = P$ itself, so $P \cdot P = I$ — swap twice and you're back to the original.

Question 3

The elimination matrix $E_{21} = \begin{bmatrix} 1 & 0 & 0 \\ -4 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ subtracts 4 times row 1 from row 2.

What is $E_{21}^{-1}$ ?

✅ Correct! Flip the sign of the multiplier — that's all it takes for elimination matrices.

❌ Not quite.

Show solution

Solution: Change $-4$ to $+4$ (flip the sign).

$E_{21}^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ +4 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

Why? $E_{21}$ subtracts 4 times row 1 from row 2. To undo this, $E_{21}^{-1}$ adds 4 times row 1 back.

Verification: The (2,1) entry of $E^{-1}E$ is: $4 \cdot 1 + 1 \cdot (-4) = 4 - 4 = 0 \;\checkmark$

⚠️ This sign-flip shortcut works only for elementary elimination matrices with a single off-diagonal entry — not for general matrices.

Question 4

How many distinct permutation matrices exist for a $3 \times 3$ system?

✅ Correct! There are $3! = 6$ ways to arrange 3 rows.

❌ Not quite.

Show solution

Solution: 6 permutation matrices.

The number of $n \times n$ permutation matrices equals $n!$ (n factorial), because each one corresponds to a unique reordering of $n$ rows.

$3! = 3 \times 2 \times 1 = 6$

The six row orderings are: $(1,2,3),\; (1,3,2),\; (2,1,3),\; (2,3,1),\; (3,1,2),\; (3,2,1)$ .

Each gives a unique permutation matrix.

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