LINEAR-ALGEBRA · Interactive Practice | Unit 4 · Video 1

Inverse of a Matrix Product & Transpose Rules

IKey Formulas

Formula	Name	Key Idea
$(AB)^{-1} = B^{-1}A^{-1}$	Product inverse rule	Order reverses
$(AB)^{T} = B^{T}A^{T}$	Product transpose rule	Order reverses
$(A^{T})^{-1} = (A^{-1})^{T}$	Transpose-inverse swap	These operations commute
$(ABC)^{-1} = C^{-1}B^{-1}A^{-1}$	Extended product inverse	Entire sequence reverses

IIVisualization 1 — Order of the Inverses

Inverting $AB$ : does the reversed order $B^{-1}A^{-1}$ pass the identity test where $A^{-1}B^{-1}$ fails?

Matrix pair

IIIVisualization 2 — Inside-Out Cancellation

Step through $(AB)(B^{-1}A^{-1})$ : the inner pair cancels first, then the outer — shoes before socks.

Step 0

💡 Keep the original order and the inner pair becomes $B\,A^{-1}$ , which does not collapse to $I$ — the reversal is what makes the inside-out cancellation work.

IVVisualization 3 — Transpose and Inverse Commute

Build any invertible $A$ : are $(A^T)^{-1}$ and $(A^{-1})^T$ always the same matrix?

a₁₁ 2 a₁₂ 1 a₂₁ 1 a₂₂ 3

VQuiz Questions

Question 1

If $A$ and $B$ are invertible matrices, what is $(AB)^{-1}$ ?

✅ Correct! The order reverses — B⁻¹ comes first, then A⁻¹.

❌ Not quite. Remember: inverting a product reverses the factor order.

Show solution

Solution:

The inverse of a product reverses the order of the factors:

$(AB)^{-1} = B^{-1}A^{-1}$

Why? Because in the product $(AB)(B^{-1}A^{-1})$ , the inner pair $BB^{-1} = I$ cancels first, then $AA^{-1} = I$ . The inside-out cancellation only works with the reversed order.

Question 2

True or False: For any invertible matrices $A$ and $B$ ,

$(AB)(A^{-1}B^{-1}) = I$

✅ Correct! Keeping the original order creates a stuck middle term BA⁻¹ that cannot simplify.

❌ Not quite. Try expanding the product — does the inner pair BA⁻¹ actually cancel?

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Solution: False

Keeping the original order gives:

$(AB)(A^{-1}B^{-1}) = A \underbrace{(BA^{-1})}_{\neq\, I} B^{-1}$

The middle term $BA^{-1}$ does not simplify to $I$ (since $B \neq A$ in general), so the product gets stuck.

The correct formula reverses the order:

$(AB)(B^{-1}A^{-1}) = A\underbrace{(BB^{-1})}_{=\,I}A^{-1} = AA^{-1} = I \;\checkmark$

Question 3

If $A$ , $B$ , and $C$ are all invertible, what is $(ABC)^{-1}$ ?

✅ Correct! Every factor inverts and the entire sequence reverses: C⁻¹B⁻¹A⁻¹.

❌ Not quite. The entire sequence must reverse — the last factor in the product becomes the first in the inverse.

Show solution

Solution:

The product inverse rule extends to any number of factors — every factor inverts and the entire sequence reverses:

$(ABC)^{-1} = C^{-1}B^{-1}A^{-1}$

Verification: $(ABC)(C^{-1}B^{-1}A^{-1})$ — the inner pair $CC^{-1} = I$ cancels first, then $BB^{-1} = I$ , then $AA^{-1} = I$ , giving $I$ .

Question 4

For an invertible matrix $A$ , which of the following equals $(A^T)^{-1}$ ?

✅ Correct! Transpose and inverse commute: (Aᵀ)⁻¹ = (A⁻¹)ᵀ. Two names for the same matrix!

❌ Not quite. Remember: inverse and transpose are operations that commute for any invertible matrix.

Show solution

Solution:

Start from $AA^{-1} = I$ and transpose both sides:

$(AA^{-1})^T = I^T$

Apply the product transpose rule (order reverses):

$(A^{-1})^T \, A^T = I$

This shows $(A^{-1})^T$ is the inverse of $A^T$ , so:

$\boxed{(A^T)^{-1} = (A^{-1})^T}$

Inverse and transpose commute — you can do them in either order.

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