Is (AB)⁻¹ equal to A⁻¹B⁻¹?

No. The inverse of a product reverses the order: (AB)⁻¹ = B⁻¹A⁻¹. Check it by multiplying: (AB)(B⁻¹A⁻¹) = A(BB⁻¹)A⁻¹ = A·I·A⁻¹ = AA⁻¹ = I, the inner pair cancels first, then the outer pair. The in-place candidate fails: (AB)(A⁻¹B⁻¹) = ABA⁻¹B⁻¹, where the A⁻¹ cannot reach the A because B sits between them and matrices do not commute (BA⁻¹ ≠ A⁻¹B). So A⁻¹B⁻¹ is not the inverse of AB.

Is AB = BA for matrices?

No, matrix multiplication does not commute in general. A concrete counterexample: take A = [[1,1],[0,1]] and B = [[1,0],[1,1]]. Then AB = [[2,1],[1,1]] but BA = [[1,1],[1,2]] — different matrices. Order matters: swapping the factors changes the answer, unlike numbers where ab = ba always.

Does (A+B)² = A² + 2AB + B² for matrices?

No. Expanding gives (A+B)² = (A+B)(A+B) = A² + AB + BA + B². Because AB ≠ BA in general, the two cross terms AB and BA are different matrices and cannot be merged into 2AB. The familiar a² + 2ab + b² only works for numbers, where ab = ba. The transpose reverses the order for the same reason: (AB)ᵀ = BᵀAᵀ.

Is a condition like PQ = QP automatic?

No. PQ = QP is a genuine restriction that selects the special matrices commuting with P; most matrices do not commute. In JEE Advanced 2025 Paper 1 Q4 the integer matrices with Q⁻¹ = Qᵀ are the 48 signed permutation matrices, but only those that also satisfy PQ = QP count, and the official answer is 16. The commutation constraint cut the count from 48 to 16 — it carried real information, it was not a free identity.

JEE Advanced Maths · Matrices & Determinants

Is (AB)⁻¹ = A⁻¹B⁻¹? Do matrices multiply like numbers?

No. Matrices do not commute: $AB \ne BA$ in general, so every rule that quietly assumed order did not matter has to be rewritten. The inverse of a product reverses the order: $(AB)^{-1} = B^{-1}A^{-1}$ , not $A^{-1}B^{-1}$ . Multiply it out to see why: $(AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} = AA^{-1} = I$ , whereas $(AB)(A^{-1}B^{-1}) = ABA^{-1}B^{-1}$ leaves $A$ and $A^{-1}$ stranded either side of $B$ .

The transpose reverses for the same reason ( $(AB)^{T} = B^{T}A^{T}$ ), and the binomial square keeps both cross terms: $(A+B)^2 = A^2 + AB + BA + B^2$ , never $A^2 + 2AB + B^2$ . The clean statement to hold is: for matrices, order matters — reverse it on an inverse or transpose, and never merge $AB$ with $BA$ . JEE Advanced punishes the scalar instinct through inverse-order identities (2021 Paper 1 Q14), matrix powers (2022 Paper 2 Q16), and commutation-constraint counting (2025 Paper 1 Q4, 2025 Paper 2 Q5).

Ready to fix this? The Matrices & Determinants lesson works through this misconception and the others in Matrices & Determinants, one altitude at a time.

How to spot it in your own work

You wrote " $(AB)^{-1} = A^{-1}B^{-1}$ " — inverting each factor without reversing the order.
You assumed $AB = BA$ , treating matrix multiplication as if it commuted.
You wrote " $(A+B)^2 = A^2 + 2AB + B^2$ ", merging the two cross terms into $2AB$ .
You treated a condition like $PQ = QP$ as automatic, so it carried no information.

An exam question that triggers it

Here is a question that tests this misconception head-on, in the spirit of JEE Advanced 2021 Paper 1 Q14:

$A$ and $B$ are invertible matrices. A classmate writes $(AB)^{-1} = A^{-1}B^{-1}$ . Multiply $AB$ by that candidate. Do you get $I$ ?

The misconception reasons "invert each factor, same as numbers." The correct move is to multiply it out and watch the middle. Keeping the order fixed (no swapping, because matrices do not commute):

(AB)(A^{-1}B^{-1}) = A\,B\,A^{-1}\,B^{-1}.

The $A^{-1}$ cannot reach back to cancel $A$ — $B$ sits between them, and you cannot slide it out of the way. So this is not $I$ . Now reverse the candidate:

(AB)(B^{-1}A^{-1}) = A\,(B\,B^{-1})\,A^{-1} = A\,I\,A^{-1} = AA^{-1} = I.

The inner pair cancels first, then the outer pair, giving $(AB)^{-1} = B^{-1}A^{-1}$ — the order reversed, not left in place.

Why students fall for this

For $1 \times 1$ matrices — ordinary numbers — multiplication commutes, the inverse of a product is the product of the inverses, and $(a+b)^2 = a^2 + 2ab + b^2$ . Students absorb these as "multiplication rules" and carry them wholesale into matrices, where the steps look identical. The seductive part is that the algebra is formally the same right up until the moment two factors need to commute — and matrices refuse.

Wawro et al. (2014) and Aydın (2014), studying linear-algebra reasoning, find that over-transferring scalar commutativity to matrix products — $AB = BA$ , $(AB)^{-1} = A^{-1}B^{-1}$ , $(A+B)^2 = A^2 + 2AB + B^2$ — is among the most persistent misconceptions. Being told "matrices don't commute" rarely dislodges it. What dislodges it is multiplying $AB$ by the claimed inverse and watching the inner factors fail to meet, then reversing the order and watching the cancellation succeed — and meeting a concrete pair with $AB \ne BA$ by hand.

The fix: For matrices, order matters

Matrices do not commute: $AB \ne BA$ in general. So the inverse and transpose of a product reverse the order: $(AB)^{-1} = B^{-1}A^{-1}$ and $(AB)^{T} = B^{T}A^{T}$ . The reversal is forced — it is the only arrangement that lets each factor sit next to its own inverse, so the cancellation nests from the inside out.

The same non-commutativity reshapes the binomial square. Expanding $(A+B)(A+B)$ in order gives $A^2 + AB + BA + B^2$ , and since $AB \ne BA$ the cross terms cannot collapse into $2AB$ . And a condition like $PQ = QP$ is a genuine restriction, not a free identity: it selects the special matrices that commute with $P$ , which is why JEE examiners impose it to cut down a count.

Worked example

Multiply $AB$ by each inverse candidate, keeping the order fixed.

The student's rule. Test $(AB)(A^{-1}B^{-1}) = A\,B\,A^{-1}\,B^{-1}$ . The neighbours are $A\text{–}B$ , $B\text{–}A^{-1}$ , $A^{-1}\text{–}B^{-1}$ — none of those pairs are inverses of each other, so nothing cancels.
Why it jams. For $A^{-1}$ to cancel $A$ they must be adjacent, but $B$ blocks them and you cannot commute it past ( $BA^{-1} \ne A^{-1}B$ ). The product is not $I$ .
Reverse the order.
$(AB)(B^{-1}A^{-1}) = A\,(B\,B^{-1})\,A^{-1} = A\,I\,A^{-1} = AA^{-1} = I.$
The inner pair $B, B^{-1}$ meet first, then the outer pair $A, A^{-1}$ — socks off, then shoes.
Conclude. $(AB)^{-1} = B^{-1}A^{-1}$ , order reversed; the transpose does the same, $(AB)^{T} = B^{T}A^{T}$ .

Now break commutativity by hand. Take $A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$ . Then

AB = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}, \qquad BA = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix},

so $AB \ne BA$ . Because the order changes the answer, $(A+B)^2 = A^2 + AB + BA + B^2$ , and the cross terms $AB$ and $BA$ cannot merge into $2AB$ — here $AB + BA = \begin{pmatrix} 3 & 2 \\ 2 & 3 \end{pmatrix}$ while $2AB = \begin{pmatrix} 4 & 2 \\ 2 & 2 \end{pmatrix}$ , which are not equal.

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Common questions

Why does $(AB)^{-1}$ reverse to $B^{-1}A^{-1}$ rather than $A^{-1}B^{-1}$ ?: Because cancellation needs each factor adjacent to its own inverse. With $B^{-1}A^{-1}$ the product $(AB)(B^{-1}A^{-1})$ nests as $A(BB^{-1})A^{-1} = AA^{-1} = I$ — inner pair first, then outer. With $A^{-1}B^{-1}$ the $A^{-1}$ is blocked by $B$ and cannot reach $A$ , so it never reduces to $I$ . The reversal is the only order that works.
Do matrices ever commute?: Sometimes — a matrix commutes with itself, with the identity, and with its own powers and inverse, and certain special families commute. But in general $AB \ne BA$ , as $A = \begin{pmatrix}1&1\\0&1\end{pmatrix}$ , $B = \begin{pmatrix}1&0\\1&1\end{pmatrix}$ show. That is exactly why a stated condition $PQ = QP$ is informative: it is true only for the special matrices that happen to commute with $P$ .
Is the transpose of a product also reversed?: Yes. $(AB)^{T} = B^{T}A^{T}$ , the same order reversal as the inverse, and for the same structural reason. So $(ABC)^{T} = C^{T}B^{T}A^{T}$ and $(ABC)^{-1} = C^{-1}B^{-1}A^{-1}$ — reverse the whole chain.
How does this trap show up in counting questions?: JEE Advanced 2025 Paper 1 Q4 asks how many $3\times3$ invertible integer matrices $Q$ satisfy $Q^{-1} = Q^{T}$ and $PQ = QP$ . The integer orthogonal matrices are the 48 signed permutation matrices, but the commutation constraint $PQ = QP$ is a real filter — it cuts the count to 16. If you treated $PQ = QP$ as automatic you would have kept all 48 and missed the answer.

Related misconceptions

A zero determinant does not mean no solutiondet A = 0 means the solution is not unique, not that there is none — a consistency check decides infinitely many versus none.
The determinant is not a linear mapdet(kA) = kⁿ·det A and det(A + B) ≠ det A + det B — the determinant multiplies, it does not add.

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