Why does det(AB) equal det(A) times det(B) for square matrices?
answer
- determinants measure volume scaling
- AB means apply B, then A
- stacking two maps stacks their factors
- scaling factors multiply, they do not add
- apply the rule to A times A inverse
basics
~10 sBecause the determinant is a volume scaling factor, and applying B and then A scales volume by det(B) and then by det(A). Scaling factors multiply, so det(AB) = det(A)det(B).
solid answer
~50 sThe determinant of a square matrix is the factor by which the map scales volume. The product `AB` means "apply `B`, then apply `A`", so volume is scaled first by `det(B)` and then by `det(A)`, and the two factors simply multiply. Check it concretely: a rotation `R` has `det(R) = 1` because rotating changes no areas, and the diagonal scaling `D = [[3, 0], [0, 2]]` has `det(D) = 6`; the composite `RD` has determinant `1 * 6 = 6` in either order. Two corollaries fall straight out. From `A A^-1 = I` and `det(I) = 1` you get `det(A^-1) = 1 / det(A)`, which also shows an inverse can only exist when `det(A)` is non-zero. And since `det(AB) = det(A)det(B) = det(B)det(A) = det(BA)`, the two products have equal determinants even though `AB` and `BA` are usually different matrices.
go deeper
Know the identity itself and be able to use it: given det(A) and det(B), report det(AB) as the product, and remember det(I) equals 1.
Justify it with the composition-of-scalings argument and derive det(A inverse) equals 1 over det(A) on the spot from A times A inverse equals I.
Show fluency with the corollaries an interviewer probes next: determinants of triangular matrices, of orthogonal matrices being plus or minus one, and the c to the n behaviour under scalar multiplication.
Own the framing that the determinant is a lossy one-number summary of a matrix, and be able to say what it does and does not license you to conclude in a design discussion.
## The claim For square matrices `A` and `B` of the same size, `det(AB) = det(A) det(B)`. This is one of the few identities in linear algebra that is both easy to state and genuinely load-bearing, and interviewers ask for the reason rather than the statement. ## The reason, in one picture Read each matrix as a linear map, and recall that `det` is the factor by which such a map scales volume (area in two dimensions, n-dimensional volume in general). The product `AB` denotes composition applied right to left: `(AB)x = A(Bx)`, so `B` acts first and `A` acts second. Now track a region of volume `V` through the composition. After `B` it has volume `|det(B)| * V`. After `A` it has volume `|det(A)| * |det(B)| * V`. Composition multiplies scaling factors, so the composite scales volume by `|det(A)| |det(B)|`. The signs behave the same way: composing two orientation-reversing maps restores orientation, exactly as the product of two negatives is positive. Putting magnitude and sign together gives `det(AB) = det(A) det(B)`. ## A concrete check Let `D = [[3, 0], [0, 2]]`, a diagonal matrix that stretches the horizontal direction by 3 and the vertical by 2. Its determinant is `3*2 - 0*0 = 6`, matching the picture: the unit square becomes a 3-by-2 rectangle of area 6. Let `R` be a rotation. A rotation moves shapes without resizing them, so `det(R) = 1`. Composing them, `det(RD) = det(R) det(D) = 1 * 6 = 6`, and equally `det(DR) = 6`. The matrices `RD` and `DR` are genuinely different — matrix multiplication does not commute — yet their determinants agree, because the same two scaling factors are multiplied either way. ## Corollaries worth having ready **Inverses.** Apply the rule to `A A^-1 = I`. The identity scales nothing, so `det(I) = 1`, giving `det(A) det(A^-1) = 1` and therefore `det(A^-1) = 1 / det(A)`. Two things follow. First, the determinant of the inverse is the reciprocal, which matches the picture: if `A` triples volume, undoing `A` must divide volume by three. Second, the equation is unsolvable when `det(A) = 0`, which is another route to the fact that a zero determinant rules out an inverse. **Powers.** Repeated application gives `det(A^k) = det(A)^k` for a positive integer `k`, since a power is just a product of copies. **Transpose.** `det(A^T) = det(A)`, so transposing never changes the scaling factor. **Triangular and diagonal matrices.** If every entry below the diagonal is zero (upper triangular) or every entry above it is zero (lower triangular), the determinant is simply the product of the diagonal entries. So `[[2, 7], [0, 5]]` has determinant `2 * 5 = 10`, and the `7` is irrelevant. A diagonal matrix is a special case, and the identity, being diagonal with all ones, has determinant 1. **Orthogonal matrices.** A square matrix `Q` is orthogonal when its columns are mutually perpendicular unit vectors, which is equivalent to `Q^T Q = I`, that is `Q^-1 = Q^T`. Applying the product rule and `det(Q^T) = det(Q)` gives `det(Q)^2 = 1`, so `det(Q) = +1` or `det(Q) = -1`. Geometrically that is exactly right: orthogonal maps are rigid motions about the origin, preserving lengths and volumes, with `+1` for rotations and `-1` for those that also mirror. Inverting one is free, since you only transpose. **Scaling the whole matrix.** Multiplying an n-by-n matrix by a scalar `c` is the same as multiplying by `cI`, and `det(cI) = c^n`, so `det(cA) = c^n det(A)`. ## What the rule does not say The determinant is multiplicative over products and **not** additive over sums: `det(A + B)` has no general relationship to `det(A)` and `det(B)`. The 2x2 counterexample is quick: with `A = I` and `B = -I`, both determinants are 1 while `A + B` is the zero matrix with determinant 0. Similarly, `det` interacts with scalar multiplication through the power `c^n`, not linearly. ## Answering in the room Lead with composition: determinants are volume scaling factors, composition stacks the factors, factors multiply. Then show one concrete pair — a rotation at determinant 1 against a diagonal scaling at determinant 6 — and finish with the corollary `det(A^-1) = 1 / det(A)` derived on the spot from `A A^-1 = I`. Deriving the corollary live is the part that signals you understand the identity rather than remember it.
- If det(A) = 4, what is det(A^-1)?It is `1/4`. From `A A^-1 = I` and `det(I) = 1`, the product rule gives `det(A) det(A^-1) = 1`, so `det(A^-1) = 1 / det(A)`. Geometrically, if `A` multiplies volume by 4 then undoing it must divide volume by 4. The same equation shows no inverse can exist when `det(A) = 0`.
- What is the determinant of an orthogonal matrix Q, where Q^T = Q^-1?Either `+1` or `-1`. From `Q^T Q = I` and `det(Q^T) = det(Q)`, the product rule gives `det(Q)^2 = 1`. That matches the geometry: orthogonal matrices are rigid, preserving lengths and volumes, with `+1` for rotations and `-1` when a mirror is included. A practical bonus is that inverting `Q` costs only a transpose.
- Does det(AB) = det(BA) even though AB and BA are usually different matrices?Yes for square `A` and `B` of the same size. Both equal `det(A) det(B)`, and scalars commute, so the determinants agree even when the product matrices do not. This is a useful reminder that the determinant discards almost all of the information in a matrix and keeps only the volume scaling.
saying these in an interview costs you the question
- Claims det(A + B) equals det(A) plus det(B)
- Says det(AB) differs from det(BA) because AB is not BA
- Thinks det(cA) equals c times det(A) for an n by n matrix
- Says det(A inverse) equals minus det(A)
- Includes off-diagonal entries when taking a triangular determinant