The determinant \det(A) is the signed volume of the parallelepiped spanned by A's columns — a single number that reports how much a linear map scales volume, and whether it flips orientation while doing it. \det(A) = 0 exactly when A's columns are dependent, which is exactly when A collapses some dimension to nothing. Every fact this course adds about determinants — as an alternating multilinear form in chapter 13, as the product of eigenvalues in chapter 06 — is a reinterpretation of this one volume-scaling fact, never a replacement for it.
Imagine every linear map as a machine. It resizes a lump of dough by some factor, and it might flip that dough over like a pancake. The determinant is that resizing factor, with a minus sign attached exactly when a flip happened. Run the dough through two machines in a row. The total resizing is just the two factors multiplied. It doesn't matter what shape the dough started as. Now suppose a machine flattens the dough down to a pancake with zero thickness. There's no way to run it backward and recover the original lump. That flattening is exactly what a zero determinant means, and exactly why that machine has no inverse.
\det(AB) = \det(A)\det(B) is not a computational curiosity to memorize separately — it's forced by the volume interpretation. Apply B first, scaling volume by \det(B). Then apply A, scaling the already-scaled volume by \det(A). The total scaling factor for the composite map AB has to be the product. The same reasoning makes \det(A^{-1}) = 1/\det(A) obvious rather than a formula to look up: undoing a volume-scaling by \det(A) takes scaling by its reciprocal. It also explains why \det(A) \ne 0 is the invertibility test. A map that collapses even one dimension to zero squashes the whole parallelepiped's volume to zero along with it. A volume of zero can never be un-collapsed back to something positive. So no inverse map can exist. A square matrix is invertible exactly when some positive volume always survives.
Theorem
Multiplicativity of the Determinant
also known as: det(AB) = det(A)det(B)
Recalled from Linear Algebra — Vectors, Matrices, and the Geometry of Data · Products, transposes, powers, and inverses — first taught there.
For any two n \times n matrices A and B, \det(AB) = \det(A)\det(B).
Fix A and treat \det(AB) as a function of B alone, writing B = \begin{pmatrix} \mathbf{b}_1 & \mathbf{b}_2 & \cdots & \mathbf{b}_n \end{pmatrix} in terms of its columns, so that AB = \begin{pmatrix} A\mathbf{b}_1 & A\mathbf{b}_2 & \cdots & A\mathbf{b}_n \end{pmatrix}.
Case 1: B is singular. Then \det B = 0, and the columns of B are linearly dependent — some \mathbf{b}_k is a linear combination of the others. Because A acts linearly, A\mathbf{b}_k is that same combination of the columns A\mathbf{b}_1, \dots, A\mathbf{b}_n, so the columns of AB are dependent too. A matrix with dependent columns has determinant 0, so \det(AB) = 0 = \det(A) \cdot 0 = \det(A)\det(B). The identity holds in this case regardless of what A is.
Case 2: B is invertible. Then B is a product of elementary matrices, B = E_1 E_2 \cdots E_k, since row-reducing B to the identity records a sequence of elementary row operations, and undoing them (in reverse) rebuilds B from I one elementary factor at a time. Each elementary matrix E scales the determinant of whatever it multiplies by a fixed factor c_E depending only on which operation E performs — a row swap gives c_E = -1, scaling a row by r gives c_E = r, and adding a multiple of one row to another gives c_E = 1 — and in every case that same factor satisfies \det E = c_E \cdot \det I = c_E. So for any matrix M, \det(EM) = c_E \det M = (\det E)(\det M). Applying this one elementary factor at a time, \det(AB) = \det(A E_1 E_2 \cdots E_k) = (\det E_k)\det(AE_1\cdots E_{k-1}) = \cdots = \det(A)\det(E_1)\det(E_2)\cdots\det(E_k). Running the same one-factor-at-a-time collapse on B itself (with A replaced by I) gives \det(B) = \det(E_1)\det(E_2)\cdots\det(E_k), so the product of the \det(E_i)'s appearing above is exactly \det(B), giving \det(AB) = \det(A)\det(B).
Together the two cases cover every B, so the identity holds in general.
Sanity check: take A = B = I_n. Both sides read \det(I_n) = 1 \cdot 1, which checks out. Take B singular and A = I_n: the left side is \det(I_n B) = \det B = 0, matching \det(I_n)\det(B) = 1 \cdot 0.
Theorem
Determinant Test for Invertibility
Recalled from Linear Algebra — Vectors, Matrices, and the Geometry of Data · Determinant and invertibility — first taught there.
An n\times n matrix A is invertible if and only if \det A \ne 0. When A is invertible, \det\left(A^{-1}\right) = \frac{1}{\det A}.
Both directions lean on one algebraic fact borrowed from determinant theory: for any two n\times n matrices A and B, \det(AB) = (\det A)(\det B).
If A is invertible, then \det A \ne 0. By definition AA^{-1} = I_n. Taking the determinant of both sides and applying the product rule above, \det(A)\det\left(A^{-1}\right) = \det I_n = 1. A product of two real numbers can only equal 1 if neither factor is 0 (anything times 0 is 0, never 1), so \det A \ne 0 — and this same equation already isolates \det\left(A^{-1}\right) = \frac{1}{\det A}, which is the stated corollary.
If \det A \ne 0, then A is invertible. Row-reduce A to an echelon form U. Each elementary row operation used along the way changes \det A by a known nonzero factor — a row swap multiplies the determinant by -1, scaling a row by a nonzero constant c multiplies the determinant by c, and adding a multiple of one row to another leaves the determinant unchanged — so \det U is \det A times a nonzero constant. Since \det A \ne 0 by hypothesis, this forces \det U \ne 0 as well.
U is upper triangular, so \det U is exactly the product of its diagonal entries. For that product to be nonzero, every diagonal entry of U must be nonzero — none of them collapsed to a zero pivot during elimination. An n\times n triangular matrix with n nonzero diagonal entries has a pivot in every row and every column, so U row-reduces further, by scaling each row to make its pivot 1 and clearing every entry above each pivot, all the way down to I_n. Since A was row equivalent to U and U is row equivalent to I_n, A is row equivalent to I_n.
Every elementary row operation is itself invertible — undoing a swap is another swap, undoing a scaling by c\ne0 is scaling by 1/c, undoing a row addition is subtracting the same multiple back off — and each can be realized as left-multiplication by an elementary matrix E_k. Chaining the operations that carried A to I_n gives E_k \cdots E_1 A = I_n, so the product B = E_k\cdots E_1 satisfies BA = I_n. Since B is itself a product of invertible elementary matrices, B is invertible with inverse B^{-1} = E_1^{-1}\cdots E_k^{-1}; multiplying BA = I_n on the left by B^{-1} gives A = B^{-1}, so A is invertible with A^{-1} = B.
Sanity check: for the 1\times 1 case A = [a], \det A = a, and the theorem reduces to the familiar fact that a single number is invertible (has a reciprocal) exactly when it is nonzero — the determinant test is the n\times n generalization of that one observation.
A 2D fact and a 3D fact, side by side
Edit an entry — watch the left panel's parallelogram area and the right panel's parallelepiped volume change together, and both flip color when a sign changes.
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Determinant as the live area-scaling readout
Nudge a, b, c, or d until \det = 0: the grid visibly collapses onto a single line exactly as the readout hits zero.
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Worked example
A repeated row forces determinant zero, without expanding
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1spotting a dependency between rows is faster than any expansion when one is visible
Without cofactor-expanding, decide whether A = \begin{pmatrix}1&2&3\\2&4&6\\0&1&-1\end{pmatrix} is invertible. \text{Row 2} = 2\times\text{Row 1}
spotting a dependency between rows is faster than any expansion when one is visible
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2
A row operation that subtracts 2\timesrow 1 from row 2 doesn't change the determinant's value, and it turns row 2 into all zeros. \det(A) = \det\begin{pmatrix}1&2&3\\0&0&0\\0&1&-1\end{pmatrix} = 0
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3\boxed{\det(A) = 0,\ \text{so } A \text{ is not invertible}}
The rows are dependent (row 2 is twice row 1), so the columns can't span all of \mathbb R^3 either — a zero determinant is the volume-collapse signature of exactly that dependency.
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Where this shows up: the determinant as the only alternating multilinear form
Chapter 13 answers a question this section never quite asks: why does this particular formula — alternating in sign, linear in each column separately — deserve to be called "the" determinant rather than just "a" volume-like number? The answer is that it's the unique function with those two properties (up to a scalar), which is a genuinely different way to arrive at the same object this section builds computationally.
Where this goes next
The spectral theorem and its converse
The characteristic polynomial \det(A-\lambda I) — a determinant with a variable subtracted along the diagonal — is where every eigenvalue this course ever computes ultimately comes from. Chapter 06's guarantee that a complex matrix always has at least one eigenvalue is, underneath, a guarantee that this determinant, as a polynomial in \lambda, always has at least one complex root — the Fundamental Theorem of Algebra doing the actual work.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Linear Algebra — Vectors, Matrices, and the Geometry of Data
Compute \det\begin{pmatrix} 5 & 2 \\ 1 & 3 \end{pmatrix}, the all-positive-entry matrix.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
\det = ad-bc = 5(3) - 2(1) = 15-2=13. Cross-multiply the diagonals and subtract.
Correct answer
Also accepted:
Why
From Linear Algebra — Vectors, Matrices, and the Geometry of Data
\mathbf{u}=(k,2) and \mathbf{v}=(3,4) span a parallelogram with \det = 14. Find k.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
\det = k(4) - 3(2) = 4k-6. Setting 4k-6=14 gives 4k=20, so k=5.
Correct answer
Also accepted:
Why
Write det in terms of k using ad - bc, set it equal to 14, and solve the resulting linear equation for k.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Linear Algebra — Vectors, Matrices, and the Geometry of Data
A linear map has \det B = -6. A region of area 5 is mapped by B. Area is always reported as a nonnegative number -- what is the area of the image?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
Area scales by |\det B| = |-6| = 6, so the image has area 6\cdot 5 = 30. The negative sign says the map reverses orientation -- it does not make the area negative.
Correct answer
Also accepted:
Why
From Linear Algebra — Vectors, Matrices, and the Geometry of Data
Linear maps S and T have \det S = 3 and \det T = -2. Applying T then S (the composite map S\circ T) to a region of area 5 produces an image of what area, and does the composite preserve or reverse orientation?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
\det(S\circ T) = \det S \cdot \det T = 3\cdot(-2) = -6 is negative, so the composite reverses orientation, not preserves it.
Composing two maps multiplies their scaling effects; a composite map does not leave area unchanged just because it's built from two steps.
1 would come from treating \det S and \det T as canceling rather than multiplying -- their product is -6, not close to 1.
Why
The determinant of a composite map is the product of the individual determinants: \det S\cdot\det T = 3(-2) = -6. The scaling factor is |-6|=6, so the image area is 6\cdot 5 = 30, and the negative product means orientation is reversed.
Apply each map's scaling factor in the order they're applied -- first T's factor, then S's -- and separately track whether each map flips orientation.