The Determinants page started from a formula, cofactor expansion, and then proved that row operations change it in three tidy ways. This page runs that story backward. Start from the three ways, the rules any honest notion of signed volume has to follow, and ask which functions obey them. There's exactly one, and it's $\det$. Turning the story around buys a lot: a fast way to compute, the reason $\det$ is linear in each row, the area of every ellipse, and where the formula came from in the first place.
Forget the formula for a minute. Suppose all you knew about the volume of a box was how it reacts when you push the box around: slide a face, stretch an edge, swap two edges. This page shows that those few reactions pin the volume down completely, and the number they force is the determinant. Along the way, the box pictures hand us a fast way to compute it, the area of an ellipse, and what a negative volume really means.
On the Determinants page, a swap flipped the sign, scaling a row by $k$ scaled $\det$ by $k$, and adding a multiple of a row changed nothing. Those were theorems there. Here they're the starting point.
Think of the rows of a square matrix as the edges of a box: two rows give a parallelogram in the plane, three rows give a slanted box in space. (The Determinants page drew the box on the columns; since $\det(A^t) = \det(A)$, the box on the rows has the same signed volume.) Whatever "signed volume" means, it should react to the three row operations the way boxes do.
A determinant function is a function $D$ from $n \times n$ matrices to $\R$ such that (1) adding a multiple of one row to another row leaves $D$ unchanged, (2) swapping two rows multiplies $D$ by $-1$, (3) multiplying one row by a number $k$ multiplies $D$ by $k$, and (4) $D(I_n) = 1$.
A determinant function is a rule $D$ that gives each $2 \times 2$ (or each $3 \times 3$) matrix a number, its signed volume, such that (1) sliding one edge along another (adding a multiple of one row to another) leaves $D$ unchanged, (2) swapping two edges multiplies $D$ by $-1$, (3) stretching one edge by $k$ multiplies $D$ by $k$, and (4) the unit square (or unit cube) has $D = 1$.
Rule (1) is a shear: the top of the box slides along, the base and the height stay put, so the volume can't change. Rule (3) is a stretch. Rule (2) is the strange one, and it's where the sign comes from: swapping two edges keeps the same box but turns it over. Watch the rules carry a value from the unit square all the way to a parallelogram, without ever measuring anything.
Every step there was forced. That's the whole idea of the page, so try it yourself with a matrix that needs a swap.
Using only the four rules, find $D\begin{pmatrix} 0 & 2 \\ 3 & 0 \end{pmatrix}$.
Swap the rows (rule 2): $D\begin{pmatrix} 0 & 2 \\ 3 & 0 \end{pmatrix} = -D\begin{pmatrix} 3 & 0 \\ 0 & 2 \end{pmatrix}$. Pull the $3$ out of row $1$ and the $2$ out of row $2$ (rule 3): $-D\begin{pmatrix} 3 & 0 \\ 0 & 2 \end{pmatrix} = -3 \cdot 2 \cdot D(I_2) = -6$ (rule 4). And $ad - bc = 0 - 6 = -6$, as it should be.
The four rules aren't even all needed.
Rule (2) follows from the other three. To swap rows $1$ and $2$: $R_1 \to R_1 + R_2$, then $R_2 \to R_2 - R_1$ (row $2$ is now $-$ the old row $1$), then $R_1 \to R_1 + R_2$ (row $1$ is now the old row $2$), then multiply row $2$ by $-1$. Three free moves and one factor of $-1$. It's listed anyway because the swap is the move you actually make.
Now build some boxes yourself. Each mission names a target; reach it from the unit square with row operations, and keep an eye on what the rules say $D$ must be.
Notice you never needed $ad - bc$. The rules did all the work, and the answer matched every time. That's not luck.
Two quick consequences first, because the main theorem leans on them. A box with two equal edges, or with an edge of length zero, is flat.
Using only the rules, show that $D$ is $0$ for a matrix with two equal rows, and for a matrix with a row of zeros.
Swap the two equal rows. The matrix doesn't change, so neither does $D$; but rule (2) says $D$ flips sign. The only number equal to its own negative is $0$.
For a zero row, multiply that row by $0$ (rule 3 allows $k = 0$). The matrix doesn't change, but $D$ gets multiplied by $0$. So $D = 0$.
With those in hand, here is the payoff: the rules leave no freedom at all.
For each $n$ there is exactly one determinant function on $n \times n$ matrices, and it is $\det$.
There is one. $\det$ obeys rules (1), (2), and (3) by the row-operation theorem, and $\det(I_n) = 1$ because $I_n$ is triangular with $1$s on its diagonal.
There is only one. Let $D$ be any determinant function and $A$ any $n \times n$ matrix. Row reduce $A$ to its reduced row echelon form $R$, using additions, swaps, and scalings by nonzero numbers $c_1, \ldots, c_m$, with $s$ swaps in all. By the rules, $D(R) = (-1)^s c_1 \cdots c_m \, D(A)$. If $R$ has a zero row, $D(R) = 0$ (the example above), so $D(A) = 0$. Otherwise $R = I_n$, and $D(A) = \dfrac{(-1)^s}{c_1 \cdots c_m}$. Either way the value of $D(A)$ was computed from the reduction alone, without knowing anything else about $D$. So any two determinant functions give the same number on every matrix, and since $\det$ is one of them, $D = \det$. $\blacksquare$
For $2 \times 2$ matrices, and for $3 \times 3$ matrices, the four rules allow exactly one determinant function: $D = \det$, the signed area or signed volume.
This is a visual argument, not a formal proof. Any box that isn't flat can be built from the unit cube by stretches and slides (and maybe a swap), and each move tells you exactly what happens to $D$. So by the time the box is built, $D$ has been decided. A flat box gets $D = 0$, like the two-equal-edges box above. Here the cube becomes a slanted box with edges $(2, 0, 0)$, $(1, 1.5, 0)$, $(1, 1.5, 1)$: two stretches make the volume $3$, two slides keep it $3$.
The full proof is in the n version.
So the formula on the Determinants page wasn't an arbitrary invention. It's the only function with the right reactions, and any other recipe that obeys the rules (cofactors along a different row, the permutation formula at the end of this page) must give the same number. The proof also hides an algorithm.
So $ad - bc$, and the $3 \times 3$ formula, weren't arbitrary. They're the only numbers that react to pushes the way volume does. The picture also hides a fast way to compute.
Row reduce until the matrix is triangular, keeping score: additions are free, each swap costs a sign.
Suppose row reduction carries $A$ to an upper triangular matrix $U$ using only additions of multiples of rows and $s$ swaps. Then
$$\det(A) = (-1)^s\, U_{11} U_{22} \cdots U_{nn}.$$So $\det(A)$ is plus or minus the product of the pivots, and $\det(A) = 0$ when a column has no pivot.
Each addition leaves the determinant alone and each swap multiplies it by $-1$, so $\det(U) = (-1)^s\det(A)$, that is, $\det(A) = (-1)^s\det(U)$. A triangular matrix's determinant is the product of its diagonal (row-operation theorem, part (e)). If a column of $U$ has no pivot, that diagonal entry is $0$ and so is the product. $\blacksquare$
This is a visual argument, not a formal proof. Additions are slides and swaps are flips, so only the triangular box at the end needs measuring. And a triangular box is easy: more slides (adding multiples of the lower rows to the upper ones) stand it up straight, until its edges lie along the axes with lengths $|U_{11}|$, $|U_{22}|$, $|U_{33}|$. A straight box's volume is length times width times height. Below, the box from the next example stands up into a $1 \times 1 \times 3$ brick.
The full proof is in the n version.
Here's one where the reduction gets stuck without a swap.
Compute $\det(A)$ for $A = \begin{pmatrix} 1 & 1 & 2 \\ 2 & 2 & 1 \\ 0 & -1 & 1 \end{pmatrix}$ by elimination.
$R_2 \to R_2 - 2R_1$ is free and gives row $2 = (0, 0, -3)$. Now the middle spot is $0$, so swap rows $2$ and $3$, which costs a sign:
$$\det(A) = \det\begin{pmatrix} 1 & 1 & 2 \\ 0 & 0 & -3 \\ 0 & -1 & 1 \end{pmatrix} = -\det\begin{pmatrix} 1 & 1 & 2 \\ 0 & -1 & 1 \\ 0 & 0 & -3 \end{pmatrix} = -\big(1 \cdot (-1) \cdot (-3)\big) = -3.$$In pictures: the slide doesn't change the box, the swap turns it over, and the triangular box has volume $3$. The sign says the original rows were a left-handed triple.
The method doesn't care about size. A $4 \times 4$ cofactor expansion is four $3 \times 3$ determinants; elimination is a handful of row operations.
Compute $\det(A)$ for $A = \begin{pmatrix} 1 & 2 & 0 & 1 \\ 2 & 4 & 1 & 3 \\ 0 & 1 & 1 & 0 \\ 1 & 2 & 0 & 3 \end{pmatrix}$.
$R_2 \to R_2 - 2R_1$ gives $(0, 0, 1, 1)$, and $R_4 \to R_4 - R_1$ gives $(0, 0, 0, 2)$. Column $2$ needs a pivot in row $2$, so swap rows $2$ and $3$:
$$\det(A) = -\det\begin{pmatrix} 1 & 2 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 2 \end{pmatrix} = -(1 \cdot 1 \cdot 1 \cdot 2) = -2.$$One more move is legal, as long as you pay it back.
Scaling a row is allowed during elimination (to clear a fraction, say), but it changes the determinant: if you multiply a row by $c$, the new matrix has $c$ times the determinant, so divide by $c$ at the end. The most common slip is to multiply by $c$ a second time instead.
Your turn. Each question shows the reduction; you keep the score. In free response, type one number.
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Rule (3) says a number comes out of one row at a time. Sums do too, as long as only one row changes.
A function $D$ of the rows of a square matrix is multilinear if it is linear in each row when the other rows are held fixed: $D(\ldots, u + v, \ldots) = D(\ldots, u, \ldots) + D(\ldots, v, \ldots)$ and $D(\ldots, ku, \ldots) = k\,D(\ldots, u, \ldots)$, with the same row changing on both sides.
The scaling half is rule (3). The sum half is new, and it's true.
$\det$ is multilinear. If three square matrices agree except in row $i$, and row $i$ of the third is the sum of row $i$ of the other two, then its determinant is the sum of their determinants.
Call the matrices $A$, $B$, $C$, with row $i$ of $C$ equal to row $i$ of $A$ plus row $i$ of $B$. Expand all three along row $i$ (cofactor expansion). Deleting row $i$ removes the only place they differ, so $\tilde{A}_{ij} = \tilde{B}_{ij} = \tilde{C}_{ij}$ for every $j$, and
$$\det(C) = \sum_j (-1)^{i+j}(A_{ij} + B_{ij})\det(\tilde{C}_{ij}) = \det(A) + \det(B).$$The scaling half is part (b) of the row-operation theorem. $\blacksquare$
This is a visual argument, not a formal proof. Hold the other rows fixed: they make the base. The area (or volume) is base times height, and the height of $u + v$ above the base is the height of $u$ plus the height of $v$, signs included (an edge pointing below the base has negative height). Shown in the plane, where the base is one edge:
The full proof is in the n version.
Splitting a row into simpler pieces is a legitimate way to compute.
Split row $1$ of $\begin{pmatrix} 2 & 1 \\ 4 & 3 \end{pmatrix}$ as $(2, 1) = (2, 0) + (0, 1)$ and compute the determinant that way.
And directly, $2 \cdot 3 - 1 \cdot 4 = 2$.
Multilinearity says more than "you may split rows". If one row is a variable point $(x, y, z)$ and the other two are fixed, the determinant is a linear function of that point, and setting it to $0$ gives a plane.
Describe all points $(x, y, z)$ with $\det\begin{pmatrix} x & y & z \\ 1 & 0 & 1 \\ 0 & 1 & 2 \end{pmatrix} = 0$.
Expand along row $1$: $x(0 - 1) - y(2 - 0) + z(1 - 0) = -x - 2y + z$. So the points are the plane $x + 2y - z = 0$. It goes through $0$ and contains both fixed rows: $1 + 0 - 1 = 0$ and $0 + 2 - 2 = 0$. That makes sense: the box is flat exactly when the first edge lies in the plane of the other two.
Watch the box as the first edge wanders: its volume is a ruler measuring how far $(x, y, z)$ sits off that plane, and it flattens to $0$ every time the edge lands in it.
Linear in each row is not the same as linear in the whole matrix.
$\det(A + B)$ is almost never $\det(A) + \det(B)$: adding two matrices changes every row at once. For example, $\det(I_2 + I_2) = \det(2I_2) = 4$, while $\det(I_2) + \det(I_2) = 2$. The sum rule works only when the matrices agree in every row but one.
Your turn: no matrix entries, just the rule. In free response, type one number.
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The area theorem said the map $x \mapsto Ax$ multiplies every area by $|\det(A)|$, because any shape is a pile of tiny grid squares and each one is stretched by the same factor. Polygons were the easy case. The pile-of-squares argument works just as well for a disc.
So an ellipse is just a circle that's been stretched, and it keeps its $\pi$ the whole time. Somewhere a circle is very proud of that.
A disc goes to an oval, and the oval's area is the disc's times $|\det(A)|$. Ellipses are the simplest case.
The ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} \le 1$ (with $a, b > 0$) has area $\pi ab$. If $A$ is a $2 \times 2$ matrix, the image of the disc of radius $r$ under $x \mapsto Ax$ has area $\pi r^2|\det(A)|$; if $A$ is $3 \times 3$, the image of the ball of radius $r$ has volume $\frac{4}{3}\pi r^3|\det(A)|$.
The disc of radius $r$ has area $\pi r^2$ and the ball has volume $\frac{4}{3}\pi r^3$, and $x \mapsto Ax$ multiplies every area or volume by $|\det(A)|$ (area theorem). For the ellipse, take $A = \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix}$, which sends $(s, t)$ to $(x, y) = (as, bt)$. Then $s^2 + t^2 \le 1$ exactly when $\frac{x^2}{a^2} + \frac{y^2}{b^2} \le 1$, so the ellipse is the image of the unit disc, with area $\pi \cdot 1^2 \cdot ab$. $\blacksquare$
This is a visual argument, not a formal proof. It's the figure above: every grid square inside the disc becomes a parallelogram $|\det(A)|$ times as big, so the whole pile grows by that factor. For the ellipse, stretch the unit disc $a$ times sideways and $b$ times up: that's the matrix with $a$ and $b$ on its diagonal, determinant $ab$, so the area goes from $\pi$ to $\pi ab$.
The full proof is in the n version.
Put an ellipse in that form first, then read off $a$ and $b$.
(a) Find the area inside $4x^2 + 9y^2 = 36$. (b) Find the area of the image of the disc $x^2 + y^2 \le 4$ under $A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$.
(a) Divide by $36$: $\frac{x^2}{9} + \frac{y^2}{4} \le 1$, so $a = 3$, $b = 2$, and the area is $6\pi$.
(b) The disc has radius $2$, so area $4\pi$, and $\det(A) = 4 - 1 = 3$. The image has area $3 \cdot 4\pi = 12\pi$. (That's the figure above.)
So much for the size. The sign of $\det(A)$ says something too.
An invertible $n \times n$ matrix $A$ (or the map $x \mapsto Ax$) is orientation-preserving if $\det(A) > 0$ and orientation-reversing if $\det(A) < 0$.
An invertible $3 \times 3$ matrix $A$ is orientation-preserving if it sends every right-handed triple of arrows to a right-handed triple, and orientation-reversing if it sends right-handed triples to left-handed ones. In the plane, the same words mean keeping or reversing counterclockwise turns.
In $\R^2$ and $\R^3$ this matches the pictures: a positive determinant keeps counterclockwise turns counterclockwise (and right-handed triples right-handed), and a negative one turns the plane, or space, over like a mirror. A rotation preserves orientation; a reflection reverses it.
And the determinant tells you which: preserving when $\det(A) > 0$, reversing when $\det(A) < 0$. A rotation is preserving; a mirror is reversing.
Does $x \mapsto -x$ preserve orientation in $\R^2$? In $\R^3$?
In $\R^2$, $\det(-I_2) = (-1)(-1) = 1 > 0$: it preserves orientation. It's a half turn, and a turn never flips anything. In $\R^3$, $\det(-I_3) = (-1)^3 = -1 < 0$: it reverses orientation. Each axis flips, and three flips make a flip.
Your turn: curved shapes. Answers are a number times $\pi$; in free response, type something like 6pi or 8/3 pi.
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Multilinearity can take a determinant all the way apart. Go back to the split of $\begin{pmatrix} 2 & 1 \\ 4 & 3 \end{pmatrix}$ and split row $2$ as well.
Split both rows of $\begin{pmatrix} 2 & 1 \\ 4 & 3 \end{pmatrix}$ into pieces with one entry each, and add up.
The first and last pieces have both entries in one column, so their rows are parallel and the box is flat. Only the pieces with one entry in each row and each column survive, and each is a shuffled diagonal matrix: $+2 \cdot 3$, and $-1 \cdot 4$ (it takes one swap to straighten). That's $ad - bc$.
The same thing happens in every size. A surviving piece picks one entry from each row, in different columns: row $i$ uses column $\sigma(i)$ for some rearrangement $\sigma$ of $1, \ldots, n$. Its sign is the determinant of the shuffled identity.
For a $3 \times 3$ the same split makes $27$ pieces, and all but six are flat. The six survivors pick one entry from each row and each column, and the sign of each says whether straightening it takes an even or odd number of swaps.
Let $\sigma$ be a permutation of $\{1, \ldots, n\}$: a function from $\{1, \ldots, n\}$ to itself that takes each value exactly once. The sign of $\sigma$ is $\sgn(\sigma) = \det(P_\sigma)$, where $P_\sigma$ is the permutation matrix with a $1$ in row $i$, column $\sigma(i)$, for each $i$.
Row swaps turn $P_\sigma$ into $I_n$, so $\sgn(\sigma)$ is $+1$ when that takes an even number of swaps and $-1$ when it takes an odd number. (The Symmetric Groups page calls these even and odd permutations.)
For every $n \times n$ matrix $A$,
$$\det(A) = \sum_{\sigma} \sgn(\sigma)\, A_{1\sigma(1)} A_{2\sigma(2)} \cdots A_{n\sigma(n)},$$the sum running over all $n!$ permutations $\sigma$ of $\{1, \ldots, n\}$.
Write row $i$ of $A$ as $A_{i1}e_1 + \cdots + A_{in}e_n$ (the $e_j$ as rows). Multilinearity in row $1$, then row $2$, and so on, expands $\det(A)$ into $n^n$ terms, one for each choice of columns $j_1, \ldots, j_n$:
$$\det(A) = \sum_{j_1, \ldots, j_n} A_{1j_1} \cdots A_{nj_n} \det\begin{pmatrix} e_{j_1} \\ \vdots \\ e_{j_n} \end{pmatrix}.$$If two of the $j$'s are equal, that matrix has two equal rows and determinant $0$. Otherwise $i \mapsto j_i$ is a permutation $\sigma$, the matrix is $P_\sigma$, and its determinant is $\sgn(\sigma)$. $\blacksquare$
For a $3 \times 3$ matrix $A$,
$$\det(A) = A_{11}A_{22}A_{33} + A_{12}A_{23}A_{31} + A_{13}A_{21}A_{32} - A_{13}A_{22}A_{31} - A_{11}A_{23}A_{32} - A_{12}A_{21}A_{33}.$$This is a visual argument, not a formal proof. Each product picks one entry from each row and each column, a pattern like a rook placement on a $3 \times 3$ board. Straighten the pattern onto the diagonal with row swaps: no swaps or two give $+$, one swap gives $-$. Here are the six patterns on the matrix from the elimination example; they add to $-3$, the same answer.
The full proof is in the n version.
That's the formula for $\det$ with no recursion in it, and it's what other books often take as the definition. It's beautiful and hopeless for computing: $n!$ terms is $3{,}628{,}800$ for a $10 \times 10$ matrix, while elimination takes a few hundred operations.
Those six products are the "diagonals" pattern for $3 \times 3$ determinants, and now you know why there are six and where the signs come from. A $4 \times 4$ would need $24$ products; elimination is the better tool there.
Group the terms of the expansion by which entry of row $1$ they use. Each group is that entry times a smaller determinant, with a sign. That's cofactor expansion, and the next page takes it apart.
The toolkit for this page: $\det$ is the one function that ignores slides, flips sign on a swap, scales with a row, and gives $\det(I) = 1$. Compute it by eliminating to triangular and multiplying the diagonal (a sign per swap). It's linear in each row separately, never in the whole matrix. Curved shapes scale by $|\det(A)|$ too ($\pi ab$ for an ellipse), and the sign of $\det(A)$ says whether $A$ preserves orientation.
Every question so far came right after its idea, which is a hint a quiz never gives you. Down here the hint is gone. First, true or false: a claim from anywhere on this page, and you decide whether it holds.
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Next, a mixed set: every question type on this page, shuffled. Type one number when it asks you to.
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Made it this far? Time for something with a little heat: a matrix with a parameter in one row, and a ball it stretches to a given volume. In free response, type one number.
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