You could have invented the determinant

I'm going to talk about how a geometric intuition becomes the definition of the determinant.

As many of you already know, the determinant of a matrix is the (hyper)volume of the (hyper)parallelepiped whose edges are the (row or column) vectors of the matrix. This page on Wolfram gives you an interactive visualization of a parallelogram or a parallelepiped while showing the corresponding determinant.

There are abundant sources on the internet (many on math.stackexchange.com) that give more-or-less the same explanation as I just said above, but most of them don't explain why. That's what I want to discuss in this post.

First goal: Define area in 2D

I'll motivate this discussion with a simple question: how to define volume in a Euclidean space?

I'm kidding. That question is actually not that simple, and people have gone to great lengths to answer that. Let me simplify it a bit more, quite gratuitously, so that we can cover it in one blog post:

How do we define the "area" of a parallelogram in a 2-dimensional space?

(I'm being intentionally vague here, but stay with me.)

First, let's agree that a parallelogram is defined by two vectors, like this:

In this picture, $u$ and $v$ are vectors in the 2-dimensional space, which I'll call $\mathbb R^2$ from now on. We're looking for a way to define the shaded area inside the parallelogram, i.e., a function $A$ that takes two vectors and gives out a number representing the area: \begin{gather*} A(u, v) = \text{ area of the parallelogram with $u$ and $v$ as edges }. \end{gather*}

It will turn out that defining a signed area will be simpler than a non-negative area, so we'll allow $A$ to go negative, and we'll try to reason what that means. Now, let's think about some properties that $A$ should have.

Property 1: Multilinear

The following picture should illustrate quite clearly why we want \begin{gather} A(u, v) + A(u, w) = A(u, v + w)\label{eq:add} \end{gather}

Obviously we want both shaded regions to have the same area. That translates directly into \eqref{eq:add}.

Note that the picture only captures the case where $v$ and $w$ both point upwards relative to $u$, which is pointing to the right. However, if $w$ actually points downward, \eqref{eq:add} would mean that there's some cancellation, i.e., $A(u, v)$ and $A(u, w)$ would have different signs. This is where the notion of signed area helps us: we don't need to split \eqref{eq:add} into cases.

If you're skeptical, you could say that the notion of signed area is forced by our artificial desire to satisfy \eqref{eq:add}. That may be the case, but it also suggests that the signed area could be a more natural definition than the non-negative area. In fact, this is just like what negative numbers mean in counting: borrowing.

(This is a common theme in algebra. Sometimes, "natural" definitions don't seem very natural at first, but they are indeed more natural once we broaden our minds to accept what "natural" means.)

For algebraic completeness, I should also mention that \begin{gather} \lambda A(u, v) = A(u, \lambda v)\label{eq:scalar} \end{gather} for any scalar $\lambda \in \mathbb R$. It should be pretty obvious to see this from the same picture above: just make $w$ parallel to $v$.

The two equations \eqref{eq:add} and \eqref{eq:scalar} can be combined into one:

\begin{gather} A(u, v + \lambda w) = A(u, v) + \lambda A(u, w)\label{eq:linear} \end{gather}

where $\lambda \in \mathbb R$. This simply says that the function $v \mapsto A(u, v)$, with $u$ fixed, is linear.

If we swap the parameters in $A$, the same argument above still applies, and we can conclude that

\begin{gather} A(u + \lambda v, w) = A(u, w) + \lambda A(v, w).\label{eq:linear2} \end{gather} In words, the function $u \mapsto A(u, w)$, with $w$ fixed, is linear.

There's a word for linearity in each parameter of a multi-parameter function (when all the other parameters are fixed): multilinearity.

Property 2: Alternating

Another observation we can make is that sheering the parallelogram does not change its area.

We can express this as an equation for $A$: \begin{gather} A(u, v + \lambda u) = A(u, v) \label{eq:sheer} \end{gather} for any scalar $\lambda \in \mathbb R$. In words, adding any multiples of $u$ to $v$ doesn't change $A(u, v)$ because it contributes no area. The special case where $v = 0$ gives us \begin{gather} A(u, u) = 0. \label{eq:alternating} \end{gather} Somehow people like to call a function with this property "alternating".

Note that we can also get

\begin{gather*} A(u + \lambda v, v) = A(u, v) \end{gather*}

from the same argument applied to $A$ with parameters swapped, or from \eqref{eq:linear2} and \eqref{eq:alternating} like this:

\begin{align} A(u + \lambda v, v) = A(u, v) + \lambda A(v, v) = A(u, v). \label{eq:sheer2} \end{align}

One convenient formula that we can derive from \eqref{eq:sheer2}, \eqref{eq:sheer} and \eqref{eq:linear2} is the following:

\begin{align*} A(u, v) & = A(u - v, v) & & \text{; from \eqref{eq:sheer2}}\\ & = A(u - v, v + u - v) & & \text{; from \eqref{eq:sheer}}\\ & = A(u - v, u) \\ & = A(-v, u) & & \text{; from \eqref{eq:sheer2}}\\ & = -A(v, u) & & \text{; from \eqref{eq:linear2}}. \end{align*}

This means swapping the order of the parameters of $A$ will just flip the sign, which is consistent with our notion of "borrowing" when the signed area goes negative.

The property that $A(u, v) = -A(v, u)$ is called skew-symmetry, which is synonymous with being alternating in the 2-dimensional case.

Remark: Even though I derived \eqref{eq:alternating} from \eqref{eq:sheer}, the reverse direction is also possible. (In fact, it is simpler too.) We can start with a geometric intuition of \eqref{eq:alternating}: $A(u, u) = 0$ means if the two edges of a parallelogram are parallel, the area is 0. Then, we proceed to derive \eqref{eq:sheer} with one application of multilinearity.

\begin{align*} A(u, v + \lambda u) & = A(u, v) + \lambda \underbrace{A(u, u)}_0 = A(u, v). \end{align*}

We actually did this already, when we derived \eqref{eq:sheer2}.

Base case

Knowing that $A$ has to be multilinear and alternating is almost enough to determine $A$ completely. We only need to define some more base cases. (As you may notice, \eqref{eq:linear}, and \eqref{eq:linear2} are, in a sense, recursive definitions. \eqref{eq:alternating} gives us some base cases, but we don't have any base cases that will give us any non-zero values yet.)

Suppose $e_1$ and $e_2$ are the standard coordinate basis vectors of $\mathbb R^2$:

\begin{gather*} e_1 = \begin{pmatrix}1\\0\end{pmatrix}, \qquad e_2 = \begin{pmatrix}0\\1\end{pmatrix}. \end{gather*}

Pictorially,

One natural choice for the base case of $A$ is

\begin{gather} A(e_1, e_2) = 1\label{eq:unit} \end{gather}

which says that a 1-by-1 square has area 1. (This is not the only choice that we could pick, but it's quite natural.)

Consequences

After picking the basis vectors and the non-zero base case of $A$, we are now ready to compute $A(u, v)$ for any vectors $u$ and $v$. Let's write out $u$ and $v$ as

\begin{gather*} u = \begin{pmatrix}a\\b\end{pmatrix} = ae_1 + be_2, \qquad v = \begin{pmatrix}c\\d\end{pmatrix} = ce_1 + de_2. \end{gather*}

Then,:

\begin{align*} A(u, v) & = A(ae_1 + be_2, v) \\ & = aA(e_1, v) + bA(e_2, v) & & \text{; from \eqref{eq:linear2}} \\ & = aA(e_1, ce_1 + de_2) + bA(e_2, ce_1 + de_2) \\ & = acA(e_1, e_1) + adA(e_1, e_2) + bcA(e_2, e_1) + bdA(e_2, e_2) & & \text{; from \eqref{eq:linear}}\\ & = adA(e_1, e_2) + bcA(e_2, e_1) & & \text{; from \eqref{eq:alternating}} \\ & = ad - bc & & \text{; from \eqref{eq:unit}.} \end{align*}

Note that I have used $A(e_2, e_1) = -A(e_1, e_2) = 1$ in the last line.

It probably hasn't been such a secret that our $A$ is the determinant. (I didn't intend for it to be a secret anyway.) The derivation above can be written with the usual determinant notation like this:

\begin{align} \begin{vmatrix} a & c\\b & d \end{vmatrix} & = a\begin{vmatrix}1 & c\\0 & d\end{vmatrix} + b\begin{vmatrix}0 & c\\1 & d\end{vmatrix}\nonumber\\ & = ac \underbrace{ \begin{vmatrix} 1 & 1\\0 & 0 \end{vmatrix}}_0 + ad \underbrace{ \begin{vmatrix} 1 & 0\\0 & 1 \end{vmatrix}}_1 + bc \underbrace{ \begin{vmatrix} 0 & 1\\1 & 0 \end{vmatrix}}_{-1} + bd \underbrace{ \begin{vmatrix} 0 & 0\\1 & 1 \end{vmatrix}}_0 \nonumber\\ & = ad - bc. \label{eq:formulafor2} \end{align}

Conclusion

We started with an attempt to define the area of a parallelogram, and we observed that our geometric intuition of area leads to the two key properties: multilinearity and skew-symmetry. These two key properties plus the "base case" choice (that $A(e_1, e_2) = 1$) are all we need to define $A$ for any pairs of vectors. If we write vectors as coordinates, $A$ becomes the usual determinant.

Addendum: Consistency of properties

How do we ensure that \eqref{eq:linear}, \eqref{eq:linear2}, \eqref{eq:alternating} and \eqref{eq:unit} are consistent, i.e., they do not contradict with each other?

We can prove the consistency by starting from the formula

\begin{gather} A(ae_1 + be_2, ce_1 + de_2) = ad - bc,\label{eq:formula} \end{gather}

which is well-defined for all pairs of vectors, then deriving \eqref{eq:linear}, \eqref{eq:linear2}, \eqref{eq:alternating} and \eqref{eq:unit} from \eqref{eq:formula}. I'm not going to show the derivation here though because it's very straightforward and easily searchable online (and in textbooks too).

This reverse way of developing math may seem a bit silly, but it is not pointless. Quite a lot of math is done the same way: we use intuition to write down properties, use those properties to make up definitions, then use those definitions to prove those properties. This ensures that our definitions capture our intuition correctly.

Higher Dimensions

Everything discussed so far extends to higher dimensions easily, but let me flesh out the detail for you a little bit for $\mathbb R^3$. Let $V(x, y, z)$ be the signed volume of a parallelepiped with sides $x$, $y$ and $z$.

Property 1: Multilinear

As depicted below, we would want that

\begin{gather*} V(x, y, u + v) = V(x, y, u) + V(x, y, v). \end{gather*}

In words, if we fix $2$ parameters in $V$, we get a function with $1$ parameter that is linear.

The situation is similar for an $n$-dimensional space. If we fix any $n - 1$ parameters, we get a function with $1$ parameter that is linear. This is the actual definition of multilinearity.

Property 2: Alternating

As depicted below, we would want that

\begin{gather*} V(x, y, u + \lambda x + \mu y) = V(x, y, u) \end{gather*}

for any scalars $\lambda, \mu \in \mathbb R$. I chose $\lambda = \mu = \frac 12$ in the picture, but they can be any scalar values. This is consistent with the intuition that shearing the parallelepiped does not change its volume.

Similar to the 2-dimensional case, the alternating property can be derived from this formula, or can be used to derive this formula. The geometric interpretation of the alternating property is that if two vectors coincide, the resulting parallelepiped will have zero volume. As equations:

\begin{align*} V(u, u, v) & = 0 \\ V(u, v, u) & = 0 \\ V(v, u, u) & = 0 \end{align*}

for all vectors $u$ and $v$.

The generalization to an $n$-dimensional space is straightforward: if any two of the $n$ vectors are equal, the resulting (hyper)parallelepiped will have zero (hyper)volume.

Base case

The most obvious choice for the base case for $\mathbb R^3$ is

\begin{align*} V(e_1, e_2, e_3) = 1. \end{align*}

The way to extend this to $\mathbb R^n$ is also obvious:

\begin{align*} V(e_1, e_2, \ldots, e_n) = 1. \end{align*}

Interlude: Permutations

As a shortcut for computing the determinant, let me reiterate the skew-symmetry property again:

\begin{gather*} A(u, v) = -A(v, u) \end{gather*}

for $u, v \in \mathbb R^2$. This generalizes to $n$-dimensions as follows:

\begin{gather*} V(\ldots, u, \ldots, v, \ldots) = -V(\ldots, v, \ldots, u, \ldots) \end{gather*}

where the "$\ldots$" in the same position on both sides of the equation represents the same sequence of vectors. This can be proved from the alternating property and multilinearity, just like how we proved it for the 2-dimensional case. I'll just repeat the proof here (with additional "$\ldots$") for ease of reference:

\begin{align*} V(\ldots, u, \ldots, v, \ldots) & = V(\ldots, u - v, \ldots, v, \ldots) \\ & = V(\ldots, u - v, \ldots, v + u - v, \ldots) \\ & = V(\ldots, u - v, \ldots, u, \ldots) \\ & = V(\ldots, -v, \ldots, u, \ldots) \\ & = -V(\ldots, v, \ldots, u, \ldots). \end{align*}

In matrix-based terms, this means swapping columns changes the sign of the determinant. This mental shortcut can simplify the way we think about the signed volume of permuted basis vectors. For example, if we want to compute

\begin{gather*} V(e_2, e_3, e_1), \end{gather*}

we only need to find how many swaps need to be done to sort $(2, 3, 1)$. In this case, we need 2:

\begin{gather*} (2, \underline{3}, \underline{1}) \\ \Downarrow \\ (\underline{2}, \underline{1}, 3) \\ \Downarrow \\ (1, 2, 3) \end{gather*}

This is not the only way to sort $(2, 3, 1)$ though. There are other ways that use 2, 4, 6, or any even number of swaps. (For example, I can swap $1$ and $2$ twenty times before I actually try to sort it.) But there will be no way to use an odd number of swaps to achieve this! Why so? That could be a topic for another post, but for now you can read more here.

Note: It is tempting to say that we have just proved that the parity of a permutation is well-defined because the determinant is well-defined, e.g., $V(e_2, e_3, e_1)$ can have only one value; but we have not really proved that the determinant is well-defined yet in the general case. To do that, we will have to state the definition of the determinant in the form like \eqref{eq:formula} first, then use that definition to prove the two properties we stipulated. The problem is that writing out the definition of the determinant for a general $n$ will involve signs of permutations already.

As an example of how this shortcut helps in calculating the determinant, let me derive the formula for the case $n = 3$:

\begin{eqnarray*} \begin{vmatrix} a & d & g\\b & e & h\\c & f & i \end{vmatrix} & = & aei \underbrace{\begin{vmatrix} 1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1 \end{vmatrix}}_{\text{no swaps}} + ahf \underbrace{\begin{vmatrix} 1 & 0 & 0\\0 & 0 & 1\\0 & 1 & 0 \end{vmatrix}}_{\text{1 swap}} \\ & & + dbi \underbrace{\begin{vmatrix} 0 & 1 & 0\\1 & 0 & 0\\0 & 0 & 1 \end{vmatrix}}_{\text{1 swap}} + dhc \underbrace{\begin{vmatrix} 0 & 0 & 1\\1 & 0 & 0\\0 & 1 & 0 \end{vmatrix}}_{\text{2 swaps}} \\ & & + gbf \underbrace{\begin{vmatrix} 0 & 1 & 0\\0 & 0 & 1\\1 & 0 & 0 \end{vmatrix}}_{\text{2 swaps}} + gec \underbrace{\begin{vmatrix} 0 & 0 & 1\\0 & 1 & 0\\1 & 0 & 0 \end{vmatrix}}_{\text{1 swap}} \\ & = & aei - ahf - dbi + dhc + gbf - gec. \end{eqnarray*}

Note: I skipped abundant applications of the alternating property that would result in zero summands, like $adg\begin{vmatrix}1 & 1 & 1\\0 & 0 & 0\\0 & 0 & 0\end{vmatrix} = 0$.

General definition

It should be quite obvious now how to extend the definition of the determinant to an arbitrary dimension $n$. The formula, called the Leibniz formula, looks like this

\begin{gather} V(x_1, \ldots, x_n) = \sum_{\sigma \in S_n} \text{sgn}(\sigma) \prod_{i=1}^n x_{i}^{\sigma(i)} \label{eq:leibniz} \end{gather}

where each $x_i^j$ is the $j$-th coordinate of $x_i \in \mathbb R^n$, $S_n$ is the set of permutations on $\{1, \ldots, n\}$, and $\text{sgn}(\sigma)$ is the sign of $\sigma$, which is defined as $1$ if $\sigma$ is expressible as an even number of swaps, or $-1$ otherwise.

Note that by saying that $\sigma$ is a permutation, we already eliminate summands with repeated basis vectors that would appear if we expand everything using multilinearity like we did in \eqref{eq:formulafor2}. This reduces the number of summands from $n^n$ to $n!$, yet it's not good enough for practical compution. There are many algorithms for computing the determinant that is way more efficient than using \eqref{eq:leibniz} directly. The algorithmic aspect of the determinant is out of scope of this post though, so I'll stop right here.

Extra: $\det(AB) = \det(A)\det(B)$

We defined the determinant based on the base case that was

\begin{gather*} V(e_1, e_2, \ldots, e_n) = 1, \end{gather*}

where $(e_1, \ldots, e_n)$ is the standard ordered basis of $\mathbb R^n$. Obviously any ordered basis of $\mathbb R^n$ other than the standard one can be chosen, and we'll get a different volume function $V$ for each ordered basis. The number on the right-hand side does not have to be $1$ either. So, let's look a bit deeper into the Leibniz formula to see what these variations might lead to.

In the Leibniz formula shown above \eqref{eq:leibniz}, $x_i^j$ are coordinates w.r.t. the ordered basis $e = (e_1, \ldots, e_n)$, and the result was based on the choice that $V(e_1, \ldots, e_n) = 1$. Since we will vary the base case, we'll be more specific with the definition of $V$ now by adding a subscript $e$ and a superscript $\nu$ to indicate that the volume function is based on these choices:

\begin{gather} V_e^\nu(e_1, \ldots, e_n) = \nu. \label{eq:pickingbasecase} \end{gather}

Since we did not really use the fact that $e_i$'s are the standard ordered basis of $\mathbb R^n$ (we only used the fact that they form a basis), any ordered basis $b = (b_1, \ldots, b_n)$ can substitute for $e$ and everything discussed so far still applies. But what about the case $\nu \ne 1$?

Non-unit base case value

Recall that in the computation of $V$ in $\mathbb R^3$, we used multilinearity to get a summation in which each summand is of the form

\begin{gather*} (\text{some product of coordinates}) \cdot V(e_i, e_j, e_k). \end{gather*}

We simplified each term by using the alternating property to get $0$, or the swapping logic to get

\begin{gather*} \pm(\text{some product of coordinates}) \cdot V(e_1, e_2, e_3). \end{gather*}

Each summand is proportional to $V(e_1, e_2, e_3)$, hence the sum is also proportional to $V(e_1, e_2, e_3)$. If we pick $V(e_1, e_2, e_3) = \nu$ that is different from $1$, we will arrive at a slightly more general Leibniz formula:

\begin{gather} V_b^\nu(x_1, \ldots, x_n) = \left(\sum_{\sigma \in S_n} \text{sgn}(\sigma) \prod_{i=1}^n (x_i)^{\sigma(i)}_b\right) \cdot \nu \label{eq:generalleibniz} \end{gather}

where $(x_i)_b$ are the coordinate vectors of $x_i$ w.r.t. ordered basis $b = (b_1, \ldots, b_n)$, and $(x_i)_b^j$ are their components, i.e., $x_i = \sum_{j=1}^n (x_i)_b^j b_j$.

If we let $x_b$ be the matrix whose columns are $(x_i)_b$, \eqref{eq:generalleibniz} can be written as

\begin{gather} V_b^\nu(x_1, \ldots, x_n) = \det(x_b) \cdot \nu, \label{eq:generalvolume}\\ \text{ where } \det(a) = \sum_{\sigma \in S_n} \text{sgn}(\sigma)\prod_{i=1}^n a_i^{\sigma(i)}\label{eq:det} \end{gather}

for any matrix $a$ with components $a_i^j$. $\det$ here is like $V_e^1$, but it is defined as a function of coordinate values---that is $\det$ takes a matrix (a 2-dimensional array of numbers) while $V$ takes a list of vectors.

Notation regarding change of basis

Suppose $c = (c_1, \ldots, c_n)$ is another ordered basis, each $c_i$ has $(c_i)_b$ as its coordinate vector w.r.t. to $b$, and $I^c_b$ is the matrix whose columns are $(c_i)_b$. I write $I^c_b$ to hint that it is the matrix representation of the identity map w.r.t. $c$ and $b$, i.e., it takes coordinate vectors w.r.t. $c$ to coordinate vectors w.r.t. $b$. Then, for any vector $x$, we have the following relationship:

\begin{gather} x_b = I^c_b x_c. \label{eq:changeofbasis} \end{gather}

where $x_b$ is the coordinate vector of $x$ w.r.t. $b$, and $x_c$ is the coordinate vector of $x$ w.r.t. $c$.

If we have many vectors, say $x_1, \ldots, x_m$ ($m$ doesn't have to be equal to $n$), we can put the coordinate vectors $(x_i)_c$ together as columns of a matrix $x_c$, and similarly for $x_b$. Equation \eqref{eq:changeofbasis} still holds, but now $x_b$ and $x_c$ can be matrices too.

This coordinate transformation works in the reverse direction too: we get $x_c = I^b_c x_b$, where the columns of $I^b_c$ are coordinate vectors of $b_i$ w.r.t. to $c$. The matrix $I^b_c$ is the inverse matrix of $I^c_b$, as it is apparent that $I^b_c I^c_b x_c = x_c$ and $I^c_b I^b_c x_b = x_b$ for any matrices $x_b$ and $x_c$. We won't need this in this post though.

Equivalent volume functions

If we have two ordered bases $b$ and $e$, is it possible to define $V_b^\nu$ such that

\begin{gather} V_b^\nu(x_1, \ldots, x_n) = V_e^1(x_1, \ldots, x_n) \text{ ?} \label{eq:equalvolumes} \end{gather}

Let's try to do this. Start by plugging in the special value for $x_i$: picking $x_i = b_i$ gives us

\begin{gather*} V_b^\nu(b_1, \ldots, b_n) = \nu \end{gather*}

by the definition of $V_b^\nu$ \eqref{eq:pickingbasecase}. Now, \eqref{eq:generalvolume} gives us

\begin{gather*} V_e^1(b_1, \ldots, b_n) = \det(I^b_e), \end{gather*}

which can be substituted into \eqref{eq:equalvolumes} to yield

\begin{gather*} \nu = \det(I^b_e). \end{gather*}

This is indeed an intuitive value that we could have guessed because if we use $V_e^1$ as the base definition of volume, we would expect the parallelepiped whose sides are $b_i$ to have volume $\det(I^b_e)$. (Recall columns of $I^b_e$ are coordinate vectors of $b_i$ w.r.t. $e$.) Now that we know what the base value $\nu$ should be, $V_b^\nu$ is completely determined. We are left to show that \eqref{eq:equalvolumes} holds.

A common method for a formal proof of this kind is to compute the difference: let $D = V_e^1 - V_b^\nu$. Then,

\begin{gather} D(b_1, \ldots, b_n) = \nu - \nu = 0. \label{eq:basecaseofdifference} \end{gather}

From $D = V_e^1 - V_b^\nu$, it is straightforward to see that $D$ is multilinear and alternating. Equation \eqref{eq:basecaseofdifference} gives the base case of $D$, which coincides with the base case of $V_b^0$, thus $D = V_b^0$. Since $V_e^1 - V_b^\nu = D = V_b^0 = 0$, we have proved that $V_e^1 = V_b^\nu$.

Consequences

The consequence of \eqref{eq:equalvolumes} is that the choice $\nu = \det(I^b_e)$ makes

\begin{align} V_e^1(x_1, \ldots, x_n) & = V_b^{\nu}(x_1, \ldots, x_n) \nonumber\\ & = \det(x_b) \cdot \nu & & \text{; from \eqref{eq:generalvolume}}\nonumber\\ & = \det(x_b) \cdot \det(I^b_e). \label{eq:volumeinsecondbasis} \end{align}

However, we can also apply \eqref{eq:generalvolume} to $V_e^1$ directly:

\begin{align} V_e^1(x_1, \ldots, x_n) & = \det(x_e) \nonumber\\ & = \det(I^b_e x_b). \label{eq:volumeinfirstbasis} \end{align}

We now have two expressions that are equal to $V_e^1(x_1, \ldots, x_n)$: \eqref{eq:volumeinsecondbasis} and \eqref{eq:volumeinfirstbasis}. Equating them gives us the main result we are after: multiplicativity of $\det$!

\begin{align} \det(I^b_e x_b) & = \det(I^b_e) \cdot \det(x_b). \label{eq:multiplicativityofdet} \end{align}

Although this equation seems to restrict the matrices to be of the forms $I^b_e$ and $x_b$, it is in fact a general formula because both $I^b_e$ and $x_b$ can be chosen arbitrarily.

Note: Equation \eqref{eq:multiplicativityofdet} is a formula about the $\det$ function itself, not the signed volume $V$. The multiplicative property wouldn't even make sense for $V$ because $V$ takes a list of vectors, not a matrix.