# Proving the Leibniz formula for determinants

tfintI'm interested in the discovery of the determinant: Say I want to find the general solution of the system \begin{aligned} &a_{11} x_{1}+a_{12} x_{2}+a_{13} x_{3}=b_{1} \\ &a_{21} x_{1}+a_{22} x_{2}+a_{23} x_{3}=b_{2} \\ &a_{31} x_{1}+a_{32} x_{2}+a_{33} x_{3}=b_{3} \end{aligned} given that the coefficients are such that one doesn't divide by zero when solving it. One quickly finds that the denominators of the general solution are given by $$a_{11} a_{22} a_{33}-a_{12} a_{21} a_{33}-a_{11} a_{23} a_{32}-a_{13} a_{22} a_{31}+a_{12} a_{23} a_{31}+a_{13} a_{32} a_{21} .$$ Having already dealt with the simple $$2\times 2$$ case, one might conjecture a formula (as the one Leibniz developed). How would one proof by induction that the denominators of an $$n \times n$$ system of equations is given by the Leibniz formula?

My approach would be to eliminate all coefficients $$a_{k1}$$ for $$k \in \{2, 3, \ldots, n\}$$ and then apply the formula to the $$(n-1)\times (n-1)$$ submatrix of the transformed matrix, neglecting the first row and column. Does this approach sound reasonnable?

Proof: I want to inductively proof that the denominator of the general solutions of an $$n \times n$$ system is given by the Leibnitz formula for the determinant \begin{align*} \sum_{\sigma \in S_{n}}^{} \text{sgn}\left(\sigma\right) \prod_{i = 1}^{n} a_{i, \sigma(i)} .\end{align*}

For the induction step, suppose our matrix $$\mathbf{A}\in \mathbb{R}^{n+1, n+1}$$ is given by $$( \mathbf{A})_{i, j} = a_{i, j}$$. My idea is now to eliminate the first column of the matrix (apart from $$(\mathbf{A})_{1, 1}$$) so that after this transformation its entries would look like \begin{align*} \overline{a}_{i, j} = a_{i, j} - \frac{a_{1, j}\cdot a_{i, 1}}{a_{1, 1}} = \frac{a_{i, j}\cdot a_{1, 1} - a_{1, j}\cdot a_{i, 1}}{a_{1, 1}}, \quad i, j > 1 .\end{align*} Now we can consider the $$(n\times n)$$ submatrix $$\widetilde{\mathbf{A}}$$ with \begin{align*} (\widetilde{\mathbf{A}})_{i, j} =\overline{a}_{i+1, j+1} = \frac{a_{i+1, j+1}\cdot a_{1, 1} - a_{1, j+1}\cdot a_{i+1, 1}}{a_{1, 1}} .\end{align*} To this matrix we can apply the formula we assume to be true, so what I want to show is \begin{align*} a_{1, 1}\sum_{\sigma \in S_{n}}^{} \text{sgn}\left(\sigma\right) \prod_{i = 1}^{n} (\widetilde{\mathbf{A}})_{i, \sigma(i)} = \sum_{\sigma \in S_{n+1}}^{} \text{sgn}\left(\sigma\right) \prod_{i = 1}^{n+1} a_{i, \sigma(i)} .\end{align*} (The additional factor $$a_{1, 1}$$ comes from the fact that for getting the denominator of the general solution we have to bring it to the other site and there it will cancel out, but this we don't consider when applying the formula directly to the submatrix. In simple terms we just exploit the block matrix properties of the determinant). However, I didn't come very far in my proof \begin{align*} a_{1, 1}\sum_{\sigma \in S_{n}}^{} \text{sgn}\left(\sigma\right) \prod_{i = 1}^{n} (\widetilde{\mathbf{A}})_{i, \sigma(i)} &= a_{1, 1}\sum_{\sigma \in S_{n}}^{} \text{sgn}\left(\sigma\right) \prod_{i = 1}^{n} \left( \frac{a_{i+1, \sigma(i)+1}\cdot a_{1, 1}- a_{1, \sigma(i)+1}\cdot a_{i+1, 1}}{a_{1, 1}} \right) \\[5pt] &= \sum_{\sigma\in S_{n}}^{}\frac{\text{sgn}\left(\sigma\right)}{a_{1, 1}^{n-1}} \prod_{i = 1}^{n} \left( a_{i+1, \sigma(i)+1}\cdot a_{1, 1}- a_{1, \sigma(i)+1}\cdot a_{i+1, 1} \right) \end{align*}

The proof that I know thought about is somehow algorithmic, and its correctness follows immediatly from its construction. The idea is to partition the sum \begin{align*} \sum_{\sigma\in S_{n+1}}^{} \text{sgn}\left(\sigma\right)\prod_{i=1}^{n+1} a_{i, \sigma(i)} \end{align*} into partial sums, of which the $$k$$th covers all permutations with $$\sigma(k) = 1$$, i.e. \begin{align*} \sum_{\sigma\in S_{n+1}}^{} \text{sgn}\left(\sigma\right)\prod_{i=1}^{n+1} a_{i, \sigma(i)} = \sum_{\substack{\sigma \in S_{n+1}, \\ \sigma(1) = 1 }}^{} \text{sgn}\left(\sigma\right)\prod_{i=1}^{n+1} a_{i, \sigma(i)} + \cdots + \sum_{\substack{\sigma \in S_{n+1}, \\ \sigma(n+1) = 1 }}^{} \text{sgn}\left(\sigma\right)\prod_{i=1}^{n+1} a_{i, \sigma(i)} .\end{align*} For the sake of illustration I will denote $$\sigma \in S_{n+1}$$ with $$\sigma(k) = 1$$ by $$\sigma_{k}$$ Moreover, we define $$P_{n+1}$$ to be the set of permutations of the sequence $$(2, 3, \ldots, n+1)$$. It remains to identify each of these partial sums in the sum \begin{align*} &\frac{1}{{a_{1, 1}^{n-1}}}\sum_{\sigma\in S_{n}}^{}{\text{sgn}\left(\sigma\right)} \prod_{i = 1}^{n} \left( a_{i+1, \sigma(i)+1}\cdot a_{1, 1}- a_{1, \sigma(i)+1}\cdot a_{i+1, 1} \right) \\ = &\frac{1}{a_{1, 1}^{n-1}}\sum_{\pi \in P_{n+1}}^{} \text{sgn}\left(\pi\right) \prod_{i = 1}^{n} (a_{i+1, \pi(i)} \cdot a_{1, 1} - a_{1, \pi(i)}\cdot a_{i+1, 1}) .\end{align*} The first we can get by only taking the first term in the product when multiplying (since the product will split into a large sum), which results in \begin{align*} \frac{1}{a_{1, 1}^{n-1}}\prod_{i = 1}^{n} (a_{i+1,\pi(i)}\cdot a_{1, 1}) = a_{1, 1}\prod_{i = 1}^{n} a_{i+1, \pi(i)} .\end{align*} This product now exactly equals \begin{align*} \prod_{i = 1}^{n+1} a_{i, \sigma_{1}(i)} \end{align*} whereby we simply append a $$1$$ before every permutation of $$P_{n+1}$$, which does not change $$\text{sgn}\left(\pi\right)$$. In a similar way we can obtain the second partial sum, namely by first selecting the second summand and afterwards only the first, i.e. \begin{align*} -\frac{1}{a_{1, 1}^{n-1}} \cdot (a_{1, \pi(1)} \cdot a_{2, 1}) \prod_{i = 2}^{n} (a_{i+1, \pi(i)} \cdot a_{1, 1}) = -\prod_{i = 1}^{n+1} a_{i, \sigma _{2}(i)} .\end{align*} The factor of $$-1$$ is consistent as inserting the $$1$$ between the first and second element of $$\pi$$ causes a change in parity of our permutation. In exactly this manner we obtain all other partial sums, namely the $$k$$th by choosing first $$k-2$$ times the first term, then once the second term and afterwards only the first term. This will always incur an additional factor of $$-1$$ to account for the change in parity. However, there needs to be done some reassignment of permutations, which we can observe for the kase $$k = 3$$: \begin{align*} -a_{2, \pi(1)} \cdot (a_{1, \pi(2)} \cdot a_{3, 1}) \prod_{i = 3}^{n} a_{i+1, \pi(i)} = -a_{2, \sigma_{3}(1)} \cdot (a_{1, \sigma_{3}(2)} \cdot a_{3, \sigma_{3}(3)}) \prod_{i = 4}^{n+1} a_{i, \sigma_{3} (i)} .\end{align*} As one can observe, the indice up to the point where we inserted the $$1$$ in the permutation do not match the desired form. Yet, this can be fixed by "reassigning" permutations, which means in the above example that we look at $$\pi(1)$$ and $$\pi(2)$$ respectively (which should be $$\pi(2)$$ and $$\pi(1)$$ respectively) and find the permutation where those values are switched, which will be our new permutation $$\sigma_{3}$$ after having inserted the additional $$1$$ at position $$3$$. Quickly thinking about this one will recognise that it only involves bijections between sets of permutations. With this procedure we can "build up" all the partial sums.

Lastly, we have to show that the remaining terms, i.e. those that have a denominator $$a_{1, 1}^{j}$$ with $$j > 0$$, cancel out. Lets start with an example, namely choosing only the second term in the product, which yields \begin{align*} \frac{\text{sgn}\left(\pi\right) (-1)^{n}}{a_{1, 1}^{n-1}} \prod_{i = 1}^{n} (a_{1, \pi(i)}\cdot a_{i+1, 1}) .\end{align*} We now have to find a permutation that has the opposite signature but produces the same product (up to the opposite sign of course). For the above we find that any permutation with opposite signature will yield the desired result, and since the number of permutations with odd and even signature respectively in $$S_{n}$$ are equal, everything cancels out as expected.

Next, when taking the first term $$1 \le \alpha \le n-2$$ times at positions $$j_{1}< j_{2}< \ldots< j_{\alpha}$$, meaning our resulting product takes the form \begin{align*} \frac{\text{sgn}\left(\pi\right)(-1)^{n - \alpha}}{a_{1, 1}^{n -1- \alpha}} &\left(\prod_{i = 1}^{j_{1} - 1} (a_{1, \pi(i)} a_{i+1, 1})\right) \cdot a_{j_{1} + 1, \pi(j_{1})} \cdots \\[5pt] &\left(\prod_{i = j_{\alpha - 1}+ 1}^{j_{\alpha} - 1} (a_{1, \pi(i)} \cdot a_{i+1, 1})\right) a_{j_{\alpha} + 1, \pi(j_{\alpha})} \left(\prod_{i = j_{\alpha }+ 1}^{n} (a_{1, \pi(i)} \cdot a_{i+1, 1})\right) .\end{align*} In the entire sum we have $$n!$$ such terms with the same choice of $$j_{k}$$ (one for each permutation), of which $$(n - \alpha)!$$ are identical up to a sign, since inspecting the above product we find that one can only achieve an identical product up to a sign when finding permutations $$\pi_{k}$$ that satisfy $$\pi_{k}(j_{1}) = \pi(j_{1}), \; \ldots, \; \pi_{k}(j_{\alpha}) = \pi(j_{\alpha}).$$ Using again the fact that the number of even and odd permutations in this scenario is equal concludes the proof.

• The Leibniz formula is more by definition of alternating multilinear than by proof, but a derivation can be found here: en.wikipedia.org/wiki/Leibniz_formula_for_determinants Commented Jan 1, 2022 at 22:48
• It is the Laplace expansion formula which is harder to prove Commented Jan 1, 2022 at 22:48
• @FShrike yes, but you don't get what I mean (I know the proof from a functional analysis point of view). Leibniz derived his formula in the way I mention above, and for me it's natural to understand this way in order to better understand the determinant. Commented Jan 1, 2022 at 22:51
• Leibnitz, like all mathematicians at that time was a virtuoso algebra computer : therefore, he must have done the computations by has in the $3 \times 3$ case, the $4 \times 4$ case, even the the $5 \times 5$ case (But I don't think it is him who discovered the general case). Commented Jan 1, 2022 at 23:44
• @JeanMarie do you happen to know the source of an inductive proof of the formula, i.e. for the denominators of the general solution of a $n\times n$ matrix? Commented Jan 2, 2022 at 13:54

To your question "is this method sounding reasonable ?", the answer is "yes".

Indeed, in the framework of a recursive proof where you make the assumption that the Cramer formulas $$x_1=\det/\det, x_2=\det/\det...$$ are valid till dimension $$n$$; let us prove that it's possible to jump to the validity in dimension $$n+1$$.

For the sake of simplicity, we will take here $$n=3$$.

Let us consider a generic $$4 \times 4$$ linear system with the fourth unknown placed in the RHSides (we have taken the fourth one, but it could have been any of the unknowns):

$$\begin{cases} &a_{11} x_{1}+a_{12} x_{2}+a_{13} x_{3}&=&b_{1}\color{red}{-a_{14}x_4} \\ &a_{21} x_{1}+a_{22} x_{2}+a_{23} x_{3}&=&b_{2}\color{red}{-a_{24}x_4} \\ &a_{31} x_{1}+a_{32} x_{2}+a_{33} x_{3}&=&b_{3}\color{red}{-a_{34}x_4} \\ &\color{red}{a_{31} x_{1}+a_{32} x_{2}+a_{33} x_{3}}&=&\color{red}{b_4-a_{44}x_4} \end{cases} \tag{1}$$

The recursion assumption allows us to solve the parametric system made of the first three equations of (1) in the following way:

$$x_1=\dfrac{1}{D}\begin{vmatrix}\color{red}{(b_1-a_{14}x_4)}&a_{12}&a_{13}\\ \color{red}{(b_2-a_{24}x_4)}&a_{22}&a_{23}\\ \color{red}{(b_3-a_{34}x_4)}&a_{32}&a_{33}\end{vmatrix} \text{where} \ D=\begin{vmatrix}a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33}\end{vmatrix}\tag{2}$$

and similar relationships for $$x_2$$ and $$x_3$$.

The idea now is to plug these expressions of $$x_1,x_2,x_3$$ into the 4th equation of (1) that hasn't yet been used, and extract $$x_4$$. By expanding the determinants, for example the expression in (2) into:

$$\begin{vmatrix}b_1-a_{14}x_4&a_{12}&a_{13}\\ b_2-a_{24}x_4&a_{22}&a_{23}\\ b_3-a_{34}x_4&a_{32}&a_{33}\end{vmatrix}=\begin{vmatrix}b_1&a_{12}&a_{13}\\ b_2&a_{22}&a_{23}\\ b_3&a_{32}&a_{33}\end{vmatrix}-x_4\begin{vmatrix}a_{14}&a_{12}&a_{13}\\ a_{24}&a_{22}&a_{23}\\ a_{34}&a_{32}&a_{33}\end{vmatrix}$$

the fourth equation in (1) will become an expression of the form:

$$\frac{1}{D}(a_{31}(P-x_4Q)+a_{32}(R-x_4S)+a_{33}(T-x_4U)=b_4-a_{44}x_4$$

where $$P,Q,R,S,T,U$$ are $$3 \times 3$$ determinants.

If relationship (3) is multiplied by $$D$$ and $$x_4$$ extracted as a quotient $$x_4=\frac{N'}{D'}$$, one recognizes in its numerator and denominator 2 Laplace expansions of $$4 \times 4$$ determinants : you have therefore established the Cramer/Leibniz formula $$x_4=\det(...)/\det(...)$$ for dimension $$4$$.

What about the other variables ? They follow evidently the same rules (it could have been for example $$x_1$$ "sent" in the RHSides instead of $$x_4$$)...

• Thanks a lot, this approach is much better than mine! Commented Feb 17, 2022 at 14:51
• could you still help me with my induction proof? Commented Feb 17, 2022 at 21:20
• Yes, but in which way ? Commented Feb 17, 2022 at 21:23
• In the above attempt I came to a point in my inductive proof where I don't know how to proceed, so I would like to know how I can transform this expression so that I arrive at the correct induction assumption. How do I get from $$\sum_{\sigma\in S_{n}}^{}\frac{\text{sgn}\left(\sigma\right)}{a_{1, 1}^{n-1}} \prod_{i = 1}^{n} \left( a_{i+1, \sigma(i)+1}\cdot a_{1, 1}- a_{1, \sigma(i)+1}\cdot a_{i+1, 1} \right)$$ to $$\sum_{\sigma \in S_{n+1}}^{} \text{sgn}\left(\sigma\right) \prod_{i = 1}^{n+1} a_{i, \sigma(i)}$$ Commented Feb 17, 2022 at 21:31
• I have a hard time trying to say pertinent things. I have been looking at proofwiki.org/wiki/Cauchy-Binet_Formula/… but it is useless here... Commented Feb 18, 2022 at 8:02