Proof of Cofactor Expansion of the Determinant along a Row or a Column, and the Adjugate Identity

Each result cited is universally quantified over the data in its own statement.

Conventions. We work in Routine Facts Used Without Citation: Arithmetic, Maps, Predicativity and Indices. For finite sets $D$ and $E$ and $\ast$ the addition or the multiplication of $\mathbb{R}$ we use: (O1) if $g:E\to D$ is a bijection and $f:D\to\mathbb{R}$, then $\mathop{\ast}\limits_{y\in E}f(g(y))=\mathop{\ast}\limits_{x\in D}f(x)$, by Iterated Operations over Finite Sets: Singletons, Disjoint Unions, Reindexing, Products of Sets, Termwise Combination, Homomorphisms and Intervals §reindexing; (O2) if $D\cap E=\emptyset$ and $f:D\cup E\to\mathbb{R}$, then $\mathop{\ast}\limits_{x\in D\cup E}f(x)=\bigl(\mathop{\ast}\limits_{x\in D}f(x)\bigr)\ast\bigl(\mathop{\ast}\limits_{x\in E}f(x)\bigr)$, by Iterated Operations over Finite Sets: Singletons, Disjoint Unions, Reindexing, Products of Sets, Termwise Combination, Homomorphisms and Intervals §disjoint-union, and $\mathop{\ast}\limits_{x\in\{y\}}f(x)=f(y)$ by Iterated Operations over Finite Sets: Singletons, Disjoint Unions, Reindexing, Products of Sets, Termwise Combination, Homomorphisms and Intervals §singleton; (O3) for $h:D\times E\to\mathbb{R}$, $\sum_{(x,y)\in D\times E}h(x,y)=\sum_{x\in D}\sum_{y\in E}h(x,y)$, by Iterated Operations over Finite Sets: Singletons, Disjoint Unions, Reindexing, Products of Sets, Termwise Combination, Homomorphisms and Intervals §product; (O4) for $f:D\to\mathbb{R}$ and $c\in\mathbb{R}$, $c\sum_{x\in D}f(x)=\sum_{x\in D}c\,f(x)$, by Sums over Finite Sets in a Commutative Ring and in an Ordered Field: Distributivity, Products of Sums, Vanishing Terms, Sums over Pairs, Expanding Products of Sums, Counting, Comparison, Monotonicity and the Triangle Inequality §distributive. For $k\in\mathbb{N}$ the set $S_{k}$ of permutations of $[k]$ is finite by Finite Sets: Removing an Element, Injective Self-Maps, Infinite Sets, Sets of Permutations, Pairs and Extended Tuples §permutations, and subsets and Cartesian products of finite sets are finite by Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice §subset and Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice §union. Entries of products, transposes and scalar multiples of matrices are given by Product of Real Matrices §product, Transpose of a Real Matrix §transpose and Real Matrices: the Set of Real m x n Matrices with Their Entrywise Operations §operations, two matrices of the same size are equal when all their entries agree by Real Matrices: the Set of Real m x n Matrices with Their Entrywise Operations §equality, and $\delta_{ik}=(I_{n})_{ik}$, so that $\delta_{ii}=1$ and $\delta_{ik}=0$ for $i\neq k$, by Identity Matrix and Standard Basis Vectors §identity. For $k\in\mathbb{N}$ and $Y\in\mathcal{M}_{k}(\mathbb{R})$, $\det Y=\sum_{\sigma\in S_{k}}t_{Y}(\sigma)$ with $t_{Y}(\sigma)=\operatorname{sgn}(\sigma)\prod_{p\in[k]}Y_{p\sigma(p)}$, by Determinant of a Real Square Matrix §determinant, The Determinant Map and the Matrices Formed from a Square Matrix by Reindexing Rows, Replacing a Row and Deleting a Row and a Column Exist and Are Unique §determinant and The Determinant Map and the Matrices Formed from a Square Matrix by Reindexing Rows, Replacing a Row and Deleting a Row and a Column Exist and Are Unique §terms with $k$ in place of $n$.

Step 1: the maps $s_{j}$ and the bijection $\Phi$. Let $m=n-1$; then $m\in\mathbb{N}$ by The Index Map Skipping One Element of [n] §difference. For $j\in[n]$, $s_{j}:[m]\to[n]$ is the deletion index map, as in The Determinant Map and the Matrices Formed from a Square Matrix by Reindexing Rows, Replacing a Row and Deleting a Row and a Column Exist and Are Unique §deletion. By The Index Map Skipping One Element of [n] §range with $j$ in place of $i$: (D) for every $j\in[n]$ and $k\in[m]$, $s_{j}(k)\neq j$, and $s_{j}$, as a map $s_{j}:[m]\to[n]\setminus\{j\}$, is a bijection. Fix $i\in[n]$, and let $\Phi:[n]\times S_{m}\to S_{n}$ be the map $\Phi_{i}$ of Extending a Permutation of [n-1] to a Permutation of [n] with a Prescribed Value, and Its Sign §map, which is a bijection by Extending a Permutation of [n-1] to a Permutation of [n] with a Prescribed Value, and Its Sign §bijection. Thus, for all $j\in[n]$, $\tau\in S_{m}$ and $k\in[m]$, $\Phi(j,\tau)(i)=j$ and (E) $\Phi(j,\tau)(s_{i}(k))=s_{j}(\tau(k))$; and, by Extending a Permutation of [n-1] to a Permutation of [n] with a Prescribed Value, and Its Sign §sign, for all $j\in[n]$ and $\tau\in S_{m}$,

$$\operatorname{sgn}(\Phi(j,\tau))=(-1)^{i+j}\operatorname{sgn}(\tau).\tag{1}$$

Step 2: the product of a term. Let $Y\in\mathcal{M}_{n}(\mathbb{R})$, $j\in[n]$, $\tau\in S_{m}$ and $\sigma=\Phi(j,\tau)$. We claim

$$\prod_{p\in[n]}Y_{p\sigma(p)}=Y_{ij}\prod_{k\in[m]}(Y^{(i,j)})_{k\tau(k)}.\tag{2}$$

As $[n]$ is the disjoint union of $\{i\}$ and $[n]\setminus\{i\}$, (O2) gives $\prod_{p\in[n]}Y_{p\sigma(p)}=Y_{i\sigma(i)}\prod_{p\in[n]\setminus\{i\}}Y_{p\sigma(p)}$, and $Y_{i\sigma(i)}=Y_{ij}$. By (O1) with the bijection $s_{i}:[m]\to[n]\setminus\{i\}$ of (D), by (E) and by The Determinant Map and the Matrices Formed from a Square Matrix by Reindexing Rows, Replacing a Row and Deleting a Row and a Column Exist and Are Unique §deletion with $Y$ in place of $A$,

$$\prod_{p\in[n]\setminus\{i\}}Y_{p\sigma(p)}=\prod_{k\in[m]}Y_{s_{i}(k)\sigma(s_{i}(k))}=\prod_{k\in[m]}Y_{s_{i}(k)s_{j}(\tau(k))}=\prod_{k\in[m]}(Y^{(i,j)})_{k\tau(k)}.$$

Step 3: expansion along a row. Let $Y\in\mathcal{M}_{n}(\mathbb{R})$ and $i\in[n]$, with $\Phi$ as in Step 1 for this $i$. The set $[n]\times S_{m}$ is finite. By (O1) with the bijection $\Phi$, (O3), (1), (2) and (O4), and with $t_{Y^{(i,j)}}$ the map of The Determinant Map and the Matrices Formed from a Square Matrix by Reindexing Rows, Replacing a Row and Deleting a Row and a Column Exist and Are Unique §terms for $Y^{(i,j)}\in\mathcal{M}_{m}(\mathbb{R})$ and $m$ in place of $n$,

$$\det Y=\sum_{\sigma\in S_{n}}t_{Y}(\sigma)=\sum_{(j,\tau)\in[n]\times S_{m}}t_{Y}(\Phi(j,\tau))=\sum_{j\in[n]}\sum_{\tau\in S_{m}}Y_{ij}(-1)^{i+j}\,t_{Y^{(i,j)}}(\tau)=\sum_{j\in[n]}Y_{ij}(-1)^{i+j}\det Y^{(i,j)}.$$

By Minor, Cofactor, and Adjugate of a Real Square Matrix §minor and Minor, Cofactor, and Adjugate of a Real Square Matrix §cofactor with $Y$ in place of $A$, $(-1)^{i+j}\det Y^{(i,j)}=C_{ij}(Y)$, so

$$\det Y=\sum_{j=1}^{n}Y_{ij}C_{ij}(Y)\qquad(Y\in\mathcal{M}_{n}(\mathbb{R}),\ i\in[n]).\tag{3}$$

Clause row is (3) with $Y=A$.

Step 4: transposes. Let $Y\in\mathcal{M}_{n}(\mathbb{R})$ and $i,j\in[n]$. For $k,l\in[m]$, by The Determinant Map and the Matrices Formed from a Square Matrix by Reindexing Rows, Replacing a Row and Deleting a Row and a Column Exist and Are Unique §deletion applied to $Y^{\top}$ and to $Y$,

$$((Y^{\top})^{(j,i)})_{kl}=(Y^{\top})_{s_{j}(k)s_{i}(l)}=Y_{s_{i}(l)s_{j}(k)}=(Y^{(i,j)})_{lk}=((Y^{(i,j)})^{\top})_{kl},$$

so $(Y^{\top})^{(j,i)}=(Y^{(i,j)})^{\top}$. By Rules for the Determinant: Rows, Scalar Multiples and the Transpose §transpose with $m$ in place of $n$, $\det(Y^{\top})^{(j,i)}=\det Y^{(i,j)}$, so, by Minor, Cofactor, and Adjugate of a Real Square Matrix §minor and Minor, Cofactor, and Adjugate of a Real Square Matrix §cofactor, applied to $Y^{\top}$ with $j,i$ in place of $i,j$ and to $Y$,

$$C_{ji}(Y^{\top})=(-1)^{j+i}\det(Y^{\top})^{(j,i)}=(-1)^{i+j}\det Y^{(i,j)}=C_{ij}(Y).\tag{4}$$

Consequently, by Minor, Cofactor, and Adjugate of a Real Square Matrix §adjugate, $\operatorname{adj}(Y^{\top})_{ij}=C_{ji}(Y^{\top})=C_{ij}(Y)=\operatorname{adj}(Y)_{ji}=(\operatorname{adj}(Y)^{\top})_{ij}$ for all $i,j\in[n]$, so

$$\operatorname{adj}(Y^{\top})=\operatorname{adj}(Y)^{\top}.\tag{5}$$

Clause column. Let $j\in[n]$. By Rules for the Determinant: Rows, Scalar Multiples and the Transpose §transpose, (3) for $Y=A^{\top}$ and row $j$, and (4),

$$\det A=\det(A^{\top})=\sum_{i=1}^{n}(A^{\top})_{ji}C_{ji}(A^{\top})=\sum_{i=1}^{n}A_{ij}C_{ij}(A).$$

Step 5: $Y\operatorname{adj}(Y)=(\det Y)I_{n}$ for every $Y\in\mathcal{M}_{n}(\mathbb{R})$. Let $i,k\in[n]$. By Minor, Cofactor, and Adjugate of a Real Square Matrix §adjugate,

$$(Y\operatorname{adj}(Y))_{ik}=\sum_{l=1}^{n}Y_{il}\operatorname{adj}(Y)_{lk}=\sum_{l=1}^{n}Y_{il}C_{kl}(Y).$$

If $k=i$, this is $\det Y$ by (3). Let $k\neq i$, let $r$ be the point of $\mathbb{R}^{n}$ with $r_{l}=Y_{il}$ for $l\in[n]$, and let $Y'=Y[k\leftarrow r]\in\mathcal{M}_{n}(\mathbb{R})$ be the matrix of The Determinant Map and the Matrices Formed from a Square Matrix by Reindexing Rows, Replacing a Row and Deleting a Row and a Column Exist and Are Unique §row-replacement with $Y$ in place of $A$ and $k$ in place of $i$. For $l\in[n]$, $Y'_{kl}=r_{l}=Y_{il}$ and, as $i\neq k$, $Y'_{il}=Y_{il}$; so rows $i$ and $k$ of $Y'$ agree and $\det Y'=0$ by Rules for the Determinant: Rows, Scalar Multiples and the Transpose §equal-rows. For $l\in[n]$ and $p,q\in[m]$, $s_{k}(p)\neq k$ by (D), so $(Y'^{(k,l)})_{pq}=Y'_{s_{k}(p)s_{l}(q)}=Y_{s_{k}(p)s_{l}(q)}=(Y^{(k,l)})_{pq}$; hence $Y'^{(k,l)}=Y^{(k,l)}$ and $C_{kl}(Y')=C_{kl}(Y)$, by Minor, Cofactor, and Adjugate of a Real Square Matrix §minor and Minor, Cofactor, and Adjugate of a Real Square Matrix §cofactor, applied to $Y'$ and to $Y$ with $k,l$ in place of $i,j$. By (3) for $Y'$ and row $k$,

$$0=\det Y'=\sum_{l=1}^{n}Y'_{kl}C_{kl}(Y')=\sum_{l=1}^{n}Y_{il}C_{kl}(Y).$$

So $(Y\operatorname{adj}(Y))_{ik}=\det Y\cdot\delta_{ik}=((\det Y)I_{n})_{ik}$ for all $i,k\in[n]$, and $Y\operatorname{adj}(Y)=(\det Y)I_{n}$.

Clause adjugate-identity. Step 5 with $Y=A$ gives $A\operatorname{adj}(A)=(\det A)I_{n}$. Step 5 with $Y=A^{\top}$, (5) and Rules for the Determinant: Rows, Scalar Multiples and the Transpose §transpose give $A^{\top}\operatorname{adj}(A)^{\top}=(\det A)I_{n}$. Taking transposes, by Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes §transpose,

$$\operatorname{adj}(A)A=(\operatorname{adj}(A)^{\top})^{\top}(A^{\top})^{\top}=\bigl(A^{\top}\operatorname{adj}(A)^{\top}\bigr)^{\top}=\bigl((\det A)I_{n}\bigr)^{\top}=(\det A)I_{n}^{\top}=(\det A)I_{n}.$$

Hence $A\operatorname{adj}(A)=\operatorname{adj}(A)A=(\det A)I_{n}$.