Show that $\det(\mathrm{adj}(A)) = \det(A)^3$ for a $(4 \times 4)$-matrix $A$

Let $A_{ik}$ be the cofactor of $a_{ik}$ in the determinant $$d = \left| \begin{matrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \\ a_{41} & a_{42} & a_{43} & a_{44} \\ \end{matrix} \right|.$$ Let $D$ be the corresponding determinant with $a_{ik}$ replaced by $A_{ik}.$ Prove $D = d^3.$

Book solution

Let $\alpha$ be the matrix of the given determinant with elements $a_{ik}$ and let $\beta$ be the matrix of the cofactors $A_{ik},$ and let $\gamma$ be the transpose of $\beta.$ Then the product matrix $\alpha \gamma$ is a diagonal matrix with all entries on the main diagonal equal to $d$.

Thus, $\det{\alpha \gamma} = d^4 = (\det{\alpha}) (\det{\gamma}) = (\det{\alpha})(\det{\beta}) = dD.$ The equation $$dD = d^4 \tag{1}$$ is an identity between polynomials in the $16$ matrix entries regard as independent determinants. Since there certainly exists a $4 \times 4$ matrix whose determinant is not zero, $d$ is non-zero in the polynomial ring. Since the polynomial ring is an integral domain, the result $$D = d^3$$ follows from $(1)$.

My questions

How is $\alpha \gamma$ a diagonal matrix with all entries on the main diagonal equal to $d$? Does this come from some linear algebra rule? Then how did they get $(\det{\alpha}) (\det{\gamma}) = (\det{\alpha})(\det{\beta}) = dD$? Does $AA^{T} = AA$? Finally I am not really understanding the last paragraph so if someone could explain that that would help.

General Solution

Let $K$ be our ground field and $L = K(x_{11}, \dotsc, x_{44})$ the field of rational functions in the 16 variables $x_{ij}$ with $1 \leq i,j \leq 4$. We will work over $L$ instead of $K$ from the very start, to avoid confusion and the taste of handwaving at later points.

We slightly change the definitions, to make everything work, but still see how this applies to the exercise: Let $\alpha \in \mathrm{M}(4 \times 4, L)$ be defined as $$\alpha = \begin{pmatrix} x_{11} & x_{12} & x_{13} & x_{14} \\ x_{21} & x_{22} & x_{23} & x_{24} \\ x_{31} & x_{32} & x_{33} & x_{34} \\ x_{41} & x_{42} & x_{43} & x_{44} \\ \end{pmatrix}.$$ Let $\beta$ be the adjugate matrix of $\alpha$, i.e. the $(i,j)$-the entry of $\beta$ is the $(i,j)$-th cofactor of $\alpha$, and let $\gamma = \beta^T$. Let $d = \det \alpha$ and $D = \det \beta$. Notice that by Leibniz’ formula the entries of $d$ and $D$ are polynomials in the variables $x_{11}, \dotsc, x_{44}$. For example \begin{align*} d &= \left| \begin{matrix} x_{11} & x_{12} & x_{13} & x_{14} \\ x_{21} & x_{22} & x_{23} & x_{24} \\ x_{31} & x_{32} & x_{33} & x_{34} \\ x_{41} & x_{42} & x_{43} & x_{44} \\ \end{matrix} \right| \\ &= \sum_{\sigma \in S_4} \mathrm{sgn}(\sigma) x_{1\sigma(1)} x_{2\sigma(2)} x_{3\sigma(3)} x_{4\sigma(4)}. \end{align*} is a polynomial of degree $4$ (notice that the monomials $x_{1\sigma(1)} x_{2\sigma(2)} x_{3\sigma(3)} x_{4\sigma(4)}$ with $\sigma \in S_4$ are pairwise different and thus cannot cancel out). In particular $d \neq 0$.

First notice that $$(\alpha \gamma)_{ik} = \left( \alpha \beta^T \right)_{ik} = \sum_{j=1}^4 \alpha_{ij} \beta^T_{jk} = \sum_{j=1}^4 \alpha_{ij} \beta_{kj}.$$ For $i = k$ this is just the expansion of $\alpha \gamma$ by its $k$-th row. Thus the diagonal entries of $\alpha \gamma$ are $\det \alpha = d$.

For $i \neq k$ consider the matrix $\alpha^{(k)}$, where we replace the $k$-th row of $\alpha$ by its $i$-th row. Then the $i$-th and $k$-th rows of $\alpha^{(k)}$ coincide, so $\det \alpha^{(k)} = 0$. Expanding $\alpha^{(k)}$ by its $k$-th row yields the right hand side of the above formula. So the non-diagonal entries of $\alpha \gamma$ are zero.

Now $\alpha \gamma$ is a diagonal matrix, so $\det(\alpha \gamma)$ is just the product of the diagonal entries, all four of which are $d$. Thus $\det(\alpha \gamma) = d^4$.

That $(\det \alpha)(\det \gamma) = (\det \alpha)(\det \beta)$ follows from $\det \gamma = \det \beta^T = \det \beta$, which follows from the determinant being invariant under transposition. That $(\det \alpha)(\det \beta) = dD$ follows from the definitions of $d$ and $D$.

For the last paragraph: We have now shown that $d^4 = dD$. Because $d \neq 0$ we can divide by $d$ (which is a unit in the field $L$ we are working over) to get $d^3 = D$. Evaluating this at $a_{11}, \dotsc, a_{44} \in K$ gives $$d^3(a_{11}, \dotsc, a_{44}) = D(a_{11}, \dotsc, a_{44}),$$ which is what we wanted to show in the first place.

Over infinite fields

If our ground field $K$ is infinite, we can also argue without using the field of rational functions. Let $\alpha$, $\beta$, $\gamma$, $d$ and $D$ be defined as originally, i.e. as in the books solution.

We still get in the same way that $\alpha \gamma$ is diagonal with all diagonal entries $d$, and thus $\det \alpha = d^4$, and that $(\det \alpha) (\det \gamma) = (\det \alpha)(\det \beta) = dD$. So $d^4 = dD$.

For the last paragraph we argue simlilary to the general case: Consider the following polynomials in the 16 variables $x_{11}, x_{12}, x_{13}, x_{14}, x_{21}, \dotsc, x_{44}$, i.e. elements of the polynomial ring $K[x_{11}, \dotsc, x_{44}]$: \begin{align*} p(x_{11}, \dotsc, x_{44}) &= \left| \begin{matrix} x_{11} & x_{12} & x_{13} & x_{14} \\ x_{21} & x_{22} & x_{23} & x_{24} \\ x_{31} & x_{32} & x_{33} & x_{34} \\ x_{41} & x_{42} & x_{43} & x_{44} \\ \end{matrix} \right| \\ &= \sum_{\sigma \in S_4} \mathrm{sgn}(\sigma) x_{1\sigma(1)} x_{2\sigma(2)} x_{3\sigma(3)} x_{4\sigma(4)}. \end{align*} and \begin{align*} &\, q(x_{11}, \dotsc, x_{44}) \\ =&\, \left| \begin{matrix} y_{11}(x_{11}, \dotsc, x_{44}) & \cdots & y_{14}(x_{11}, \dotsc, x_{44}) \\ \vdots & \ddots & \vdots \\ y_{41}(x_{11}, \dotsc, x_{44}) & \cdots & y_{44}(x_{11}, \dotsc, x_{44}) \end{matrix} \right| \\ =&\, \sum_{\sigma \in S_4} \mathrm{sgn}(\sigma) y_{1\sigma(1)}(x_{11}, \dotsc, x_{44}) \dotsm y_{4\sigma(4)}(x_{11}, \dotsc, x_{44}). \end{align*} where for all $1 \leq i,j \leq 4$ we define $y_{ij}(x_{11}, \dotsc, x_{44})$ to be the $(i,j)$-th cofactor of the matrix. $$\begin{pmatrix} x_{11} & x_{12} & x_{13} & x_{14} \\ x_{21} & x_{22} & x_{23} & x_{24} \\ x_{31} & x_{32} & x_{33} & x_{34} \\ x_{41} & x_{42} & x_{43} & x_{44} \\ \end{pmatrix}.$$

By definition of $p$ and $q$ we have $$d = p(a_{11}, \dotsc, a_{44}) \quad\text{and}\quad D = q(a_{11}, \dotsc, a_{44}).$$ Therefore we have shown so far that $$p(a_{11}, \dotsc, a_{44})q(a_{11}, \dotsc, a_{44}) = p^4(a_{11}, \dotsc, a_{44})$$ for all $a_{11}, \dotsc, a_{44} \in K$. So the polynomial $p^4-pq = p(p^3-q)$ evaluates zero everywhere. If our field is infinite we can now follows that already $p(p^3-q) = 0$, i.e. not only evaluates $p(p^3-q)$ zero everywhere, but it is already the zero polynomial.

But we can find $b_{11}, \dotsc, b_{44}$ such that $p(b_{11}, \dotsc, b_{44}) \neq 0$ (i.e. the ($4 \times 4$)-matrix with entries $b_{ij}$ is invertible). So $p \neq 0$. Because the polynomial ring $K[x_{11}, \dotsc, x_{44}]$ is an integral domain we can follow from $p \neq 0$ and $p(p^3-q) = 0$ that already $p^3 - q = 0$, i.e. $p^3 = q$. Evaluating these polynomials, we get that $$d^3 = p^3(a_{11}, \dotsc, a_{44}) = q(a_{11}, \dotsc, a_{44}) = D$$ for all $a_{11}, \dotsc, a_{44} \in K$, which is what we wanted to prove in the first place.

PS: Also notice that in the same way(s) we can show that for every $n \times n$ square matrix $A$ we have $$\det(\mathrm{adj}(A)) = \det(A)^{n-1},$$ where $\mathrm{adj}(A)$ denotes the adjugate of $A$.