Determinants and Adjugates

Question

If the determinant of a matrix $A$ is zero, then is the product $$A \cdot \operatorname{adj}(A)$$ also zero? Is there an explanation that does not involve eigenvalues?

Answer 1

To address the question writer's question, I derive from the well-known Laplace expansion formula. In this way, OP will really understand how $\operatorname{adj}(A)$ interacts with $A$ through matrix multiplication to give a diagonal matrix.

Denote $C_{ij}$ as the $(i,j)$-th entry of the cofactor matrix, which is the transpose of $\operatorname{adj}(A)$. I hope following classical argument is accessible to any interested high school students.

By the well-known Laplace expansion formula for calculating determinants, for any $i \in \lbrace1,\dots,n\rbrace$

\begin{align} \det(A) &= \sum_{j=1}^n (-1)^{i+j} a_{ij} \det(M_{ij}) \\ &= \sum_{j=1}^n a_{ij} C_{ij} \label{1}\tag{1} \end{align}

But a matrix product of $AB$ has entries of the form $$(AB)_{ij} = \sum_{k=1}^n a_{ik} b_{kj} \label{2}\tag{2}.$$

To make \eqref{1} resembles more \eqref{2}, we consider the transpose of the cofactor matrix, whose $(j,i)$-th entry is $C_{ij}$. For any $i$ fixed,

$$ \det(A) = \sum_{j=1}^n a_{ij} (C^T)_{ji} \label{1'}\tag{1'} $$

We change the $j$ in \eqref{1'} to $k$. For each $i$ fixed,

$$ \det(A) = \sum_{k=1}^n a_{ik} (C^T)_{ki} \label{1''}\tag{1''} $$

\eqref{1''} represents any diagonal entry of $AC^T$. To show that $AC^T$ is a diagonal matrix, one needs to justify that the non-diagonal entries of $AC^T$ equals zero.

For each $(i,j)$ fixed with $i\ne j$, $$ (AC^T)_{ij} = \sum_{k=1}^n a_{ik} (C^T)_{kj} = \sum_{k=1}^n \color{red}{a_{ik}} C_{jk} \label{3}\tag{3} $$ This is in fact the Laplace expansion of the determinant $$ \begin{vmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots& \vdots & \ddots & \vdots \\ a_{i1} & a_{i2} & \cdots & a_{in} \\ \vdots& \vdots & \ddots & \vdots \\ \color{red}{a_{i1}} & \color{red}{a_{i2}} & \cdots & \color{red}{a_{in}} \\ \vdots& \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{vmatrix} $$ Note that the $\color{red}{j\text{-th row}}$ of the above matrix is replaced by the $i$-th row of $A$ as the Laplace expansion formula \eqref{3} suggests. Since any matrix with two identical rows has zero determinant, we conclude the following useful formula. $$\bbox[2px, border: 1px solid red]{\det(A)I_n=AC^T=C^TA}$$ Uptil this step, this works for entries defined on any commutative ring (equipped with addition and abelian multiplication, I don't know whether this works for non-commutative rings.)

To answer the question, we set $\det(A) = 0$, so that $A \operatorname{adj}(A) = 0$.

Remark: We assume nothing on the existence of multiplicative inverse in the commutative ring.

Answer 2

From the formula for matrix inverse, you can find that the determinant is equal precisely to the expression you're asking about. So determinant being zero is equivalent to this expression being zero.

Answer 3

Note that $$A \times \operatorname{adj}(A)= \det(A) I$$ where $I$ is the identity matrix.

Thus $$ \det(A)=0 \implies A \times \operatorname{adj}(A)=0$$