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?


3 Answers 3


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.

  • 1
    $\begingroup$ Very nice! ${}{}$ $\endgroup$
    – copper.hat
    Apr 3, 2018 at 21:22
  • $\begingroup$ Hey, GNU, I don't quite understand why the remaining entries are zero's. It could be the notation. I do understand if there are two row vectors that are equivalent in a matrix, then the determinant of that matrix is zero. $\endgroup$ Apr 5, 2018 at 20:06
  • 1
    $\begingroup$ @RafaelVergnaud Equation \eqref{3'} describes the off-diagonal entries in the matrix product $AC^T$. $$(AC^T)_{ij} = \sum_{k=1}^n a_{ik} (C^T)_{kj} = \sum_{k=1}^n \color{red}{a_{ik}} C_{jk} \label{3'}\tag{3'}$$ $C_{jk}$ (row $j$ column $k$) with $k$ running from $1$ to $n$ means a row of cofactors $C_{jk}$ with the $j$-th row of $A$ deleted. The factor $a_{ik}$, $k = 1,\dots,n$ replaces $a_{jk}$ in the above Laplace expansion, giving a zero determinant. This holds for $\binom{n}{2}$ pairs of $(i,j)$. $\endgroup$ Apr 5, 2018 at 20:17

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

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


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.

  • $\begingroup$ Hey. That's actually what I'm trying to prove. I'm assuming that A is not invertible and trying to show that A * adj(A) = det(A) * I_n = 0. But, just wanted to make sure. Thank you! $\endgroup$ Apr 3, 2018 at 19:38
  • $\begingroup$ @RafaelVergnaud If you want to be sure about this (over any commutative ring), I suggest you to take a look at my answer. $\endgroup$ Apr 3, 2018 at 19:48

Not the answer you're looking for? Browse other questions tagged .