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Let $A$ be a $n \times n$ matrix over $\mathbb{R}$ or $\mathbb{C}$.

Let $\text{adj}(A)$ denote the classical adjoint matrix of $A$ (or adjugate matrix of $A$).

It is known that $A \cdot \text{adj}(A) = \text{det}(A) I_n = \text{adj}(A) \cdot A$.

Thus, if $A$ is invertible, then $\text{det}(A) \neq 0$, and so $\text{adj}(A)$ is also invertible,

since $\frac{A}{\text{det}(A)} \cdot \text{adj}(A) = I_n = \text{adj}(A) \cdot \frac{A}{\text{det}(A)}$.

My question :

Is there a case that $A$ is not invertible and $\text{adj}(A)$ is invertible?

with-forest
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1 Answers1

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Suppose $\operatorname{adj}(A)$ is invertible but $A$ is not. Then $\det(A)=0$, and $A\operatorname{adj}(A)=\det(A)I$ implies $A=\det(A)\left(\operatorname{adj}(A)\right)^{-1}$, i.e., $A$ is the zero matrix. In this case it's easy to see $\operatorname{adj}(A)$ is also the zero matrix (all of the cofactors of $A$ are zero), contradicting the assumption.

pancini
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