Is there any non zero matrix whose adjoint is a zero matrix Just wanted to know whether their exits a non zero matrix whose adjoint is a zero matrix. And if so what would be inverse of a matrix whose adjoint as well as determinant is zero, as $$A^{-1}=\dfrac{1}{ |A|} adj(A)$$
 A: The matrix
$$A=\begin{pmatrix}1&0&0\\0&0&0\\0&0&0\end{pmatrix} $$
satisfies $adj(A)=0$ and $\det(A)=0$. The inverse does not exist if the determinent is zero.
This answers your question!
A: Let $A$ be $n\times n$. Since each entry of $\operatorname{adj}(A)$ is a signed multiple of a $(n-1)$-rowed minor, $\operatorname{adj}(A)=0$ if and only if $\operatorname{rank}(A)\le n-2$.
It follows that $A\ne0=\operatorname{adj}(A)$ if and only if $n\ge3$ and $0<\operatorname{rank}(A)\le n-2$.
A: If $A$ is an $n\times n$ matrix with $n>2$, and all entries are equal, then the adjoint (adjugate) matrix will be zero.
A matrix with a $0$ adjugate matrix cannot have an inverse, as $A\times\operatorname{adj}(A) = \det(A)\cdot I$. And if this product turns out to be $0$, then that means that the determinant of $A$ is zero, which implies that $A$ is not invertible.
A: I'm not sure if the proof is concise enough, but here is my attempt. Every remark is welcome!
$\text{LaPlace formulae}$
Let $A\in M_n(\mathbb N).$

$$A_{ij}:=\sum_{p\in S_n\\p(i)=j}^n(-1)^{I(p)}a_{1p(1)}\cdot\ldots
 a_{i-1,p(i-1)}a_{i+1,p(i+1)}\cdot\ldots\cdot a_{np(n)}$$

Note (reminder): there is no $a_{ip(i)}$, that's why $A_{ij}$ is called $\text{the algebraic complement}$ of the element $a_{ij}.$
Now we can simplify the initial formula:
$$\det(A)=\sum_{j=1}a_{ij}A_{ij}\leftarrow\text{development by the i-th row}$$
As colleagues have already mentioned:$$ A_{ij}=0,\forall\; i,j\in\{1,\ldots,n\}\implies\det(A)=0$$
We can also write: $$A_{ij}=(-1)^{i+j}\Delta_{ij}$$
More conrete example:
$$B=\begin{bmatrix}A_{11}&A_{12}&\ldots &A_{1,n-1}&A_{1n}\\A_{21}&A_{22}&\ldots&A_{2,n-1}&A{2n}\\\vdots&\vdots&\ddots&\vdots&\vdots\\A_{n1}&A_{n2}&\ldots&A_{n,n-1}&A_{nn}\end{bmatrix}$$
If the matrix $B$ above were to represent what happens when we calculate $\det(A)$ with all the algebraic complements: $A_{ij}=0\;\forall\;i,j\in\{1,\ldots,n\}$, it would be  $\text{the null-matrix}$. 
Here comes the initial matrix: 

$$A=(a_{ij}),\;a_{ij}=x,\;\forall\;i,j\in\{1,\ldots,n\},x\in\mathbb F$$

This is the example @Arthur suggested.
We can also prove @Arthur's argument that tells us $\text{the matrix must be singular}.$
I can't think of other cases rather than Arthur's, but even if they were the only ones, it would be enough for the statement $A=0_n$ to fall.
