If $\det(\mathrm{adj}\,A) \ne 0$, then $\det(A) \ne 0$. I'm trying to understand the reason why $A$ is invertible only if $\mathrm{adj}\,A$ is invertible.
That's what I have right now: $A\, \mathrm{adj}\,A = |A|\cdot I$.
So if we take $\det$ of both sides we get: $|A\,\mathrm{adj}\,A| = ||A|\cdot I|$
and then: $|A| \cdot |\mathrm{adj}\,A| = |A|^n$
but now I'm stuck...
Appreciate your help.
 A: For simplicity put $\,B:=adj\, A\,$ , so:
$$AB=|A|\cdot I\Longrightarrow |A||B|=|A|^n$$
We're done, since 
$$|B|=0\Longrightarrow |A|^n=0\Longrightarrow |A|=0$$
A: Suppose $\det{(\operatorname{adj}{A})} \neq 0$ but $\det{A} = 0$. Since $A \operatorname{adj} A = 0$ and $\operatorname{adj}{A}$ is invertible, we have $A=0$, so $\operatorname{adj}{A} = 0$, giving $\det{(\operatorname{adj}{A})} = 0$ which is a contradiction.
A: This is my answer, but in advance, I apologize if any anomaly is found regarding the wording, because I speak Spanish and I do not speak English perfectly.
By the counterreciprocal:
If $\det A=0$, then $\det (\text{adj }A)=.0$
For the determinant of $A$ to be zero, there are necessary conditions that $A$ has:

*

*A row of zeros, or

*A row that is a multiple of another row.

Both propositions 1 and 2 are equivalent, let's see:
Suppose that the $j$-th row of $A$ is a multiple of the $i$-th row of $A$, and we know that if we multiply the $i$-th row of $A$ by that same multiple and add it to the $j$-th, this last row will be a row of zeros, and the determinant of $A$ does not change; that is, it is zero.
Suppose that the $i$-th row of $A$ is a row of zeros. Then the minors associated with the $k$-th row different from the $i$-th row will be zeros, since they will have a row of zeros.
Finally, since the components of the adjoint matrix of $A$ is the transpose of the matrix of cofactors associated with the matrix $A$, then the adjoint matrix of $A$ will have a column of zeros, the $k$-th column, therefore, the determinant of the adjoint matrix of $A$ is zero.
