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.

  • 4
    $\begingroup$ Is $A$ a $5\times 5$ matrix? $\endgroup$ Oct 13, 2012 at 14:22
  • $\begingroup$ What is $\mathrm{adj} A$? $\endgroup$
    – Norbert
    Oct 13, 2012 at 14:31
  • $\begingroup$ @Norbert en.wikipedia.org/wiki/Adjugate_matrix $\endgroup$
    – anon
    Oct 13, 2012 at 14:38
  • $\begingroup$ @anon thanks!${}$ $\endgroup$
    – Norbert
    Oct 13, 2012 at 14:40

3 Answers 3


For simplicity put $\,B:=adj\, A\,$ , so:

$$AB=|A|\cdot I\Longrightarrow |A||B|=|A|^n$$

We're done, since

$$|B|=0\Longrightarrow |A|^n=0\Longrightarrow |A|=0$$

  • $\begingroup$ I don't understand why did you set |B| = 0. My question may not be clear. I mean if we know that |adjA| isn't zero how can we assert that |A| isn't zero. Now in your proof you set |B|=|adjA|=0 and you show that |A|=0. how does it help me? $\endgroup$ Oct 13, 2012 at 15:13
  • 1
    $\begingroup$ @user44471: You should change "equal" to "not equal" in the title then, so that the title matches the actual question. $\endgroup$ Oct 13, 2012 at 15:29
  • $\begingroup$ Yes my mistake... sorry. $\endgroup$ Oct 13, 2012 at 15:48

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.


By contrapositive:

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:

  1. A row of zeros, or
  2. 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.


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