# Does $\mathrm{adj}(A)=\mathrm{adj}(B)$ imply $A=B$?

If $A$ and $B$ are any two square matrices of same order and if $\mathrm{adj}(A)=\mathrm{adj}(B)$, does it imply $A=B$?

I am pretty sure if $A$ and $B$ are invertible and if $A^{-1}=B^{-1}$, then $A=B$.

So is it true for Adjoint?

• @Widawensen made a mistake, I'm deleting that – Omnomnomnom Jun 6 '17 at 13:46

Indeed

if $A^{-1}=B^{-1}$ then $AA^{-1}B=AB^{-1}B$ i.e. $B=A$.

If $A,B$ are non-invertible then $\det(A)=\det(B)=0$ and $\text{adj}(A)A=\det(A)I=0$ and $\text{adj}(B)B=\det(B)I=0$.

This can be satisfied without $A=B$.

Hence counterexamples for supposed theorem are, for example

$\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \ \end{bmatrix}$, $\begin{bmatrix} 0 & -1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \ \end{bmatrix}$.

Here of course $\text{adj}(A)=\text{adj}(B)$.

• @EkaveeraKumarSharma with non-singular matrices when $A=kB$ the only possibility is $k=1$ because it follows that they have equal determinants (as Farrukh has shown) - so their inverses are equal and $A=B$ . – Widawensen Jun 7 '17 at 12:22
• @EkaveeraKumarSharma However I have now some little doubts about Farukh's completness of the answer for case $k=-1$ which can arise when $n$ is odd ..Because :$\text{adj}(- {A}) = (-1)^{n - 1}\text{adj}( {A})$ then $\text {adj}(A)=\text{adj}(B)$ for odd n.. in this case evidently it can follow also $A=-B$.. – Widawensen Jun 7 '17 at 13:05
• @EkaveeraKumarSharma So my final conclusion is that for odd $n$ (dimension) from equality of adjugates follows $A=B$ or $A=-B$. For even $n$ only $A=B$.. – Widawensen Jun 7 '17 at 13:24
• @EkaveeraKumarSharma The easiest way to check above statement is to take identity matrix $I$ and $-I$. You will see that adjugates of these matrices are equal when dimension is odd. – Widawensen Jun 7 '17 at 13:39
• @EkaveeraKumarSharma So you see that in this case from adj(A)=adj(B) we have $A=B$ or ("or" doesn't mean "only") $A=-B$. – Widawensen Jun 8 '17 at 10:22

No.

The adjugate matrix is basically the wedge power $\wedge^{n-1} A$ if $n$ is the dimension of the underlying vector space. If the underlying field has nontrivial $(n-1)$-th roots of unity, then we can multiply $A$ by one of them without changing the adjugate matrix. Also, the adjugate matrix of any matrix of rank $< n-1$ will be $0$.

Not necessarily. It is given: $adj(A)=A^{-1}|A|=adj(B)=B^{-1}|B|$. Multiply both sides by $A$, then $B$ to get: $A=\frac{|A|}{|B|}B$. It implies: $$|A|=\left(\frac{|A|}{|B|}\right)^n|B| \Rightarrow \left(\frac{|A|}{|B|}\right)^{n-1}=1\Rightarrow \begin{cases} |A|=|B|, \ \ \ \ \ if \ n \ is \ even \\ |A|=\pm|B|, \ \ if \ n \ is \ odd \end{cases}.$$

• Well done Farrukh, so we have equality of determinants and following equality of inverses for invertible matrices... – Widawensen Jun 6 '17 at 14:30
• However it seems that for $n$ odd as a final conclusion can be also made $\vert A \vert = -\vert B \vert$ – Widawensen Jun 7 '17 at 13:11
• @Widawensen, thank you. Updated. – farruhota Jun 7 '17 at 13:42
• Ok. Farrukh, now we have a complete answer for Ekkaveera, it seems.. – Widawensen Jun 7 '17 at 13:43