Difficulties with adjoint matrices

Question

I'm trying to find a proof or a counter-example to the next statement:

Let A be a matrix such that $A\in M_n(\mathbb{R})$.

Prove or disprove that $adj(-A)=(-1)^{n-1}adj(A)$.

I know that if A is invertible, then $adj(cA)=c^{n-1}adj(A)$ and taking c=-1 we get that $adj(-A)=(-1)^{n-1}adj(A)$.

What about the case A is not invertible? can we find a counter-example?

Please help, thanks!

score 1 · Answer 1 · answered Dec 23 '13 at 09:51

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Recall the construction of $\operatorname{adj}(A)$: the entry $(i,j)$ is $(-1)^{i+j}$ the determinant of $A$ without the line $i$ and the column $j$, denoted $D_{i,j}(A)$. This is the definition for any $A$, invertible or not.

Since $D_{i,j}(A)$ has dimension $(n-1)\times (n-1)$, we have $D_{i,j}(-A)=(-1)^{n-1}D_{i,j}(A)$.

answered Dec 23 '13 at 09:51

Davide Giraudo

172,925

First, thank you for your answer. Second, do you mean that the statement is true for any A, whether it's invertible or not? – Galc127 Dec 23 '13 at 09:54
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Yes. ${}{}{}{{}$ – Davide Giraudo Dec 23 '13 at 09:55

score 1 · Accepted Answer · answered Dec 23 '13 at 09:51

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Hint: $\mathrm{Adj}(A)\cdot A=A\cdot\mathrm{Adj}(A)=I_n\cdot\det(A)$.

answered Dec 23 '13 at 09:51

Michael Hoppe

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May you show why is it true for any matrix? – Galc127 Dec 23 '13 at 09:54
Eventually, I understand. Useful hint, thanks! – Galc127 Dec 23 '13 at 13:28

Difficulties with adjoint matrices

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