Check the proof of: $(A')^{-1}=\left(A^{-1}\right)'$.

Question

Prove: $(A')^{-1}=\left(A^{-1}\right)'$.

$$\begin{align*}\left(A^{-1}\right)'=\left(\frac{1}{\det (A)}A^*\right)'=\frac{1}{\det (A)}\left(A^*\right)'=\frac{1}{\det (A)}(A')^*=(A')^{-1}\end{align*}$$

Is it right?（This is the question...）Duplicate?

$A'$ is the transpose of A.

$A^*$ is the adjugate matrix

I think $A'$ is the transpose and $A^*$ is the matrix of cofactors. — xavierm02, Aug 03 '13 at 12:34
Let us give the OP the opportunity to explain his notation... — DonAntonio, Aug 03 '13 at 12:43
It is right, but incomplete. You need to show that $(A^)' = (A')^$, since you're using it and it's not obviously true. — Vedran Šego, Aug 03 '13 at 13:01

score 2 · Answer 1 · answered Aug 03 '13 at 12:38

2

I'm not sure $\left(A^*\right)'=\left(A'\right)^*$ is easier to prove than your statement.

You probably know $\left(AB\right)'=B'A'$. From that $I=I'=\left(AA^{-1}\right)'=\left(A^{-1}\right)'A'$ so $\left(A'\right)^{-1}=\left(A^{-1}\right)'$.

answered Aug 03 '13 at 12:38

xavierm02

7,495

score 2 · Answer 2 · answered Aug 03 '13 at 12:39

2

More simply:

$(A')\left(A^{-1}\right)'= \left(A^{-1} A\right)'=I'=I$, hence $A'$ is ionverse to $\left(A^{-1}\right)'$.

answered Aug 03 '13 at 12:39

Boris Novikov

17,470

1

I would have written this, too. But, technically it does not answer the question ("Is it right?") – GEdgar Aug 03 '13 at 12:52
@ GEdgar 9 : Yes, you are right. – Boris Novikov Aug 03 '13 at 12:53

Check the proof of: $(A')^{-1}=\left(A^{-1}\right)'$.

2 Answers2