Step 1 : Using the inverse formula to prove tha given relation.

We know that

${\mathrm{A}}^{-1}=\frac{\mathrm{adj}\mathrm{A}}{\left|\mathrm{A}\right|}$

∴ A.adjA = |A|I       .....(1)

Substituting A with adjA

adjA.adj(adjA) = |adjA|                           

From (1)

$\mathrm{adjA}=\frac{\left|\mathrm{A}\right|\mathrm{I}}{\mathrm{A}}$       .....(2)

Substituting (2) in (1)

$\frac{\left|\mathrm{A}\right|\mathrm{I}}{\mathrm{A}}\mathrm{adj}(\mathrm{adj}\mathrm{A})=|\mathrm{adj}\mathrm{A}|\mathrm{I}$      ........(3)

Taking determinant on both sides of (1)

|A.adjA| = || A | I |

⇒ |A| |adjA| = |A|ⁿ

⇒ |adjA| = |A|^n–1       ....(4)

Substituting (4) in (3)

$\frac{\left|\mathrm{A}\right|\mathrm{I}}{\mathrm{A}}\mathrm{adj}(\mathrm{adj}\mathrm{A})=|\mathrm{A}{|}^{\mathrm{n}-1}\mathrm{I}$

⇒ adj(adjA) = |A|^n–2a

Hence, we proved that adj(adjA) = |A|^n–2A.