Answer with explanation:
Part A
It is given that A is invertible.That is inverse of A exist.
Means, Determinant of A is non zero.
→ |A|≠0
⇒AA=A²
⇒|A²|=A|²
=|A|×|A|
⇒As determinant of A is non zero, so determinant of A² will also be non zero.
⇒|A|² ≠0
Which shows that A² is also invertible.
Part B
We have to prove
[tex]\Rightarrow A'=(A^{T} \times A^{-1})\times A^{T}\\\\\Rightarrow \text{Cancelling A' from both sides}\\\\\Rightarrow I=(A^{-1})\times A^{T}\\\\\Rightarrow A^{-1}= [A^{T}]^{-1}\\\\\Rightarrow \frac{Adj.A}{|A|}=\frac{Adj.[A^{T}]}{|A^{T}|}\\\\As, |A|=[|A|]^{T}\\\\Adj.A=Adj.[A^{T}]\\\\A=A^{T}[/tex]
→Matrix multiplication is associative.For three matrix A, B and C
(A B)C=A(B C)
→So, this equation is Possible if,
A=A'
