Question

2. Suppose A is invertible. a. (25 points) Explain why AA is also invertible. b. (25 points) Then show that A'=(A^T A^-1) A^T

Answer 1

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

`\Rightarrow A'=(A^(T) * A^(-1))* A^(T)\\\\\Rightarrow \text{Cancelling A' from both sides}\\\\\Rightarrow I=(A^(-1))* A^(T)\\\\\Rightarrow A^(-1)= [A^(T)]^(-1)\\\\\Rightarrow (Adj.A)/(|A|)=(Adj.[A^(T)])/(|A^(T)|)\\\\As, |A|=[|A|]^(T)\\\\Adj.A=Adj.[A^(T)]\\\\A=A^(T)`

→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'