Solution :Matrices and law of matrices and Squares
Question: 1
Since we are told that A is indepented, then we have
(I – A )2 we expand the problem and then solve
(I – A)2 = I2 + A2 – 2AI2
(I– A)2 = I + A –2A
(I – A)2 = I – A
Therefore, I – A is idempotent.
If A is invertible, we consider having A–1
Therefore, A2 = A which is AA =A,
We then multiply both sides by A–1
A–1(AA) = A–1 A
(A–1 A) A = I A= I, therefore we conclude by proofing that it is true to say that,
A= I