If `A=[(a,b),(c,d)]`, where a, b, c and d are real numbers, then prove that `A^(2)-(a+d)A+(ad-bc) I=O`. Hence or therwise, prove that if `A^(3)=O` then `A^(2)=O`

Explanation

Given, `A=[(a,b),(c,d)]`
`implies A^(2)=[(a,b),(c,d)].[(a,b),(c,d)]=[(a^(2)+bc,ab+bd),(ac+cd,bc+d^(2))]`
Hence, `A^(2)-(a+d)A+(ad-bc) I`
`=[(a^(2)+bc,ab+bd),(ac+cd,bc+d^(2))]-(a+d) [(a,b),(c,d)]+(ad-bc) [(1,0),(0,1)]`
`=[(a^(2)+bc-(a^(2)+ad)+(ad-bc),ab+bd-(ab+bd)),(ac+cd-(ac+cd),bc+d^(2)-(ad+d^(2))+(ad-bc))]`
`=[(0,0),(0,0)]=O`
given `A^(3)=O`
`implies |A|=0` or `ad-bc=0`
`implies A^(2)-(a+d)A=O`
or `A^(2)=(a+d)A` (1)
Case I : `a+d=0`
From equation (A), `A^(2)=O`.
Case II : `a+d ne 0`
Given `A^(3)=O`
`implies A^(2)A=O`
`implies (a+d)A A=O`
`implies A^(2)=O`