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If A=[{:(0,1),(1,1):}] "and" B=[{:(0,-1...

If `A=[{:(0,1),(1,1):}] "and" B=[{:(0,-1),(1,0):}]`, then show that
`(A+B)(A-B)neA^(2)-B^(2)`

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We have, `A=[{:(0,1),(1,1):}]"and" B=[{:(0,-1),(1,0):}]`
`(A+ B)=[{:(0+0,1 -1),(1+1,1+0):}]=[{:( 0,0),(2,1):}]`
`therefore (A+B)=[{:(0+0,1-1),(1+1,1+0):}]=[{:(0,0),(2,1):}]_(2xx2)`
and ` (A-B)=[{:(0-0,1+1),( 1-1,1-0):}]=[{:(0,2),(0,1):}]_(2xx2)`
Since (A+B),(A-B) is defined, if the number of columns of (A+ B) is equal to the number of rows of (A-B) so here multiplication of matrices (A+B ),(A-B) is possible
Now, `(A+B)_(2xx2).(A-B)_(2x2)=[{:(0+0,0+0),(0+0,4+1):}]=[{:(0,0),(0,5):}]`
Also `A^(2)=A,A`
`=[{:(0,1),(1,1):}],[{:(0,1),(1,1):}]`
`=[{:(0+1,0+1) ,(0+1,1+1):}]=[{:(1,1) ,(1,2):}] `
and `B^(2)=B,B =[{:(0,-1),(1,0):}][{:(0,-1),(1,0)]`
`=[{:(0-1, 0+0),(0+0,-1+0):}] =[{:(-1,0),( 0,-1):}]`
`rArr (A+B),(A-B)neA^(2)-B^(2)`
`rArr [{:( 0,0),(0,5):}] ne[{:(2,1),(1,3):}]`
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