a) Let A =

[a b]

[c d]

Given AC = CA perform the multiplications (AC and CA) and find out the relation between a,b,c,d for AC = CA to hold true . As BC = CB same relation will hold for elements of B. Under this relation you can show AB = BA

b) Again all you need to show is

[1 nx ny+(n(n-1)/2)x²]

[0 1 nx]

[0 0 1]

when multiplied by

[1 x y]

[0 1 x]

[0 0 1]

gives

[1 mx my+(m(m-1)/2)x²]

[0 1 mx]

[0 0 1]

where m=n+1