If  A=1    −12    −1,B=a    1b    −1  and (A+B)2=A2+B2 then the values of  a  and  b are

We have, A+B=a+1    0b+2    −2  A2=1−12−11−12−1=−100−1B2=a1b−1a1b−1=a2+ba−1ab−bb+1(A+B)2=a+10b+2−2a+10b+2−2=(a+1)20(a−1)(b+2)4∴(A+B)2=A2+B2⇒(a+1)20(a−1)(b+2)4=−100−1+a2+ba−1ab−bb+1⇒(a+1)20(a−1)(b+2)4=a2+b−1a−1ab−bb⇒a−1=0,b=4,(a+1)2=a2+b−1,(a−1)(b+2)=ab−b    ⇒ a=1  and b=4