If `A=[(1,-1),(2,-1)]` and `B=[(1,a),(4,b)]` and `(A+B)^(2)=A^(2)+B^(2)`.
Then a and b are respectively

Options

A. `1,-1`
B. `2,-3`
C. `-1,1`
D. `3,-2`

Answer

A

Explanation

Now `A+B=[(1,-1),(2,-1)]+[(1,a),(4,b)]=[(2,-1+a),(6,-1+b)]`
`implies(A+B)^(2)=[(2,-1+a),(6,-1+b)][(2,-1+a),(6,-1+b)]`
`=[(-2+6a, -1+a-b+ab),(6+6b,-5+6a-2b+b^(2))]`
and `A^(2)=[(1,-1),(2,-1)][(1,-1),(2,-1)]=[(1,0),(0,-1)]`
Also `B^(2)=[(1,a),(4,b)][(1,a),(4,b)]=[(1+4a,a+ab),(4+4b,4a+b^(2))]`
Given `(A+B)^(2)=A^(2)+B^(2)`
`:.[(-2+6a,-1+a-b+ab),(6+6b,-5+6a-2b+b^(2))]`
`=[(-1,0),(0,-1)]+[(1+4a,a+ab),(4+4b,4a+b^(2))]`
`implies[(-2+6a,-1+a-b+ab),(6+6b,-5+6a-2b+b^(2))]=[(4a,a+ab),(4+4b,-1+4a+b^(2))]`
On comparing both sides we get
`-2+6a=4a`
and `6+6b=4+4b`
`:.a=1` and `b=-1`