Answer 1

`A = [(-1, 1),(a, b)]`

`A^2= [(-1, 1),(a, b)][(-1, 1),(a, b)]`

`= [(1 + a, -1 + b),(-a + ab, a + b^2)]`

It is given that `A^2 = I`

`∴ [(1 + a, -1 + b),(-a + ab, a + b^2)] = [(1, 0),(0, 1)]`

Comparing the corresponding elements we get

1 + a = 1

Therefore a = 0

-1 + b = 0

Therefore b = 1