Question

If A = `[(cos α, sin α),(-sin α, cos α)]`, then find α satisfying `0 < α < π/2`, when A + A^T = `sqrt(2)  l_2` where A^T is transpose of A.

Options

• `π/2`

• `π/4`

• `π/6`

• `π/3`

Answer 1

`bb(π/4)`

Explanation:

Given, A = `[(cos α, sin α),(-sin α, cos α)]`

Also given, A + A^T = `sqrt(2)  l_2`

∴ `[(cos α, sin α),(- sin α, cos α)] + [(cos α, sin α),(- sin α, cos α)]^T = sqrt(2)[(1, 0),(0, 1)]`

`\implies [(cos α, sin α),(- sin α, cos α)] + [(cos α, - sin α),(sin α, cos α)] = [(sqrt(2), 0),(0, sqrt(2))]`

= `[(cos α + cos α, sin α - sin α),(- sin α + sin α, cos α + cos α)]`

= `[(sqrt(2), 0),(0, sqrt(2))]`

`\implies [(2 cos α, 0),(0, 2 cos α)] = [(sqrt(2), 0),(0, sqrt(2))]`

On equating the corresponding elements, we get

2 cos α = `sqrt(2)`

`\implies` cos α = `sqrt(2)/2`

`\implies` cos α = `1/sqrt(2)`

∴ α = `π/4`, which is satisfying `0 < α < π/2`.