Question

If x, y ∈ R, then the determinant ∆ = `|(cosx, -sinx, 1),(sinx, cosx, 1),(cos(x + y), -sin(x + y), 0)|` lies in the interval.

विकल्प

• `[-sqrt(2), sqrt(2)]`

• [–1, 1]

• `[-sqrt(2), 1]`

• `[-1, -sqrt(2)]`

Answer 1

If x, y ∈ R, then the determinant ∆ = `|(cosx, -sinx, 1),(sinx, cosx, 1),(cos(x + y), -sin(x + y), 0)|` lies in the interval `[-sqrt(2), sqrt(2)]`.

Explanation:

Indeed applying R₃ → R₃ – cosyR₁ + sinyR₂, we get

∆ = `|(cosx, -sinx, 1),(sinx, cosx, 1),(0, 0, siny - cosy)|`

Expanding along R₃, we have

∆ = (siny – cosy) (cos²x + sin²x)

= (siny – cosy)

= `sqrt(2)[1/sqrt(2) siny - 1/sqrt(2)  cosy]`

= `sqrt(2)[cos  pi/4  sin y - sin  pi/4  cos y]`

= `sqrt(2) sin(y - pi/4)`

Hence `-sqrt(2) ≤ ∆ ≤ sqrt(2)`.