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प्रश्न
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)]`
उत्तर
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)`.
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