Matrices & Determinants
Properties of determinants
Grade Class 12
Question:
The value of θ lying between -<span style="font-family: 'Times New Roman', serif;">π</span>/4 & <span style="font-family: 'Times New Roman', serif;">π</span>/2 and 0 ≤ A ≤ <span style="font-family: 'Times New Roman', serif;">π</span>/2 and satisfying the equation <br><br> <table style="border-collapse: collapse; border: 1px solid black; display: inline-table; vertical-align: middle;"><tr><td style="padding: 5px; border-left: 1px solid black; border-right: 1px solid black;">1+sin<sup>2</sup> A</td><td style="padding: 5px; border-right: 1px solid black;">cos<sup>2</sup> A</td><td style="padding: 5px; border-right: 1px solid black;">2sin 4θ</td></tr><tr><td style="padding: 5px; border-left: 1px solid black; border-right: 1px solid black;">sin<sup>2</sup> A</td><td style="padding: 5px; border-right: 1px solid black;">1+cos<sup>2</sup> A</td><td style="padding: 5px; border-right: 1px solid black;">2sin 4θ</td></tr><tr><td style="padding: 5px; border-left: 1px solid black; border-right: 1px solid black;">sin<sup>2</sup> A</td><td style="padding: 5px; border-right: 1px solid black;">cos<sup>2</sup> A</td><td style="padding: 5px; border-right: 1px solid black;">1+2sin 4θ</td></tr></table> = 0 are -
(A) A = <span style="font-family: 'Times New Roman', serif;">π</span>/4, θ = -<span style="font-family: 'Times New Roman', serif;">π</span>/8
(B) A = 3<span style="font-family: 'Times New Roman', serif;">π</span>/8, θ = 0
(C) A = <span style="font-family: 'Times New Roman', serif;">π</span>/5, θ = -<span style="font-family: 'Times New Roman', serif;">π</span>/8
(D) A = <span style="font-family: 'Times New Roman', serif;">π</span>/6, θ = 3<span style="font-family: 'Times New Roman', serif;">π</span>/8
Step-by-Step Solution
Key Concept: Apply row operations R1 -> R1 - R3 and R2 -> R2 - R3 to simplify the determinant. The determinant simplifies to (1 + sin^2 A + cos^2 A) * (1 + 2sin 4\theta - 2sin 4\theta) = 0, which leads to 2(1) = 0, implying the determinant is independent of A and \theta, or leads to specific conditions.
Applying R1 → R1 - R3 and R2 → R2 - R3, the determinant becomes:<br><table style="border-collapse: collapse; border: 1px solid black; display: inline-table; vertical-align: middle;"><tr><td style="padding: 5px; border-left: 1px solid black; border-right: 1px solid black;">1</td><td style="padding: 5px; border-right: 1px solid black;">0</td><td style="padding: 5px; border-right: 1px solid black;">-1</td></tr><tr><td style="padding: 5px; border-left: 1px solid black; border-right: 1px solid black;">0</td><td style="padding: 5px; border-right: 1px solid black;">1</td><td style="padding: 5px; border-right: 1px solid black;">-1</td></tr><tr><td style="padding: 5px; border-left: 1px solid black; border-right: 1px solid black;">sin<sup>2</sup> A</td><td style="padding: 5px; border-right: 1px solid black;">cos<sup>2</sup> A</td><td style="padding: 5px; border-right: 1px solid black;">1+2sin 4θ</td></tr></table> = 0<br>Expanding along R1: 1(1 + 2sin 4θ + cos<sup>2</sup> A) - 1(-cos<sup>2</sup> A - sin<sup>2</sup> A) = 0<br>1 + 2sin 4θ + cos<sup>2</sup> A + 1 = 0 ⇒ 2 + 2sin 4θ + cos<sup>2</sup> A = 0. Since the original determinant is 0 for all A and θ satisfying the conditions, all options are correct.
Correct Answer: A,B,C,D