NDA 2022
Exam NDA
Year 2022
Subject Mathematics
Chapter Determinants NDA 2022 · Mathematics · Determinants
Paragraph: Let A= ( array ccc 0 & ^2 θ & ^2 θ ^2 θ & 0 & ^2 θ ^2 θ...
Paragraph:
Let
A = A= A = \left(
0 sin 2 θ cos 2 θ cos 2 θ 0 sin 2 θ sin 2 θ cos 2 θ 0 \begin{array}{ccc}0 & \sin ^2 \theta & \cos ^2 \theta \\ \cos ^2 \theta & 0 & \sin ^2 \theta \\ \sin ^2 \theta & \cos ^2 \theta & 0\end{array} 0 cos 2 θ sin 2 θ sin 2 θ 0 cos 2 θ cos 2 θ sin 2 θ 0 \right)
a n d and an d A=P+Q
w h e r e where w h ere P
i s s y m m e t r i c m a t r i x a n d is symmetric matrix and i ssy mm e t r i c ma t r i x an d Q
i s s k e w − s y m m e t r i c m a t r i x . < b r > < s t r o n g > Q u e s t i o n : < / s t r o n g > W h a t i s t h e m i n i m u m v a l u e o f d e t e r m i n a n t o f is skew-symmetric matrix.<br><strong>Question:</strong>
What is the minimum value of determinant of i ss k e w − sy mm e t r i c ma t r i x . < b r >< s t ro n g > Q u es t i o n :< / s t ro n g > Wha t i s t h e minim u m v a l u eo fd e t er minan t o f A ?
Options A 1 4 \frac{1}{4} 4 1 B 1 2 \frac{1}{2} 2 1 C 3 4 \frac{3}{4} 4 3 D 1 1 1 Correct answer A. 1 4 \frac{1}{4} 4 1
Step-by-step solution
∣ A ∣ = [ 0 sin 2 θ cos 2 θ cos 2 θ 0 sin 2 θ sin 2 θ cos 2 θ 0 ] = 0 − sin 2 θ ( 0 − sin 4 θ ) + cos 2 θ ( cos 4 θ ) = sin 6 θ + cos 6 θ = ( sin 2 θ + cos 2 θ ) ( sin 4 θ + cos 4 θ − sin 2 θ ⋅ cos 2 θ ) = ( sin 2 θ + cos 2 θ ) 2 − 3 sin 2 θ cos 2 θ = 1 − 3 4 sin 2 2 θ = 1 4 [ 4 − 3 ( 1 − cos 4 θ ) 2 ] = 1 8 [ 5 + 3 cos 4 θ ] \begin{aligned}
& |A|=\left[\begin{array}{ccc}
0 & \sin ^2 \theta & \cos ^2 \theta \\
\cos ^2 \theta & 0 & \sin ^2 \theta \\
\sin ^2 \theta & \cos ^2 \theta & 0
\end{array}\right] \\
& =0-\sin ^2 \theta\left(0-\sin ^4 \theta\right)+\cos ^2 \theta\left(\cos ^4 \theta\right) \\
& =\sin ^6 \theta+\cos ^6 \theta \\
& =\left(\sin ^2 \theta+\cos ^2 \theta\right)\left(\sin ^4 \theta+\cos ^4 \theta-\sin ^2 \theta \cdot \cos ^2 \theta\right) \\
& =\left(\sin ^2 \theta+\cos ^2 \theta\right)^2-3 \sin ^2 \theta \cos ^2 \theta \\
& =1-\frac{3}{4} \sin ^2 2 \theta=\frac{1}{4}\left[4-3 \frac{(1-\cos 4 \theta)}{2}\right] \\
& =\frac{1}{8}[5+3 \cos 4 \theta]
\end{aligned} ∣ A ∣ = 0 cos 2 θ sin 2 θ sin 2 θ 0 cos 2 θ cos 2 θ sin 2 θ 0 = 0 − sin 2 θ ( 0 − sin 4 θ ) + cos 2 θ ( cos 4 θ ) = sin 6 θ + cos 6 θ = ( sin 2 θ + cos 2 θ ) ( sin 4 θ + cos 4 θ − sin 2 θ ⋅ cos 2 θ ) = ( sin 2 θ + cos 2 θ ) 2 − 3 sin 2 θ cos 2 θ = 1 − 4 3 sin 2 2 θ = 4 1 [ 4 − 3 2 ( 1 − cos 4 θ ) ] = 8 1 [ 5 + 3 cos 4 θ ]
Minimum value of
cos 4 θ \cos 4 \theta cos 4 θ is -1 .
∴ ∣ A ∣ min = 1 8 [ 5 − 3 ] = 2 8 = 1 4 \therefore \quad|A|_{\min }=\frac{1}{8}[5-3]=\frac{2}{8}=\frac{1}{4} ∴ ∣ A ∣ m i n = 8 1 [ 5 − 3 ] = 8 2 = 4 1
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