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Question : 6 of 160
Marks:
+1,
-0
Solution:
A=[| 5 | sin‌2θ | cos2θ |
| −sin‌2θ | −5 | 1 |
| cos2θ | 1 | 5 |
] |A|=5(−25−1)−sin‌2θ(−5sin‌2θ−cos2θ)+cos2θ(−sin‌2θ+5cos2θ)
=−130−sin‌2θ[−5‌(1−cos2θ)−cos2θ]‌+cos2θ[−sin‌2θ+5(1−sin‌2θ)]
=−130−sin‌2θ(−5+4cos2θ)+cos2θ(−6sin‌2θ+5)
=−130+5sin‌2θ−4sin‌2θcos2θ−6sin‌2θcos2θ+5cos2θ
‌=−130+5−10sin‌2θcos2θ‌‌‌|A|=−125−10sin‌2θcos2θ‌∴‌‌|Amax=−125
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