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Question : 35 of 160
Marks:
+1,
-0
Solution:
OA=−2−,OB=4+0−3 OC=+−,OD=−4−5 Given,
b=AB=OB−OA=3+2−2 c=AC=OC−OA=0+4+0 d=AD=OD−OA=−2−4 b+c=3+6−2,c+d=+2−4 d+b=4+0−6 AWKT;
[b×cc×da×b]=[bcd]2 Now, [b c d
]2=||2=(−40)2=1600 and
[b+cc+dd+b]=|| =|| (Applying
R1→R1+R2−R3) =−8(−6+16)=−8×10=−80 Hence,
‌| [b×cc×dd×b] |
| [b+cc+dd+b] |
=‌=−20
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