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Question Number 140888 by Ahmed1hamouda last updated on 13/May/21
Answered by physicstutes last updated on 14/May/21
(i)letM=[12−33−1541a2−14]fornosolutions,∣M∣=0thus|12−33−1541a2−14|=0⇒1|−151a2−14|−2|354a2−14|−3|3−141|=0⇒14−a2−5−2[3(a2−14)−20]−3(3+4)=0⇒14−a2−5−6a2+84+40−21=0⇒7a2=112⇒a2=16ora=±4
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