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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) let M = [(1,2,(−3)),(3,(−1),5),(4,1,(a^2 −14)) ] for no solutions , ∣M∣ = 0 thus determinant ((1,2,(−3)),(3,(−1),5),(4,1,(a^2 −14)))= 0 ⇒ 1 determinant (((−1),5),(1,(a^2 −14)))−2 determinant ((3,5),(4,(a^2 −14)))−3 determinant ((3,(−1)),(4,1))=0 ⇒ 14−a^2 −5 −2[3(a^2 −14)−20]−3(3+4) = 0 ⇒ 14−a^2 −5−6a^2 +84+40−21 = 0 ⇒ 7a^2 =112 ⇒ a^2 = 16 or a = ±4

(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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