D=∣12126−2−3−117∣
=20−2(25)−3(−10)
=20−50+30=0
D1=∣abc26−2−3−117∣
=20a−2(7b+11c)−3(−2b−6c)
=20a−14b−22c+6b+18c
=20a−8b−4c
=4(5a−2b−c)
D2=∣121abc−3−117∣
=7b+11c−a(25)−3(2c−b)
=7b+11c−25a−6c+3b
=−25a+10b+5c
=−5(5a−2b−c)
D3=∣12126−2abc∣
=6c+2b−2(2c−b)−10a
=−10a+4b+2c
=−2(5a−2b−c)
for infinite solution
D=D1=D2=D3=0
⇒5a=2b+c