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Show that`|1 1+p1+p+q2 3+2p1+3p+2q3 6+3p 106 p+3q|=1.`

Answer» `" Let " Delta =|{:(1,,1+p,,1+p+q),(2,,3+2p,,1+3p+2q),(3,,6+3p,,1+6p+3q):}|`
Applying `C_(2) to C_(2) -pC " and C_(3) to C_(3) -qC_(1)` we get
`Delta= |{:(1,,1,,1+p),(2,,3,,1+3),(3,,6,,1+6p):}|`
`=|{:(1,,1,,1),(2,,3,,1),(3,,6,,1):}|" ""[Applying "C_(3)to C_(3) -pC_(3)"]"`
`=|{:(0,,0,,1),(1,,2,,1),(2,,5,,1):}|" ""[Applying "C_(1)to C_(1)-C_(3),C_(2)to C_(2)-"]"`
`=5-4=1" ""[Expanding ]"`


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