“…Now, by the same method as Example 4 we observe the det (π§ π΄ ) and det (π§ π΄ π ) where π΄ π = (π 1 , β¦ , π πβ1 , π, π π+1 , β¦ , π π ) for π β {1,2,3}.π§ π΄ = ( π§ π β² π§ π β² π§ 4 π§ 1 π§ π β² π§ π β² π§ π β² π§ 0 π§ π β² ),then we get det(π§ π΄ ) = π§5 , πππ(π΄) = 5 and π πππ(π΄) = +1. π§ π β² π§4 π§ 1 π§ π β² π§ π β² π§ 0 π§ 0 π§ π β² ),then we get πππ‘(π§ π΄ 1 ) = π§5 , πππ(π΄ 1 ) = 5 and π πππ(π΄1 ) = +1. π΄ 2 = ( πβ² πβ² 4 1 1 πβ² πβ² 0 πβ² ) , π§ π΄ 2 = ( π§ π β² π§ π β² π§ 4 π§ 1 π§ 1 π§ π β² π§ 0 π§ 0 π§ π β²…”