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![[Post New]](/templates/default/images/icon_minipost_new.gif) 3 May 2007 21:25:29 IST
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wot i think is.........may be u think its silly but......det(adjA) =lAl^(n-1)
nd we can also prove det {adj(adjA)} = lAl^ (n-1)^2
sooo.........similarly det{adj(adj.........A}} = lAl^(n-1)^r
where n is the order of determinant
hey tell me if u r getting the ans frm dis
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