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If 2^N could become m^n, then you could use a recursive solution: (Updated: cleaned up slightly!)

#! perl -slw use strict; sub expand{ return @{ $_[0] } if @_ == 1; map{ my $x = $_; map{ $x * $_ } &expand; } @{ pop @_ }; } my @mat = ( [ -1, 2 ], [ 3, -4 ] ); printf +("%s " x @mat ) . '= ', map{ "[ @$_ ] " } @mat; print join ' ', expand( @mat ); @mat = ( [ 1, 2 ], [ 3, 4 ], [ 5, 6 ], ); printf +("%s " x @mat ) . '= ', map{ "[ @$_ ] " } @mat; print join ' ', expand( @mat ); @mat = ( [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ], ); printf +("%s " x @mat ) . '= ', map{ "[ @$_ ] " } @mat; print join ' ', expand( @mat ); __END__ P:\test>junk [ -1 2 ] [ 3 -4 ] = -3 6 4 -8 [ 1 2 ] [ 3 4 ] [ 5 6 ] = 15 30 20 40 18 36 24 48 [ 1 2 3 ] [ 4 5 6 ] [ 7 8 9 ] = 28 56 84 35 70 105 42 84 126 32 64 96 40 80 120 48 96 144 36 72 108 45 + 90 135 54 108 162

Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error.
Lingua non convalesco, consenesco et abolesco. -- Rule 1 has a caveat! -- Who broke the cabal?
"Science is about questioning the status quo. Questioning authority".
In the absence of evidence, opinion is indistinguishable from prejudice.

In reply to Re^3: Obtaining terms in an expansion by BrowserUk
in thread Obtaining terms in an expansion by randyk

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