### comment on

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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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