"When I removed the "\" it works just fine. I can not exaclty (sic) understand the error or the reason...."

The backslash "\" you removed was escaping the "@" sigil, identifying your variable. That prevented Perl from understanding what you wanted.

Without getting into referencing and dereferencing, observe what happens with a simple array:

#!/usr/bin/perl
use 5.016;
use strict;
use warnings;

my @foo = ("bar", "baz", "bat");
say "\@foo with the backslash is: \@foo";

say "\@foo without the slash is @foo";


=head

C:\>thanos.pl
@foo with the backslash is: @foo  # i.e., the escape makes it print as
+ a literal "@" here
@foo without the slash is bar baz bat  # but here, Perl gives you the 
+values in @foo!


=cut
[download]

In other words, be alert to the different uses of the "\" and the "/".

For your precise case (and you seem to have done most of your homework here), see:

1. perldoc perlref read down to referencing and dereferencing.
2. References quick reference and look for the slash