use awk to match rows for each column - bash

Question

How can awk be used to find values that match in row 2 for each column?
I would like to take in a tab limited file and for each column if any row below row 2 matches what is in row 2, print field with "match".
transforming this tab delimited file:
header1 | header2 | header3
1 | 1 | B
--------+---------+----------
3 | 1 | A
2 | A | B
1 | B | 1
To this:
header1 | header2 | header3
1 | 1 | B
--------+---------+----------
3 | 1 match | A
2 | A | B match
1 match | B | 1

Answer 1

I would go for something like this:
$ cat file
header1 header2 header3
1 1 B
3 1 A
2 A B
1 B 1
$ awk -v OFS='\t' 'NR == 2 { for (i=1; i<=NF; ++i) a[i] = $i }
NR > 2 { for(i=1;i<=NF;++i) if ($i == a[i]) $i = $i " match" }1' file
header1 header2 header3
1 1 B
3 1 match A
2 A B match
1 match B 1
On the second line, populate the array a with the contents of each field. On subsequent lines, add "match" when they match the corresponding value in the array. The 1 at the end is a common shorthand causing each line to be printed. Setting the output field separator OFS to a tab character preserves the format of the data.

Answer 2

Pedantically, with GNU Awk 4.1.1:
awk -f so.awk so.txt
header1 header2 header3
1 1 B
3 1* A
2 A B*
1* B 1
with so.awk:
{
if(1 == NR) {
print $0;
} else if(2 == NR) {
for(i = 1; i <= NF; i++) {
answers[i]=$i;
}
print $0;
} else {
for(i = 1; i <= NF; i++) {
field = $i;
if(answers[i]==$i) {
field = field "*" # a match
}
printf("%s\t",field);
}
printf("%s", RS);
}
}
and so.txt as a tab delimited data file:
header1 header2 header3
1 1 B
3 1 A
2 A B
1 B 1
This isn't homework, right...?