Preparing a multiple choice test Introduction
Note: this is not some kind of method to encourage cheating. As Cᴏɴᴏʀ O'Bʀɪᴇɴ already said, studying is the best solution for passing a test :3.
Consider the following answers to the multiple choice test:
ABCBCAAB

Here is a table which indicates whether the answers match:
    A B C B C A A B

A   1 0 0 0 0 1 1 0
B   0 1 0 1 0 0 0 1
C   0 0 1 0 1 0 0 0

This gives us the following numbers:
10000110, 01010001, 00101000

The challenge is to print these binary numbers. But it is important to see which letters are used in the multiple choice test. For example:
ABCDCDBCDABC

This highest letter is D, which is the 4th letter in the alphabet. Therefore, we need to output 4 different binary numbers. Namely:
100000000100, 010000100010, 001010010001, 000101001000

Note that you have to look at the highest letter. Consider the following example:
AACCACAC

Although the B is not used, we need to output the binary result for B. That means, the answer would be:
11001010, 00000000, 00110101


Task
Given the answers to a multiple choice test, output the binary numbers. You may assume that the input will be non-empty and only containing the letters [A-Z]
. Instead of ones and zeroes, you may also use true and false.

Test cases:
Input: ABBBABABA
Output: 100010101, 011101010

Input: AAAAAAAAA
Output: 111111111

Input: AFGHEEHFD
Output: 100000000 000000000 000000000 000000001 000011000 010000010 001000000 000100100

Input: Z
Output: 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1

Input: ABCDCDBCDABC
Output: 100000000100, 010000100010, 001010010001, 000101001000

This is code-golf, so the submission with the least amount of bytes wins!
Python 3, 71
Saved 22 bytes thanks to Ogaday.
Saved 3 bytes thanks to DSM.
Saved a bunch of bytes thanks to an array of bools being valid.
*k,=map(ord,input())
x=65
while x&lt;=max(k):print([v==x for v in k]);x+=1

Takes uppercase command line input.