# Trying to understand MixColumns with Hexadecimal matrix

I am trying to read and understand the textbook's exercise questions. The sample solution at the end the textbook has a final answer and it does not clarify how it got the result. I am trying to workout the question but I am not successful. Any help would be appreciated.

Questions: Show the value of State after MixColumns:

$\begin{bmatrix}7C & 6B & 01 & D7\\F2 & 30 & FE & 63\\2B & 76 & 7B & C5\\AB & 77 & 6F &67\end{bmatrix}$

$\begin{bmatrix}75 & 87 & 0F & B2\\55 & E6 & 04 & 22\\3E & 2E & B8 & 8C\\10 & 15 & 58 & 0A\end{bmatrix}$

I used the path that has been mentioned here to find $2\times 0x7C$ and $3\times 0xF2$.

$\begin{bmatrix} 2&3&1&1\\1&2&3&1\\1&1&2&3\\3&1&1&2\end{bmatrix} \times \begin{bmatrix}7C \\ F2 \\ 2B \\ AB\end{bmatrix} = \begin{bmatrix} ?\\?\\?\\?\end{bmatrix}$

    7C = 0111 1100
2 * 7C = 1111 1000

F2 = 1111 0010
2 * F2 = 1110 0100
3 * F2 = 0001 0110

2B = 0010 1011
AB = 1010 1011
-----------------------

Now find value for Cell (1,1)

2 * 7C = 1111 1000
3 * F2 = 0001 0110
2B = 0010 1011
XOR     AB = 1010 1011
--------------
0110 1110 = 0x6E <-- is not equal to 0x75


That's your problem; 2 * F2 is not F2 shifted left one. Instead, because the msbit is set, it is F2 shifted left one, xor'ed with the feedback polynomial (1B).
So, 2 * F2 = 1110 0100 $\oplus$ 0001 1011 = 1111 1111