Is Finite Field Multiplication Distributive? Moving Affine Transform in AES
From the above question $m = da_0$

At the bit level:

$m\oplus b = [m_7 \oplus b_7 , m_6 \oplus b_6 , m_5 \oplus b_5 , .... , m_0 \oplus b_0 ]$

$
\begin{align*}
2(m\oplus b) =&amp; \operatorname{xtime}(2(m\oplus b)) = \\
=&amp; [m_6 \oplus  b_6 , m_5\oplus b_5 , m_4\oplus b_4, m_3\oplus b_3 \oplus (m_7\oplus b_7), m_2\oplus b_2 \oplus (m_7\oplus b_7) ,\\&amp; m_1\oplus  b_1,m_0
\oplus b_0 \oplus (m_7\oplus b_7),m_7\oplus b_7]\\
\end{align*}
$

Where the multiplication by $02$ is denoted $\operatorname{xtime}(x)$.

$
\begin{align*}
2m =\;&amp; [m_6 ,m_5,m_4,m_3\oplus m_7,m_2\oplus m_7,m_1,m_0\oplus m_7,m_7]\\
2b =\;&amp; [b_6 ,b_5,b_4,b_3\oplus b_7,b_2\oplus b_7,b_1,b_0\oplus m_7,m_7]\\
\end{align*}
$

It follows that:

$2m\oplus 2b = 2(m\oplus b)$

because they result in the same formula at the bit level. You can see a similar result for multiplication by 3.

However, the question also assumes that it is possible to convert $2da_0 \oplus  3da_1 \oplus  da_2 \oplus  da_3$ to $d(2a_0\oplus 3da_1 \oplus  da_2 \oplus  da_3)$ this is not possible because it would require the original to be written as $d2a_0 \oplus  d3a_1 \oplus  da_2 \oplus  da_3$ but matrix multiplication is not  commutative.