<s>[INST] Conjugate of Quaternion 
The conjugate of $$\alpha=\left[ {\begin{array}{cc}
   a+bi & c+di \\
   -c+di & a-bi \\
  \end{array} } \right]$$ is$$\overline{\alpha}=\left[ {\begin{array}{cc}
   a-bi & -c-di \\
   c-di & a+bi \\
  \end{array} } \right]$$
The norm of $\alpha$ is $a^2+b^2+c^2+d^2$ and is written $\lvert\lvert\alpha\rvert\rvert$. Show directly that $$\overline{\alpha}\alpha=\alpha\overline{\alpha}=\left[ {\begin{array}{cc}
   t & 0 \\
   0 & t \\
  \end{array} } \right]$$ where $t= \lvert\lvert\alpha\rvert\rvert$.
  Conclude that the multiplicative inverse of $\alpha$ is $(1/t)\overline{\alpha}$.

So to show directly I should do matrix multiplication, so
$\overline{\alpha}\alpha=\left[ {\begin{array}{cc}
   a-bi & -c-di \\
   c-di & a-bi \\
  \end{array} } \right]\cdot \left[ {\begin{array}{cc}
    a+bi & c+di \\
   -c+di & a-bi \\
  \end{array} } \right] = \left[ {\begin{array}{cc}
    (a - bi) (a + bi) + (c - di) (c + di) & (a + bi) (-c - di) + (a + bi) (c + di) \\
   (a - bi) (c - di) + (a - bi) (di - c) & (a - bi) (a + bi) + (-c - di) (di - c) \\
  \end{array} } \right] = \left[ {\begin{array}{cc}
    a^2-(bi)^2+c^2-(di)^2 & 0 \\
   0 & a^2-(bi)^2+c^2-(di)^2 \\
  \end{array} } \right] = \left[ {\begin{array}{cc}
    a^2-(bibi)+c^2-(didi) & 0 \\
   0 & a^2-(bibi)+c^2-(didi) \\
  \end{array} } \right]$. 
EDIT: Since complex number commute I am able to conclude that $\overline{\alpha}\alpha=\alpha\overline{\alpha}= \left[ {\begin{array}{cc}
    t & 0 \\
   0 & t \\
  \end{array} } \right]$.
 [/INST] You do not need to bother with matrix multiplication being commutative.  Once you have figured out the matrix product, multiplication of the complex numbers within each component of that product is still commutative.  What doesn't always commute is multiplication of one matrix by another, which you are already done with.
So, in your product, $(bi)^2$ in the upper left corner is just a product of complex numbers, not any matrices, and that commutes, so it equals $b^2i^2=-b^2$.  Similarly for $(di)^2$ in the same component and the imaginary number squares in the lower right corner.
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