<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 did everything right except you need to go one more step:
$a,b,c,d$ are ordinary numbers and they commute multiplicatively, and they commute with $i$ as well. So $bibi = b^2i^2 = -b^2$ and
$$
a^2 - bibi + c^2 -didi = a^2+b^2+c^2+d^2 = t
$$ 
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