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Calculate the commutator ![$[A,B]$](http://images.physicslibrary.org/cache/objects/838/l2h/img1.png "$[A,B]$") of the matrices
Ans.
The commutator is calculated by definition as
Carrying out the first matrix multiplication gives
and the second multiplication is
Finally, subtracting the two yields
Since the commutator is non-zero, we say that  and  do not commute.
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![$\displaystyle A = \left[ \begin{array}{ccc} 5 & 3i & -i \ 1 & 0 & 2 \ i & 2... ...egin{array}{ccc} 1 & 0 & 1 \ -2 & 2i & 3 \ 3i & 1 & -1 \end{array} \right] $](http://images.physicslibrary.org/cache/objects/838/l2h/img2.png "$\displaystyle A = \left[ \begin{array}{ccc} 5 & 3i & -i \ 1 & 0 & 2 \ i & 2... ...egin{array}{ccc} 1 & 0 & 1 \ -2 & 2i & 3 \ 3i & 1 & -1 \end{array} \right] $")
![$\displaystyle [A,B] = AB-BA$](http://images.physicslibrary.org/cache/objects/838/l2h/img3.png "$\displaystyle [A,B] = AB-BA$")
![$\displaystyle AB = \left[ \begin{array}{ccc} 8-6i & -6-i & 5+10i \ 1+6i & 2 & -1 \ -6i & -5 & 1+7i \end{array} \right]$](http://images.physicslibrary.org/cache/objects/838/l2h/img4.png "$\displaystyle AB = \left[ \begin{array}{ccc} 8-6i & -6-i & 5+10i \ 1+6i & 2 & -1 \ -6i & -5 & 1+7i \end{array} \right]$")
![$\displaystyle BA = \left[ \begin{array}{ccc} 5+i & 5i & -1-i \ -10+5i & 0 & -3+6i \ 1+14i & -9-2i & 6 \end{array} \right].$](http://images.physicslibrary.org/cache/objects/838/l2h/img5.png "$\displaystyle BA = \left[ \begin{array}{ccc} 5+i & 5i & -1-i \ -10+5i & 0 & -3+6i \ 1+14i & -9-2i & 6 \end{array} \right].$")
![$\displaystyle [A,B] = AB-BA \left[ \begin{array}{ccc} 3-7i & -6-6i & 6+11i \ 11+i & 2 & 2-6i \ -1-20i & 4+2i & -5+7i \end{array} \right].$](http://images.physicslibrary.org/cache/objects/838/l2h/img6.png "$\displaystyle [A,B] = AB-BA \left[ \begin{array}{ccc} 3-7i & -6-6i & 6+11i \ 11+i & 2 & 2-6i \ -1-20i & 4+2i & -5+7i \end{array} \right].$")