
Notice how [A3] = 8 and [B1] = 5. Following Suit, [A1] = 5. Hold off on filling in the 3 in [C3] as I want to demonstrate the rule.
After the solutions indicated above, we are left with [B3] = (6,7), [C3] = (6,7,3) and [C1] = 3. The rule applies here as follows. If there are 2 squares the have the (6,7) combination in them making these the only two squares that can possibly have those two numbers, we can eliminate the 3 in [C3] even being an option. Now that is a little brainwashing, but it really works, as long as you stay within a row, column or box. Thus [C1] = 3.
Then filling in the obvious squares: [A4] = 5 we will conclude for today.
Here is our puzzle as of today.

Moving forward with some "Matched Pair Magic" I will be placing a 2 in [D3], and a 9 in [F7] next time.
The Doctor.
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