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How to prove that the square of third order determinant is equal to the determinant of cofactors of its terms using the properties of determinant?

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    What is "a third order determinant", anyway? – 2017-01-22
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    I meant "Determinant of order 3". A one consisting of 3 rows and 3 columns. – 2017-01-22
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    Ok, so determinant of a $\;3\times3\;$ matrix. Thanks....but then I can't see how the square of such a matrix equals "the determinant of cofactors of its terms" (I am guessing here you meant "the determinant of its cofactors")...What I honestly think is that you mean why a $\;3\times3\;$ matrix's determinant (not its square!) can be evaluated by the diagonals' trick. Anyway, do this and do the determinant developed by some row of column and check they're the same – 2017-01-22
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    @Don, I think the question is why, if you replace each entry by its cofactor (that is, by the determinant of the 2 x 2 matrix you get by deleting the row and column containing the entry), do you get the square of the determinant of the original matrix. – 2017-01-22
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    @GerryMyerson Oh, now that makes sense, Gerry. Thanks. And it even is true...writing a hint: – 2017-01-22
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    @Gerry, Thanks for the interpretation. – 2017-01-22

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Denote by $\;\mathcal B\;$ the cofactors matrix, which is just the classical adjoint (or adjugate) of $\;A\;$ . Observe that the claim is trivial if $\;A\;$ is singular $\;\iff |A|=0\iff|\mathcal B|=0\iff\mathcal B\;$ is singular , so assume $\;A\;$ it is regular, and then we have

$$A\cdot\mathcal B=|A|I\implies|A||\mathcal B|=|A|^n$$

Finish now the argument using that $\;n=3\;$ in your case...