After the 4 by 4 and 5 by 5 matrices of Pascal's triangle, we realise that the determinant of the matrix is 1.

Let $M_{m,n} = \left( \begin{array}{ccc} m \\ n \\\end{array} \right)$

Prove that for any matrix of size $a \times a$ in the form $\left( \begin{array}{ccc}M_{0,0} & M_{1,0} & . & . & M_{a,0}\\M_{1,1} & M_{2,1} & & & \\ . & & . & & .\\ . & & & . & .\\ M_{a,a} & . & . & . & M_{2a,a} \end{array} \right)=A$

$det(A)=1$

Or, if you rather...

$det \left( \begin{array}{ccc}\left( \begin{array}{ccc} 0 \\ 0 \\\end{array} \right) & \left( \begin{array}{ccc} 1 \\ 0 \\\end{array} \right) & . & . & \left( \begin{array}{ccc} a \\ 0 \\\end{array} \right) \\\left( \begin{array}{ccc} 1 \\ 1 \\\end{array} \right) & \left( \begin{array}{ccc} 2 \\ 1 \\\end{array} \right) & & & \\ . & & . & & .\\ . & & & . & .\\ \left( \begin{array}{ccc} a \\ a \\\end{array} \right) & . & . & . & \left( \begin{array}{ccc} 2a \\ a \\\end{array} \right) \end{array} \right)=1$

But, there a few methods which may help a lot and will be shared next...

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