Determinants 4.2

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×aa \times a in the form (M0,0M1,0..Ma,0M1,1M2,1......Ma,a...M2a,a)=A\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)=1det(A)=1

Or, if you rather...

det((00)(10)..(a0)(11)(21)......(aa)...(2aa))=1det \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...

#Determinant #Pascal'sTriangle #Matrix

Note by Aloysius Ng
6 years, 6 months ago

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