Summation of Determinant

Let $ \large \Delta_r = \left| \begin{array}{ccc} f(r) & a & l \\ g(r) & b & m \\ h(r) & c & n \end{array} \right| $ ; Where a, b, c, l, m and n are constants, independent of r .

Then , $ \large \Sigma_{r=1}^{ n}\Delta_r = \left| \begin{array}{ccc} \Sigma_{r=1}^{ n}f(r) & a & l \\ \Sigma_{r=1}^{ n}g(r) & b & m \\ \Sigma_{r=1}^{ n}h(r) & c & n \end{array} \right| $ Here function of r can be the elements of only one row or one column.

Illustration : Let $ \large \Delta_a = \left| \begin{array}{ccc} a-1 & n & 6 \\ (a-1)^2 & 2n^2 & 4n-2 \\ (a-1)^3 & 3n^3 & 3n^2-3n \end{array} \right| $

Show that : $\large \Sigma_{a=1}^{n} \Delta_a = 0$

Solution: By adding all the corresponding elements of C1 of all determinants Δa We have

$ \large \Sigma_{a=1}^{n} \Delta_a = \left| \begin{array}{ccc}\Sigma_{a=1}^{n}( a-1) & n & 6 \\ \Sigma_{a=1}^{n}(a-1)^2 & 2n^2 & 4n-2 \\ \Sigma_{a=1}^{n}(a-1)^3 & 3n^3 & 3n^2-3n \end{array} \right| $

$ \large = \left| \begin{array}{ccc} \frac{(n-1)n}{2} & n & 6 \\ \frac{(n-1)n(2n-1)}{6} & 2n^2 & 4n-2 \\ \frac{(n-1)^2 n^2}{4} & 3n^3 & 3n^2-3n \end{array} \right| $

By taking n(n-1)/2 as common factor from C1 and , we get

$ \large = \frac{n(n-1)}{2}\left| \begin{array}{ccc} 1 & n & 6 \\ \frac{(2n-1)}{3} & 2n^2 & 4n-2 \\ \frac{(n-1) n}{2} & 3n^3 & \frac{n(n-1)}{2} \end{array} \right| $

By taking 6 as common factor from C3

$ \large = 3 n(n-1)\left| \begin{array}{ccc} 1 & n & 1 \\ \frac{(2n-1)}{3} & 2n^2 & \frac{2n-1}{3} \\ \frac{(n-1) n}{2} & 3n^3 & 3n^2-3n \end{array} \right| $

Since C1 and C3 are identical

$\large \Sigma_{a=1}^{n} \Delta_a = 0$

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