[A] = [ ] Solution I will write S{i=1,n} the summing symbol, and P{i=1,n} the pr

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I will write S{i=1,n} the summing symbol, and P{i=1,n} the product symbol: First positive diagonal is: D1 = (x+a1)(x+a2)...(x+an) , or D1 = P{i=1,n} (x+ai) All other positive diagonals (there are n-1) are like: D2 to Dn = a1.a2.a3....an. So for i > 1, Di = P{i=1,n} ai And the total positive side of this determinant is: D+ = [P{i=1,n} (x+ai)] + [(n-1) P{i=1,n} ai] Negative diagonals all include one (x+ai) term and the rest of aj terms as : N1= - (x+a1).a2.a3....an N2= - a1.a2.a3.....(x+a[n-1])).an alternating the (x+ai) at the beginning for ODD terms and at the end of the product for EVEN terms, but in the end they all look alike and cover all the values of i. So the negative terms of the determinant will add up to: D- = S{i=1,n} [(x+ai) P{j=1,n,j not = i} aj] Det = D+ - D-

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