Equation root in determinant

Algebra Level pending

Let x 1 x_1 , x 2 x_2 , x 3 x_3 and x 4 x_4 are the roots of the equation x 4 + p x 2 + q x + r = 0 x^4+px^2+qx+r=0 . Find the value of the determinant

∣ x 1 x 2 x 3 x 4 x 2 x 3 x 4 x 1 x 3 x 4 x 1 x 2 x 4 x 1 x 2 x 3 ∣ \begin{vmatrix}\ x_1 & x_2 & x_3 & x_4 \\ \\ x_2 & x_3 & x_4 & x_1 \\ \\ x_3 & x_4 & x_1 & x_2 \\ \\ x_4 & x_1 & x_2 & x_3\ \end{vmatrix} ,


The answer is 0.

Danish Ahmed by Danish Ahmed , 24, India

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1 solution

Mark Hennings
Nov 26, 2020

Adding the second, third and fourth rows to the first, we see that ∣ x 1 x 2 x 3 x 4 x 2 x 3 x 4 x 1 x 3 x 4 x 1 x 2 x 4 x 1 x 2 x 3 ∣ = ∣ x 1 + x 2 + x 3 + x 4 x 1 + x 2 + x 3 + x 4 x 1 + x 2 + x 3 + x 4 x 1 + x 2 + x 3 + x 4 x 2 x 3 x 4 x 1 x 3 x 4 x 1 x 2 x 4 x 1 x 2 x 3 ∣ = ( x 1 + x 2 + x 3 + x 4 ) ∣ 1 1 1 1 x 2 x 3 x 4 x 1 x 3 x 4 x 1 x 2 x 4 x 1 x 2 x 3 ∣ \begin{aligned} \left| \begin{array}{cccc} x_1 & x_2 & x_3 & x_4 \\ x_2 & x_3 & x_4 & x_1 \\ x_3 & x_4 & x_1 & x_2 \\ x_4 & x_1 & x_2 & x_3 \end{array}\right| & = \; \left| \begin{array}{cccc} x_1 + x_2 + x_3 + x_4 & x_1 + x_2 + x_3 + x_4 & x_1 + x_2 + x_3 + x_4 & x_1 + x_2 + x_3 + x_4 \\ x_2 & x_3 & x_4 & x_1 \\ x_3 & x_4 & x_1 & x_2 \\ x_4 & x_1 & x_2 & x_3 \end{array}\right| \\ & = \; (x_1 + x_2 + x_3 + x_4) \left| \begin{array}{cccc} 1 & 1 & 1 & 1 \\ x_2 & x_3 & x_4 & x_1 \\ x_3 & x_4 & x_1 & x_2 \\ x_4 & x_1 & x_2 & x_3 \end{array}\right| \end{aligned} Since x 1 + x 2 + x 3 + x 4 = 0 x_1 + x_2 + x_3 + x_4 = 0 , this determinant is equal to 0 \boxed{0} .

It is interesting to note that the determinant factorises as − ( x 1 + x 2 + x 3 + x 4 ) ( x 1 + i x 2 − x 3 − i x 4 ) ( x 1 − x 2 + x 3 − x 4 ) ( x 1 − i x 2 − x 3 + i x 4 ) -(x_1 + x_2 + x_3 + x_4)(x_1 + ix_2 - x_3 -ix_4)(x_1 - x_2 + x_3 - x_4)(x_1 -ix_2 - x_3 +ix_4)

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