Then there must be x 1 (a, b, c, d)+x2 (b, -a, d, -c)+x3 (c, -d, -a, b)+x4 (d, c, -b, -a) = 0.

The conversion is ("."means multiplication):

a.x 1+b.x2+c.x3+d.x4=0

-a.x2+b.x 1+c.x4-d.x3=0

-a.x3-b.x4+c.x 1+d.x2=0

-a.x4+b.x3-c.x2+d.x 1=0

In this system of equations, because A, B, C and D are not equal to zero, there is a non-zero solution, so the corresponding determinant is equal to zero.

| x 1 x2 x3 x4|

|-x2 x 1 x4 -x3|=0

|-x3 -x4 x 1 x2|

|-x4 x3 -x2 x 1|

X 1=x2=x3=x4=0 can be solved, so a 1, A2, A3 and A4 are linearly independent.

Certificate of completion

By the way, two conclusions:

1.n unknowns The homogeneous equation of n equations has only zero solution < = > The determinant is not equal to zero.

2. Homogeneous equations with n unknowns and N equations with nonzero solutions.

I hope it helps you.

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