A triangle determinant that is always zero?

Question

How do we prove, without actually expanding, that $$\\begin{vmatrix} \\sin {2A}& \\sin {C}& \\sin {B}\\\\ \\sin{C}& \\sin{2B}& \\sin {A}\\\\ \\sin{B}& \\sin{A}& \\sin{2C} \\end{vmatrix}=0$$ where $A,B,C$ are angles of a triangle? I tried adding and subtracting from the rows and columns and I even tried using the sine rule, but to no avail.

Answer 1

Matrix:

⌈ sin(2A) sin(C) sin(B) ⌉

| sin(C) sin(2B) sin(A) |

⌊ sin(B) sin(A) sin(2C)⌋

Using the double-angle identity sin 2θ =2 sin θ cos θ gives:

⌈ 2sin(A)cos(A) sin(C) sin(B) ⌉

| sin(C) 2sin(B)cos(B) sin(A) |

⌊ sin(B) sin(A) 2sin(C)cos(C) ⌋

We can apply the law of sines:

⌈ 2sin(A)cos(A) c*sin(A)/a b*sin(A)/a ⌉

| c*sin(B)/b 2sin(B)cos(B) a*sin(B)/b |

⌊ b*sin(C)/c a*sin(C)/c 2sin(C)cos(C) ⌋

So it has determinant [sin(A)sin(B)sin(C)]x where x is the determinant of the matrix below where the rows were divided by sin(A), sin(B), sin(C).

⌈ 2cos(A) c/a b/a ⌉

| c/b 2cos(B) a/b |

⌊ b/c a/c 2cos(C) ⌋

This new matrix has determinant y/(abc)² where y is the determinant of a matrix below where the columns were multiplied by bc, ac, and ab.

⌈ 2bc*cos(A) c² b² ⌉

| c² 2ac*cos(B) a² |

⌊ b² a² 2ab*cos(C) ⌋

Now substitute the laws of cosines.

⌈ b²+c²-a² c² b² ⌉

| c² a²+c²-b² a² |

⌊ b² a² a²+b²-c² ⌋

This final matrix has determinant 0 for all real a,b,c. You can show this by simply expanding it out, or by row reduction.