Let the given homogeneous system be interpreted as
{ x + y + 3 z = 0 − x + ( tan θ ) y + 7 z = 0 x + y + ( tan θ ) z = 0 \begin{cases}
x+y+\sqrt{3}\,z=0 \\
-x+(\tan\theta)y+\sqrt{7}\,z=0 \\
x+y+(\tan\theta)z=0
\end{cases} ⎩ ⎨ ⎧ x + y + 3 z = 0 − x + ( tan θ ) y + 7 z = 0 x + y + ( tan θ ) z = 0 We need the values of θ ∈ [ − π , π ] \theta\in[-\pi,\pi] θ ∈ [ − π , π ] non-trivial solution .
For a homogeneous system, non-trivial solutions exist iff the determinant of the coefficient matrix is zero.
1. Form the coefficient matrix
The coefficient matrix is
A = ( 1 1 3 − 1 tan θ 7 1 1 tan θ ) A=\begin{pmatrix}
1 & 1 & \sqrt{3}\\
-1 & \tan\theta & \sqrt{7}\\
1 & 1 & \tan\theta
\end{pmatrix} A = 1 − 1 1 1 tan θ 1 3 7 tan θ So we require
det ( A ) = 0. \det(A)=0. det ( A ) = 0. 2. Compute the determinant
Let t = tan θ t=\tan\theta t = tan θ
det ( 1 1 3 − 1 t 7 1 1 t ) \det\begin{pmatrix}
1 & 1 & \sqrt{3}\\
-1 & t & \sqrt{7}\\
1 & 1 & t
\end{pmatrix} det 1 − 1 1 1 t 1 3 7 t Expanding along the first row:
det ( A ) = 1 ∣ t 7 1 t ∣ − 1 ∣ − 1 7 1 t ∣ + 3 ∣ − 1 t 1 1 ∣ \det(A)=1\begin{vmatrix} t & \sqrt7 \\ 1 & t \end{vmatrix}
-1\begin{vmatrix} -1 & \sqrt7 \\ 1 & t \end{vmatrix}
+\sqrt3\begin{vmatrix} -1 & t \\ 1 & 1 \end{vmatrix} det ( A ) = 1 t 1 7 t − 1 − 1 1 7 t + 3 − 1 1 t 1 Now evaluate each minor:
∣ t 7 1 t ∣ = t 2 − 7 \begin{vmatrix} t & \sqrt7 \\ 1 & t \end{vmatrix}=t^2-\sqrt7 t 1 7 t = t 2 − 7 ∣ − 1 7 1 t ∣ = ( − 1 ) t − 7 = t + ( − 7 ? ) \begin{vmatrix} -1 & \sqrt7 \\ 1 & t \end{vmatrix}=(-1)t-\sqrt7=t+(-\sqrt7?) − 1 1 7 t = ( − 1 ) t − 7 = t + ( − 7 ?) More carefully,
( − 1 ) ( t ) − ( 7 ) ( 1 ) = − t − 7 (-1)(t)-(\sqrt7)(1)=-t-\sqrt7 ( − 1 ) ( t ) − ( 7 ) ( 1 ) = − t − 7 Hence the second term is
− 1 ( − t − 7 ) = t + 7 -1(-t-\sqrt7)=t+\sqrt7 − 1 ( − t − 7 ) = t + 7 And
∣ − 1 t 1 1 ∣ = ( − 1 ) ( 1 ) − t ( 1 ) = − 1 − t \begin{vmatrix} -1 & t \\ 1 & 1 \end{vmatrix}=(-1)(1)-t(1)=-1-t − 1 1 t 1 = ( − 1 ) ( 1 ) − t ( 1 ) = − 1 − t So the third term is
3 ( − 1 − t ) = − 3 ( 1 + t ) \sqrt3(-1-t)=-\sqrt3(1+t) 3 ( − 1 − t ) = − 3 ( 1 + t ) Therefore,
det ( A ) = t 2 − 7 + t + 7 − 3 ( 1 + t ) \det(A)=t^2-\sqrt7+t+\sqrt7-\sqrt3(1+t) det ( A ) = t 2 − 7 + t + 7 − 3 ( 1 + t ) The 7 \sqrt7 7
det ( A ) = t 2 + t − 3 − 3 t \det(A)=t^2+t-\sqrt3-\sqrt3 t det ( A ) = t 2 + t − 3 − 3 t det ( A ) = t 2 + ( 1 − 3 ) t − 3 \det(A)=t^2+(1-\sqrt3)t-\sqrt3 det ( A ) = t 2 + ( 1 − 3 ) t − 3 Thus
t 2 + ( 1 − 3 ) t − 3 = 0 t^2+(1-\sqrt3)t-\sqrt3=0 t 2 + ( 1 − 3 ) t − 3 = 0 3. Solve the quadratic in t t t
Factor:
t 2 + ( 1 − 3 ) t − 3 = ( t + 1 ) ( t − 3 ) t^2+(1-\sqrt3)t-\sqrt3=(t+1)(t-\sqrt3) t 2 + ( 1 − 3 ) t − 3 = ( t + 1 ) ( t − 3 ) So
( t + 1 ) ( t − 3 ) = 0 (t+1)(t-\sqrt3)=0 ( t + 1 ) ( t − 3 ) = 0 Hence
t = − 1 or t = 3 t= -1 \quad \text{or} \quad t=\sqrt3 t = − 1 or t = 3 That is,
tan θ = − 1 or tan θ = 3 \tan\theta=-1 \quad \text{or} \quad \tan\theta=\sqrt3 tan θ = − 1 or tan θ = 3 4. Find all θ ∈ [ − π , π ] \theta\in[-\pi,\pi] θ ∈ [ − π , π ]
Case 1: tan θ = − 1 \tan\theta=-1 tan θ = − 1
General solution:
θ = − π 4 + n π \theta=-\frac\pi4+n\pi θ = − 4 π + nπ Within [ − π , π ] [-\pi,\pi] [ − π , π ]
θ = − π 4 , 3 π 4 \theta=-\frac\pi4,\ \frac{3\pi}{4} θ = − 4 π , 4 3 π Case 2: tan θ = 3 \tan\theta=\sqrt3 tan θ = 3
General solution:
θ = π 3 + n π \theta=\frac\pi3+n\pi θ = 3 π + nπ Within [ − π , π ] [-\pi,\pi] [ − π , π ]
θ = π 3 , − 2 π 3 \theta=\frac\pi3,\ -\frac{2\pi}{3} θ = 3 π , − 3 2 π So
S = { − π 4 , 3 π 4 , π 3 , − 2 π 3 } S=\left\{-\frac\pi4,\ \frac{3\pi}{4},\ \frac\pi3,\ -\frac{2\pi}{3}\right\} S = { − 4 π , 4 3 π , 3 π , − 3 2 π } 5. Compute the sum of all values
∑ θ ∈ S θ = − π 4 + 3 π 4 + π 3 − 2 π 3 \sum_{\theta\in S}\theta
= -\frac\pi4+\frac{3\pi}{4}+\frac\pi3-\frac{2\pi}{3} θ ∈ S ∑ θ = − 4 π + 4 3 π + 3 π − 3 2 π First,
− π 4 + 3 π 4 = π 2 -\frac\pi4+\frac{3\pi}{4}=\frac\pi2 − 4 π + 4 3 π = 2 π and
π 3 − 2 π 3 = − π 3 \frac\pi3-\frac{2\pi}{3}=-\frac\pi3 3 π − 3 2 π = − 3 π Thus,
∑ θ ∈ S θ = π 2 − π 3 = π 6 \sum_{\theta\in S}\theta=\frac\pi2-\frac\pi3=\frac\pi6 θ ∈ S ∑ θ = 2 π − 3 π = 6 π Now,
120 π ∑ θ ∈ S θ = 120 π ⋅ π 6 = 20 \frac{120}{\pi}\sum_{\theta\in S}\theta
=\frac{120}{\pi}\cdot\frac\pi6=20 π 120 θ ∈ S ∑ θ = π 120 ⋅ 6 π = 20 6. Final answer
20 \boxed{20} 20 So the correct option is D .