Given `x = cy + bz`, `y = az + cx`, `z = bx + ay` where `x, y, z` are not all zero, then `a^2+b^2+c^2+2abc=`

Options

A. 0
B. 1
C. 2
D. none

Answer

To solve the problem given by the equations \( x = cy + bz \), \( y = az + cx \), and \( z = bx + ay \), we will express these equations in matrix form and find the determinant to derive the required expression.

### Step-by-Step Solution:

1. **Rewrite the equations**:
We start with the given equations:
\[ x - cy - bz = 0 \quad (1) \]
\[ -cx + y - az = 0 \quad (2) \]
\[ -bx - ay + z = 0 \quad (3) \]

2. **Formulate the matrix**:
We can express the system of equations in matrix form as follows:
\[ \begin{bmatrix} 1 & -c & -b \\ -c & 1 & -a \\ -b & -a & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]

3. **Calculate the determinant**:
We need to find the determinant of the matrix:
\[ D = \begin{vmatrix} 1 & -c & -b \\ -c & 1 & -a \\ -b & -a & 1 \end{vmatrix} \]

We can calculate the determinant using cofactor expansion:
\[ D = 1 \cdot \begin{vmatrix} 1 & -a \\ -a & 1 \end{vmatrix} - (-c) \cdot \begin{vmatrix} -c & -a \\ -b & 1 \end{vmatrix} - (-b) \cdot \begin{vmatrix} -c & 1 \\ -b & -a \end{vmatrix} \]

Now calculating each of these 2x2 determinants:
\[ \begin{vmatrix} 1 & -a \\ -a & 1 \end{vmatrix} = 1 \cdot 1 - (-a)(-a) = 1 - a^2 \]
\[ \begin{vmatrix} -c & -a \\ -b & 1 \end{vmatrix} = (-c)(1) - (-a)(-b) = -c - ab \]
\[ \begin{vmatrix} -c & 1 \\ -b & -a \end{vmatrix} = (-c)(-a) - (1)(-b) = ac + b \]

Substituting these back into the determinant:
\[ D = 1(1 - a^2) + c(c + ab) + b(ac + b) \]
\[ = 1 - a^2 + c^2 + abc + abc + b^2 \]
\[ = 1 - a^2 + b^2 + c^2 + 2abc \]

4. **Set the determinant to zero**:
Since \( x, y, z \) are not all zero, the determinant must equal zero:
\[ 1 - a^2 + b^2 + c^2 + 2abc = 0 \]

5. **Rearranging the equation**:
We can rearrange this to find the expression we need:
\[ a^2 + b^2 + c^2 + 2abc = 1 \]

### Final Answer:
Thus, the value of \( a^2 + b^2 + c^2 + 2abc \) is equal to 1.