Let a,b,c, be any real number. Suppose that there are real numbers x,y,z not all zero such that x=cy+bz,y=az+cx and z=bx+ay. Then
`a^(2)+b^(2)+c^(2)` +2abc is equal to

Options

A. -1
B. 0
C. 1
D. 2

Answer

To solve the problem, we need to analyze the given equations and derive the required expression step by step.

### Step 1: Write the equations in standard form
We are given the equations:
1. \( x = cy + bz \)
2. \( y = az + cx \)
3. \( z = bx + ay \)

We can rewrite these equations in the form of a system of linear equations:
1. \( x - cy - bz = 0 \)
2. \( -cx + y - az = 0 \)
3. \( -bx + -ay + z = 0 \)

### Step 2: Set up the coefficient matrix
The system can be represented in matrix form as:
\[ \begin{bmatrix} 1 & -c & -b \\ -c & 1 & -a \\ -b & -a & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]

### Step 3: Calculate the determinant
For the system to have a non-trivial solution (where \(x, y, z\) are not all zero), the determinant of the coefficient matrix must be zero. We calculate the determinant:
\[ D = \begin{vmatrix} 1 & -c & -b \\ -c & 1 & -a \\ -b & -a & 1 \end{vmatrix} \]

### Step 4: Expand the determinant
Using the determinant expansion formula:
\[ D = 1 \cdot \begin{vmatrix} 1 & -a \\ -a & 1 \end{vmatrix} - (-c) \cdot \begin{vmatrix} -c & -a \\ -b & 1 \end{vmatrix} - (-b) \cdot \begin{vmatrix} -c & 1 \\ -b & -a \end{vmatrix} \]

Calculating the 2x2 determinants:
1. \( \begin{vmatrix} 1 & -a \\ -a & 1 \end{vmatrix} = 1 \cdot 1 - (-a)(-a) = 1 - a^2 \)

2. \( \begin{vmatrix} -c & -a \\ -b & 1 \end{vmatrix} = (-c)(1) - (-a)(-b) = -c - ab \)

3. \( \begin{vmatrix} -c & 1 \\ -b & -a \end{vmatrix} = (-c)(-a) - (1)(-b) = ac + b \)

Putting it all together:
\[ D = 1(1 - a^2) + c(c + ab) + b(ac + b) \]
\[ D = 1 - a^2 + c^2 + abc + abc + b^2 \]
\[ D = 1 - a^2 - b^2 - c^2 + 2abc \]

### Step 5: Set the determinant to zero
For a non-trivial solution:
\[ 1 - a^2 - b^2 - c^2 + 2abc = 0 \]
Rearranging gives us:
\[ a^2 + b^2 + c^2 + 2abc = 1 \]

### Final Answer
Thus, we conclude that:
\[ a^2 + b^2 + c^2 + 2abc = 1 \]