If x = cy + bz, y = az + cx and z = bx + ay , where x, y and z are not all zero, then `a^(2)+b^(2)+c^(2)` is equal to

Options

A. 1+ 2abc
B. 1− 2abc
C. 1+ abc
D. abc-1

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 value of \( a^2 + b^2 + c^2 \).

### Step-by-Step Solution:

1. **Rearranging the equations**:
We start by rewriting the equations in a standard form:
\[ -cy - bz + x = 0 \]
\[ -az - cx + y = 0 \]
\[ -bx - ay + z = 0 \]

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

3. **Finding the determinant**:
For the system to have non-trivial solutions (where \( x, y, z \) are not all zero), the determinant of the coefficient matrix must be zero:
\[ \text{Det} = \begin{vmatrix} 1 & -c & -b \\ -c & 1 & -a \\ -b & -a & 1 \end{vmatrix} = 0 \]

4. **Calculating the determinant**:
We can calculate the determinant using the formula for a 3x3 matrix:
\[ \text{Det} = 1 \cdot (1 \cdot 1 - (-a)(-b)) - (-c)(-c \cdot 1 - (-a)(-b)) - (-b)(-c(-a) - 1 \cdot (-b)) \]
Simplifying this, we get:
\[ = 1 - ab - c(1 - ab) + b(ac - 1) \]
\[ = 1 - ab - c + abc + b(ac - 1) \]
After further simplification, we arrive at:
\[ = 1 - a^2 - b^2 - c^2 + 2abc = 0 \]

5. **Solving for \( a^2 + b^2 + c^2 \)**:
From the equation we derived:
\[ 1 - a^2 - b^2 - c^2 + 2abc = 0 \]
Rearranging gives:
\[ a^2 + b^2 + c^2 = 1 + 2abc \]

### Final Answer:
Thus, the value of \( a^2 + b^2 + c^2 \) is:
\[ \boxed{1 + 2abc} \]