Key Concept: Condition for Non-Trivial Solutions of a Homogeneous System of Linear Equations

For a system of linear equations to have solutions other than the trivial solution ($x=0, y=0, z=0$), a specific condition must be met. If we have a homogeneous system of $n$ linear equations in $n$ variables, represented in matrix form as $AX = 0$, where $A$ is the coefficient matrix, $X$ is the column vector of variables, and $0$ is the zero vector: $A = \begin{pmatrix} a_{11} & a_{12} & \dots & a_{1n} \\ a_{21} & a_{22} & \dots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \dots & a_{nn} \end{pmatrix}, \quad X = \begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{pmatrix}$ This system has non-trivial solutions (i.e., solutions where not all $x_i$ are zero) if and only if the determinant of the coefficient matrix $A$ is equal to zero, i.e., $\det(A) = 0$.

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Step-by-Step Solution:

Step 1: Rearrange the Given Equations into Standard Homogeneous Form

The given equations are:

1. $x = cy + bz$
2. $y = az + cx$
3. $z = bx + ay$

To apply the concept of homogeneous systems, we need to rewrite these equations such that all terms involving $x, y, z$ are on one side and the right-hand side is zero.

Rearranging the terms, we get:

1. $-x + cy + bz = 0$
2. $cx - y + az = 0$
3. $bx + ay - z = 0$

Explanation: This step is crucial for identifying the coefficients of $x, y, z$ that will form our coefficient matrix. A homogeneous system always has $0$ on the right-hand side of all equations.

Step 2: Form the Coefficient Matrix

From the rearranged equations, we can extract the coefficients of $x, y, z$ to form a $3 \times 3$ coefficient matrix, $A$: $A = \begin{pmatrix} -1 & c & b \\ c & -1 & a \\ b & a & -1 \end{pmatrix}$ Explanation: Each row of the matrix corresponds to an equation, and each column corresponds to a variable ($x$, $y$, or $z$). For example, the first row $(-1, c, b)$ represents the coefficients of $x, y, z$ in the first equation $-x + cy + bz = 0$.

Step 3: Apply the Condition for Non-Trivial Solutions

The problem states that there exist real numbers $x, y, z$ which are not all zero that satisfy the given equations. This means the system has non-trivial solutions.

According to the key concept, for a homogeneous system to have non-trivial solutions, the determinant of its coefficient matrix must be zero. Therefore, we must have $\det(A) = 0$: $\begin{vmatrix} -1 & c & b \\ c & -1 & a \\ b & a & -1 \end{vmatrix} = 0$ Explanation: If the determinant were non-zero, the system would have a unique solution, which for a homogeneous system is always the trivial solution ($x=0, y=0, z=0$). Since the problem explicitly states that $x,y,z$ are not all zero, the determinant must be zero.

Step 4: Calculate the Determinant

We will expand the $3 \times 3$ determinant. Let's expand along the first row for clarity. The formula for expanding a $3 \times 3$ determinant $\begin{vmatrix} p & q & r \\ s & t & u \\ v & w & x \end{vmatrix}$ along the first row is $p(tx-uw) - q(sx-uv) + r(sw-tv)$.

Applying this to our determinant: $-1 \begin{vmatrix} -1 & a \\ a & -1 \end{vmatrix} - c \begin{vmatrix} c & a \\ b & -1 \end{vmatrix} + b \begin{vmatrix} c & -1 \\ b & a \end{vmatrix} = 0$ Now, calculate each $2 \times 2$ determinant:

• $\begin{vmatrix} -1 & a \\ a & -1 \end{vmatrix} = (-1)(-1) - (a)(a) = 1 - a^2$
• $\begin{vmatrix} c & a \\ b & -1 \end{vmatrix} = (c)(-1) - (a)(b) = -c - ab$
• $\begin{vmatrix} c & -1 \\ b & a \end{vmatrix} = (c)(a) - (-1)(b) = ac + b$

Substitute these back into the expansion: $-1(1 - a^2) - c(-c - ab) + b(ac + b) = 0$

Explanation: This step is a direct application of the determinant expansion formula. Pay close attention to the signs of the terms (alternating $+,-,+$ for a row/column expansion). A common mistake here is errors in sign or in the $2 \times 2$ determinant calculations.

Step 5: Simplify the Equation

Now, we perform the algebraic simplification: $-1 + a^2 + c^2 + abc + abc + b^2 = 0$ Combine the $abc$ terms: $-1 + a^2 + b^2 + c^2 + 2abc = 0$

Explanation: Carefully distribute the negative signs and coefficients. Make sure all terms are correctly accounted for.

Step 6: Isolate the Desired Expression

The question asks for the value of $a^2 + b^2 + c^2 + 2abc$. From our simplified equation, we can easily find this: $a^2 + b^2 + c^2 + 2abc = 1$

Explanation: This is the final step to present the answer in the required format.

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Summary and Key Takeaway:

This problem demonstrates a fundamental application of determinants in the context of systems of linear equations. The key takeaway is that a homogeneous system of linear equations ($AX=0$) has non-trivial solutions if and only if the determinant of its coefficient matrix is zero ($\det(A)=0$). This condition is frequently tested in competitive exams. The expression $a^2+b^2+c^2+2abc$ often appears in problems related to symmetric matrices or specific geometric contexts. Recognizing this pattern and its connection to determinant properties is a valuable skill.

The final answer is $\boxed{1}$.