This problem elegantly combines concepts from Matrices and Algebraic Identities. We will leverage the properties of orthogonal matrices and fundamental algebraic identities to establish relationships between $a, b,$ and $c$, ultimately solving for $abc$.

Key Concepts Used

1. Orthogonal Matrix: A square matrix $A$ is called an orthogonal matrix if its transpose is equal to its inverse, i.e., $A^T A = I$, where $I$ is the identity matrix. A crucial property of orthogonal matrices is that their row vectors (and column vectors) form an orthonormal set. This means:
   • The dot product of any row vector with itself is 1 (unit vectors).
   • The dot product of any two distinct row vectors is 0 (mutually orthogonal).
2. Algebraic Identities: We will extensively use two fundamental identities for three variables $x, y, z$:
   • The square of a sum: $(x+y+z)^2 = x^2+y^2+z^2+2(xy+yz+zx)$
   • The sum of cubes identity: $x^3+y^3+z^3-3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx)$

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Step-by-Step Solution

We are given that $a, b, c \in \mathbb{R}$ are non-zero, and $a^3 + b^3 + c^3 = 2$. We are also given the matrix $A = \begin{pmatrix} a & b & c \\ b & c & a \\ c & a & b \end{pmatrix}$ satisfies $A^T A = I$.

Step 1: Utilize the condition $A^T A = I$ to find relationships between $a, b, c$

The condition $A^T A = I$ implies that $A$ is an orthogonal matrix. Let's perform the matrix multiplication.

First, find the transpose of $A$: $A^T = \begin{pmatrix} a & b & c \\ b & c & a \\ c & a & b \end{pmatrix}^T = \begin{pmatrix} a & b & c \\ b & c & a \\ c & a & b \end{pmatrix}$ (Notice that in this specific case, $A$ is a symmetric matrix, so $A^T = A$).

Now, compute $A^T A$: $A^T A = \begin{pmatrix} a & b & c \\ b & c & a \\ c & a & b \end{pmatrix} \begin{pmatrix} a & b & c \\ b & c & a \\ c & a & b \end{pmatrix}$ Performing the multiplication, we calculate each element:

• The (1,1) element is $(a)(a)+(b)(b)+(c)(c) = a^2+b^2+c^2$.
• The (1,2) element is $(a)(b)+(b)(c)+(c)(a) = ab+bc+ca$.
• The (1,3) element is $(a)(c)+(b)(a)+(c)(b) = ac+ba+cb$.
• The (2,1) element is $(b)(a)+(c)(b)+(a)(c) = ba+cb+ac$.
• And so on for all elements.

Simplifying the elements, we get: $A^T A = \begin{pmatrix} a^2+b^2+c^2 & ab+bc+ca & ab+bc+ca \\ ab+bc+ca & a^2+b^2+c^2 & ab+bc+ca \\ ab+bc+ca & ab+bc+ca & a^2+b^2+c^2 \end{pmatrix}$

We are given that $A^T A = I$, where $I$ is the $3 \times 3$ identity matrix: $I = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$

Equating the corresponding elements of the matrices:

• From the diagonal elements, we must have: $a^2+b^2+c^2 = 1 \quad \text{(Equation 1)}$
• From the off-diagonal elements, we must have: $ab+bc+ca = 0 \quad \text{(Equation 2)}$

Explanation: This step is crucial because the matrix condition $A^T A = I$ directly provides us with two fundamental algebraic relationships between $a, b,$ and $c$. These relationships are essential for simplifying expressions in subsequent steps.

Step 2: Determine the value of $(a+b+c)$

We use the algebraic identity for the square of a sum: $(a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca)$

Now, substitute the values we found from Equation 1 and Equation 2 into this identity: $(a+b+c)^2 = (1) + 2(0)$ $(a+b+c)^2 = 1$

Taking the square root of both sides, we get: $a+b+c = \pm 1 \quad \text{(Equation 3)}$

Explanation: We calculate $(a+b+c)$ because it is a common factor in the sum of cubes identity, which we will use in the next step to involve $a^3+b^3+c^3$ and $abc$.

Step 3: Apply the sum of cubes identity to find $abc$

We are given $a^3+b^3+c^3=2$. We need to find a value for $abc$. The algebraic identity that connects these terms, along with $a+b+c$, $a^2+b^2+c^2$, and $ab+bc+ca$, is: $x^3+y^3+z^3-3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx)$ Substitute $x=a, y=b, z=c$: $a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-(ab+bc+ca))$

Now, substitute the values we found in Step 1 and Step 2, along with the given condition $a^3+b^3+c^3=2$:

• $a^3+b^3+c^3 = 2$ (given)
• $a^2+b^2+c^2 = 1$ (from Equation 1)
• $ab+bc+ca = 0$ (from Equation 2)
• $a+b+c = \pm 1$ (from Equation 3)

Plugging these into the identity: $2 - 3abc = (\pm 1)(1 - (0))$ $2 - 3abc = (\pm 1)(1)$ $2 - 3abc = \pm 1$

This gives us two possible cases:

Case 1: $2 - 3abc = 1$ $3abc = 2 - 1$ $3abc = 1$ $abc = \frac{1}{3}$

Case 2: $2 - 3abc = -1$ $3abc = 2 - (-1)$ $3abc = 3$ $abc = 1$

Explanation: This identity is the key to solving the problem, as it directly links the given sum of cubes ($a^3+b^3+c^3$) with the product ($abc$) and the other expressions we derived from the matrix condition.

Final Answer

From our derivations, a value of $abc$ can be either $\frac{1}{3}$ or $1$. Looking at the given options: (A) 3 (B) ${1 \over 3}$ (C) -${1 \over 3}$ (D) ${2 \over 3}$

One of the derived values, $abc = \frac{1}{3}$, matches option (B).

The final answer is $\boxed{B}$

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Tips and Common Mistakes to Avoid

• Careful Matrix Multiplication: Ensure accurate calculation of $A^T A$. A small error here will propagate through the entire solution.
• Recognize Orthogonal Matrix Properties: The condition $A^T A = I$ is very powerful. It immediately tells you that the rows (and columns) are orthonormal, leading to the equations $a^2+b^2+c^2=1$ and $ab+bc+ca=0$.
• Correct Algebraic Identities: Memorize or correctly recall the fundamental algebraic identities for sums and cubes of three variables. They are indispensable for problems like this.
• Consider Both Positive and Negative Roots: When solving $(a+b+c)^2=1$, remember to consider both $a+b+c = 1$ and $a+b+c = -1$. Both possibilities need to be followed through.
• The Determinant of A: While not used in this specific solution path, for this type of matrix $A$, its determinant squared equals 1 ($\det(A)^2 = \det(A^T A) = \det(I) = 1$). This means $\det(A) = \pm 1$. This can be a useful check or an alternative approach in related problems.

Summary/Key Takeaway

This problem demonstrates how seemingly disparate topics like matrices and algebraic identities can be intricately linked. The condition of an orthogonal matrix provides critical algebraic relationships, which are then combined with a given sum of cubes and fundamental identities to solve for the product of the variables. The key steps involve:

1. Extracting $a^2+b^2+c^2=1$ and $ab+bc+ca=0$ from $A^T A = I$.
2. Using these to find $a+b+c = \pm 1$.
3. Applying the sum of cubes identity $a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ to solve for $abc$.