<p>If \(A = \begin{bmatrix}a & b & c \\ c & a & b \\ b & c & a\end{bmatrix}\) and \(a, b, c\) are roots of the equation \(x^3 + x^2 - 4 = 0\) then \(AA^T\) is equal to</p>
Step-by-Step Solution
Key Concept: A is a circulant matrix whose determinant relates to roots via Vieta's formulas. Since a, b, c are roots of x³ + x² - 4 = 0, we have a + b + c = -1 and abc = 4. The product AA^T can be computed using the fact that for circulant matrices, eigenvalues are determined by the first row and Vieta's relationships.
<p><strong>Step 1:</strong> Recognize that A is a circulant matrix. For circulant matrices with first row [a, b, c], the eigenvalues are λₖ = a + bωᵏ + cω²ᵏ where ω = e^(2πi/3).</p><p><strong>Step 2:</strong> Compute AA^T directly for position (1,1): (AA^T)₁₁ = a² + b² + c². From Vieta's formulas for x³ + x² - 4 = 0: a + b + c = -1, ab + bc + ca = 0, abc = 4.</p><p><strong>Step 3:</strong> Calculate a² + b² + c² = (a + b + c)² - 2(ab + bc + ca) = (-1)² - 2(0) = 1.</p><p><strong>Step 4:</strong> For position (1,2): (AA^T)₁₂ = ac + ba + cb = ab + bc + ca = 0. By circulant symmetry, all off-diagonal elements equal 0.</p><p><strong>Step 5:</strong> Since AA^T is symmetric and circulant with a² + b² + c² = 1 on the diagonal and 0 elsewhere:</p><p>∴ AA^T = <strong>I</strong> or <strong>[1, 0, 0; 0, 1, 0; 0, 0, 1]</strong> (Answer: A)</p>
Correct Answer: A