Question

Let \( A = \begin{bmatrix} x & y & z \\ y & z & x \\ z & x & y \end{bmatrix} \), where \( x, y \) and \( z \) are real numbers such that \( x + y + z > 0 \) and \( xyz = 2 \). If \( A^2 = I_3 \), then the value of \( x^3 + y^3 + z^3 \) is __________.

Hint:

The determinant of a circulant matrix like this is always $-(x^3+y^3+z^3-3xyz)$. If $A$ is orthogonal ($A^2=I$), its determinant must be $\pm 1$.

Answer 1

Correct Answer: 7

Step 1: $A^2 = I \Rightarrow |A|^2 = |I| = 1 \Rightarrow |A| = \pm 1$.
Step 2: $|A| = -(x^3+y^3+z^3-3xyz)$. Since $x+y+z>0$ and $A^2=I$, the only eigenvalue sum condition and positive trace implies $|A| = -1$.
Step 3: $-(x^3+y^3+z^3-3(2)) = -1 \Rightarrow x^3+y^3+z^3-6 = 1$.
Step 4: $x^3+y^3+z^3 = 7$.