Question 5

**Setting up:** Let $R = [\,r_1 \; r_2\,]$ (columns). Then $QR = RP$ gives $$Q r_1 = 2 r_1,\qquad Q r_2 = 3 r_2 .$$ Since all entries of $R$ are non-zero, $r_1, r_2 \neq 0$, so $2$ and $3$ are eigenvalues of $Q$. **Consequences:** - Trace: $x + 4 = 2 + 3 = 5 \Rightarrow x = 1$. - Determinant: $4x - yz = 6 \Rightarrow yz = 4 - 6 = -2$. (Consistency check: eigenvector for $\lambda=2$ satisfies $-v_1 + y v_2 = 0$, giving both components non-zero since $y\neq0$; similarly for $\lambda = 3$. So such an $R$ exists.) **Checking options:** - (A) $\det(Q-2I) = (2-2)(3-2) = 0$ ✔ - (B) $\det(Q-6I) = (2-6)(3-6) = 12$ ✔ - (C) $\det(Q-3I) = (2-3)(3-3) = 0 \neq 15$ ✘ - (D) $yz = -2 \neq 2$ ✘

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No written reasoning was returned for this item.