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Count the total number of $3\times3$ matrices of the form $A=\begin{pmatrix}0&2y&1\\2x&y&-1\\2x&-y&1\end{pmatrix}$ with $x,y\in\mathbb{R},\ x\neq y$, satisfying $A^TA=3I_3$.
Compute $A^TA = 3I_3$ by equating columns to be orthogonal with magnitude $\sqrt{3}$.
Column 1 norm$^2$: $0+(2x)^2+(2x)^2=8x^2=3 \Rightarrow x^2=\frac{3}{8}$: 2 values of $x$.
Column 2 norm$^2$: $4y^2+y^2+y^2=6y^2=3 \Rightarrow y^2=\frac{1}{2}$: 2 values of $y$.
With $x\neq y$: we need to discard cases where $x=y$. Since $x^2=3/8$ and $y^2=1/2$, $x\neq\pm y$ (different magnitudes), so all $2\times2=4$ combinations are valid.
Total $= \mathbf{4}$
Given that $r > s > 0$, compare the two quantities below.
Column A: $\dfrac{rs}{r}$
Column B: $\dfrac{rs}{s}$
Simplify each expression:
- Column A: $\dfrac{rs}{r} = s$
- Column B: $\dfrac{rs}{s} = r$
Since we are given $r > s > 0$, it follows that $r > s$.
Therefore Column B ($r$) is greater than Column A ($s$).
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