1. **State the problem:** Given that $a^{-\frac{1}{2}} = 3$, find the value of $a$. 2. **Recall the rule for negative exponents:** $a^{-n} = \frac{1}{a^n}$. 3. Apply this rule to rewrite the equation: $$a^{-\frac{1}{2}} = \frac{1}{a^{\frac{1}{2}}} = 3$$ 4. This implies: $$\frac{1}{a^{\frac{1}{2}}} = 3$$ 5. Multiply both sides by $a^{\frac{1}{2}}$: $$1 = 3 a^{\frac{1}{2}}$$ 6. Divide both sides by 3: $$\frac{1}{3} = a^{\frac{1}{2}}$$ 7. Recall that $a^{\frac{1}{2}} = \sqrt{a}$, so: $$\sqrt{a} = \frac{1}{3}$$ 8. Square both sides to solve for $a$: $$a = \left(\frac{1}{3}\right)^2 = \frac{1}{9}$$ **Final answer:** $a = \frac{1}{9}$, which corresponds to option B.