1. **State the problem:** We have two matrices $$P=\begin{pmatrix} x^{2} & 3 \\ 1 & 3x \end{pmatrix}$$ and $$Q=\begin{pmatrix} 3 & 6 \\ 2 & x \end{pmatrix}$$ We are told that $P$ and $Q$ commute under multiplication, i.e., $PQ = QP$. We need to find the positive value of $x$. 2. **Recall the rule:** Two matrices $A$ and $B$ commute if and only if $AB = BA$. 3. **Calculate $PQ$:** $$PQ = \begin{pmatrix} x^{2} & 3 \\ 1 & 3x \end{pmatrix} \begin{pmatrix} 3 & 6 \\ 2 & x \end{pmatrix} = \begin{pmatrix} x^{2} \cdot 3 + 3 \cdot 2 & x^{2} \cdot 6 + 3 \cdot x \\ 1 \cdot 3 + 3x \cdot 2 & 1 \cdot 6 + 3x \cdot x \end{pmatrix} = \begin{pmatrix} 3x^{2} + 6 & 6x^{2} + 3x \\ 3 + 6x & 6 + 3x^{2} \end{pmatrix}$$ 4. **Calculate $QP$:** $$QP = \begin{pmatrix} 3 & 6 \\ 2 & x \end{pmatrix} \begin{pmatrix} x^{2} & 3 \\ 1 & 3x \end{pmatrix} = \begin{pmatrix} 3 \cdot x^{2} + 6 \cdot 1 & 3 \cdot 3 + 6 \cdot 3x \\ 2 \cdot x^{2} + x \cdot 1 & 2 \cdot 3 + x \cdot 3x \end{pmatrix} = \begin{pmatrix} 3x^{2} + 6 & 9 + 18x \\ 2x^{2} + x & 6 + 3x^{2} \end{pmatrix}$$ 5. **Set $PQ = QP$ and equate corresponding elements:** - Top-left: $3x^{2} + 6 = 3x^{2} + 6$ (always true) - Top-right: $6x^{2} + 3x = 9 + 18x$ - Bottom-left: $3 + 6x = 2x^{2} + x$ - Bottom-right: $6 + 3x^{2} = 6 + 3x^{2}$ (always true) 6. **Solve the system:** From top-right: $$6x^{2} + 3x = 9 + 18x \implies 6x^{2} + 3x - 9 - 18x = 0 \implies 6x^{2} - 15x - 9 = 0$$ From bottom-left: $$3 + 6x = 2x^{2} + x \implies 2x^{2} + x - 6x - 3 = 0 \implies 2x^{2} - 5x - 3 = 0$$ 7. **Solve each quadratic:** - For $6x^{2} - 15x - 9 = 0$: $$x = \frac{15 \pm \sqrt{(-15)^{2} - 4 \cdot 6 \cdot (-9)}}{2 \cdot 6} = \frac{15 \pm \sqrt{225 + 216}}{12} = \frac{15 \pm \sqrt{441}}{12} = \frac{15 \pm 21}{12}$$ So, $$x = \frac{15 + 21}{12} = 3$$ or $$x = \frac{15 - 21}{12} = -\frac{1}{2}$$ - For $2x^{2} - 5x - 3 = 0$: $$x = \frac{5 \pm \sqrt{(-5)^{2} - 4 \cdot 2 \cdot (-3)}}{2 \cdot 2} = \frac{5 \pm \sqrt{25 + 24}}{4} = \frac{5 \pm \sqrt{49}}{4} = \frac{5 \pm 7}{4}$$ So, $$x = \frac{5 + 7}{4} = 3$$ or $$x = \frac{5 - 7}{4} = -\frac{1}{2}$$ 8. **Check for common solutions:** Both quadratics have solutions $x=3$ and $x=-\frac{1}{2}$. Since we want the positive value, the answer is: $$\boxed{3}$$