1. **State the problem:** We are given a system of linear equations: $$\begin{cases} 3x_1 + x_2 - 4x_3 = 3 \\ -x_1 + 4x_2 + x_3 = -3 \\ -x_1 - 2x_3 = -9 \end{cases}$$ 2. **Matrix given:** A 2x3 matrix: $$\begin{bmatrix} 2 & -3 & 1 \\ 6 & -9 & 3 \end{bmatrix}$$ 3. **Goal:** Solve the system for $x_1, x_2, x_3$. 4. **Step 1: Express $x_1$ from the third equation:** $$-x_1 - 2x_3 = -9 \implies -x_1 = -9 + 2x_3 \implies x_1 = 9 - 2x_3$$ 5. **Step 2: Substitute $x_1$ into the first two equations:** First equation: $$3(9 - 2x_3) + x_2 - 4x_3 = 3$$ Simplify: $$27 - 6x_3 + x_2 - 4x_3 = 3$$ $$x_2 - 10x_3 = 3 - 27$$ $$x_2 - 10x_3 = -24$$ Second equation: $$-(9 - 2x_3) + 4x_2 + x_3 = -3$$ Simplify: $$-9 + 2x_3 + 4x_2 + x_3 = -3$$ $$4x_2 + 3x_3 = -3 + 9$$ $$4x_2 + 3x_3 = 6$$ 6. **Step 3: Solve the system for $x_2$ and $x_3$:** $$\begin{cases} x_2 - 10x_3 = -24 \\ 4x_2 + 3x_3 = 6 \end{cases}$$ Multiply the first equation by 4: $$4x_2 - 40x_3 = -96$$ Subtract the second equation: $$\cancel{4x_2} - 40x_3 - (\cancel{4x_2} + 3x_3) = -96 - 6$$ $$-40x_3 - 3x_3 = -102$$ $$-43x_3 = -102$$ $$x_3 = \frac{-102}{-43} = \frac{102}{43}$$ 7. **Step 4: Find $x_2$:** From $x_2 - 10x_3 = -24$: $$x_2 = -24 + 10x_3 = -24 + 10 \times \frac{102}{43} = -24 + \frac{1020}{43}$$ Convert $-24$ to fraction: $$-24 = \frac{-24 \times 43}{43} = \frac{-1032}{43}$$ So: $$x_2 = \frac{-1032}{43} + \frac{1020}{43} = \frac{-12}{43}$$ 8. **Step 5: Find $x_1$:** Recall $x_1 = 9 - 2x_3$: $$x_1 = 9 - 2 \times \frac{102}{43} = 9 - \frac{204}{43}$$ Convert 9 to fraction: $$9 = \frac{387}{43}$$ So: $$x_1 = \frac{387}{43} - \frac{204}{43} = \frac{183}{43}$$ **Final solution:** $$\boxed{\left(x_1, x_2, x_3\right) = \left(\frac{183}{43}, \frac{-12}{43}, \frac{102}{43}\right)}$$