1. **State the problem:** Solve the system of linear equations by matrix methods for two systems: ①\ $$\begin{cases} 2x_1 - x_2 + 3x_3 = 2 \\ x_1 + 3x_2 - x_3 = 11 \\ 2x_1 - 2x_2 - 5x_3 = 3 \end{cases}$$ ②\ $$\begin{cases} x_1 + 2x_2 - 3x_3 = 3 \\ 2x_1 - x_2 - x_3 = 11 \\ 3x_1 + 2x_2 + x_3 = -5 \end{cases}$$ 2. **Set up augmented matrices:** For system ①: $$\left[\begin{array}{ccc|c} 2 & -1 & 3 & 2 \\ 1 & 3 & -1 & 11 \\ 2 & -2 & -5 & 3 \end{array}\right]$$ For system ②: $$\left[\begin{array}{ccc|c} 1 & 2 & -3 & 3 \\ 2 & -1 & -1 & 11 \\ 3 & 2 & 1 & -5 \end{array}\right]$$ 3. **Solve system ① by Gaussian elimination:** Step 1: Use row 1 as pivot. - R2 = R2 - (1/2)R1: $$[1,3,-1,11] - 0.5*[2,-1,3,2] = [0, 3.5, -2.5, 10]$$ - R3 = R3 - R1: $$[2,-2,-5,3] - [2,-1,3,2] = [0, -1, -8, 1]$$ Matrix becomes: $$\left[\begin{array}{ccc|c} 2 & -1 & 3 & 2 \\ 0 & 3.5 & -2.5 & 10 \\ 0 & -1 & -8 & 1 \end{array}\right]$$ Step 2: Pivot on R2, second element. - R3 = R3 - (-1 / 3.5) * R2 = R3 + (2/7) * R2: Calculate: $$[0, -1, -8, 1] + \frac{2}{7} * [0, 3.5, -2.5, 10] = [0, -1 + 1, -8 - \frac{5}{7}, 1 + \frac{20}{7}] = [0, 0, -\frac{61}{7}, \frac{27}{7}]$$ Step 3: Back substitution: - From R3: $$-\frac{61}{7} x_3 = \frac{27}{7} \Rightarrow x_3 = -\frac{27}{61}$$ - Substitute into R2: $$3.5 x_2 - 2.5 x_3 = 10$$ $$3.5 x_2 = 10 + 2.5 * (-\frac{27}{61}) = 10 - \frac{67.5}{61} = \frac{610 - 67.5}{61} = \frac{542.5}{61}$$ $$x_2 = \frac{542.5}{61 \times 3.5} = \frac{542.5}{213.5} = \frac{1085}{427}$$ - Substitute $x_2, x_3$ into R1: $$2 x_1 - x_2 + 3 x_3 = 2$$ $$2 x_1 = 2 + x_2 - 3 x_3 = 2 + \frac{1085}{427} - 3 *\left(-\frac{27}{61}\right)$$ Calculate: $$3 * \left(-\frac{27}{61}\right) = -\frac{81}{61}$$ Sum: $$2 + \frac{1085}{427} + \frac{81}{61}$$ Find common denominator for fractions: 427 and 61 $427 = 7 \times 61$ So, $$\frac{1085}{427} = \frac{1085}{7 \times 61}$$ Convert all to denominator $7 \times 61 = 427$: $$2 = \frac{2 \times 427}{427} = \frac{854}{427}$$ $$\frac{81}{61} = \frac{81 \times 7}{427} = \frac{567}{427}$$ Sum numerators: $$854 + 1085 + 567 = 2506$$ So, $$2 x_1 = \frac{2506}{427}$$ $$x_1 = \frac{2506}{854} = \frac{1253}{427}$$ 4. **Solutions for system ①:** $$x_1 = \frac{1253}{427}, \quad x_2 = \frac{1085}{427}, \quad x_3 = -\frac{27}{61}$$ 5. **Solve system ② by Gaussian elimination:** - Write augmented matrix: $$\left[\begin{array}{ccc|c} 1 & 2 & -3 & 3 \\ 2 & -1 & -1 & 11 \\ 3 & 2 & 1 & -5 \end{array}\right]$$ - Eliminate first column under pivot: R2 = R2 - 2*R1 = $$[2,-1,-1,11] - 2*[1,2,-3,3] = [0,-5,5,5]$$ R3 = R3 - 3*R1 = $$[3,2,1,-5] - 3*[1,2,-3,3] = [0,-4,10,-14]$$ - New matrix: $$\left[\begin{array}{ccc|c} 1 & 2 & -3 & 3 \\ 0 & -5 & 5 & 5 \\ 0 & -4 & 10 & -14 \end{array}\right]$$ - Pivot on row 2, second element. Eliminate R3 second column: R3 = R3 - (-4/-5)*R2 = R3 - (4/5) * R2 Calculate: $$[0,-4,10,-14] - \frac{4}{5} * [0,-5,5,5] = [0,-4 +4, 10 -4, -14 -4] = [0,0,6,-18]$$ - Back substitution: From R3: $$6 x_3 = -18 \Rightarrow x_3 = -3$$ From R2: $$-5 x_2 + 5 x_3 = 5 \Rightarrow -5 x_2 + 5(-3) = 5 \Rightarrow -5 x_2 -15 = 5$$ $$-5 x_2 = 20 \Rightarrow x_2 = -4$$ From R1: $$x_1 + 2 x_2 - 3 x_3 = 3 \Rightarrow x_1 + 2(-4) - 3(-3) = 3$$ $$x_1 -8 +9 = 3 \Rightarrow x_1 +1 = 3 \Rightarrow x_1 = 2$$ 6. **Solutions for system ②:** $$x_1 = 2, \quad x_2 = -4, \quad x_3 = -3$$