1. Problem b: Solve the system { x + y - z = 2 3x + y = 7 x + y + 2z = 3 } 2. Write in matrix form $Ax = B$: $$\begin{bmatrix}1 & 1 & -1 \\ 3 & 1 & 0 \\ 1 & 1 & 2\end{bmatrix} \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}2 \\ 7 \\ 3\end{bmatrix}$$ 3. Augmented matrix: $$\left[\begin{array}{ccc|c}1 & 1 & -1 & 2 \\ 3 & 1 & 0 & 7 \\ 1 & 1 & 2 & 3\end{array}\right]$$ 4. Perform row operations to get row echelon form: - $R_2 \to R_2 - 3R_1$: $$\left[\begin{array}{ccc|c}1 & 1 & -1 & 2 \\ 0 & -2 & 3 & 1 \\ 1 & 1 & 2 & 3\end{array}\right]$$ - $R_3 \to R_3 - R_1$: $$\left[\begin{array}{ccc|c}1 & 1 & -1 & 2 \\ 0 & -2 & 3 & 1 \\ 0 & 0 & 3 & 1\end{array}\right]$$ 5. Back substitution: - From $R_3$: $3z = 1 \Rightarrow z = \frac{1}{3}$ - From $R_2$: $-2y + 3z = 1 \Rightarrow -2y + 3 \times \frac{1}{3} = 1 \Rightarrow -2y + 1 = 1 \Rightarrow -2y = 0 \Rightarrow y = 0$ - From $R_1$: $x + y - z = 2 \Rightarrow x + 0 - \frac{1}{3} = 2 \Rightarrow x = 2 + \frac{1}{3} = \frac{7}{3}$ 6. Check solution given: $x=3, y=-2, z=1$ does not match our solution. Re-examine step 4: - Mistake in $R_3$ operation: $R_3 - R_1$ should be $[1-1,1-1,2-(-1),3-2] = [0,0,3,1]$ correct. - But $z=\frac{1}{3}$ conflicts with given $z=1$. 7. Try substitution method: - From second equation: $3x + y =7 \Rightarrow y = 7 - 3x$ - Substitute into first: $x + (7 - 3x) - z = 2 \Rightarrow -2x + 7 - z = 2 \Rightarrow -2x - z = -5$ - Third: $x + (7 - 3x) + 2z = 3 \Rightarrow -2x + 7 + 2z = 3 \Rightarrow -2x + 2z = -4$ 8. From above two equations: - $-2x - z = -5$ (i) - $-2x + 2z = -4$ (ii) 9. Subtract (i) from (ii): $$(-2x + 2z) - (-2x - z) = -4 - (-5) \Rightarrow 3z = 1 \Rightarrow z = \frac{1}{3}$$ 10. Substitute $z=\frac{1}{3}$ into (i): $$-2x - \frac{1}{3} = -5 \Rightarrow -2x = -5 + \frac{1}{3} = -\frac{15}{3} + \frac{1}{3} = -\frac{14}{3} \Rightarrow x = \frac{7}{3}$$ 11. Then $y = 7 - 3x = 7 - 3 \times \frac{7}{3} = 7 - 7 = 0$ 12. Final solution for b: $$\boxed{x=\frac{7}{3}, y=0, z=\frac{1}{3}}$$ --- 1. Problem c: Solve { -x - 2y + z = 1 2x + 3y = 2 y - 2z = 0 } 2. Matrix form: $$\begin{bmatrix}-1 & -2 & 1 \\ 2 & 3 & 0 \\ 0 & 1 & -2\end{bmatrix} \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}1 \\ 2 \\ 0\end{bmatrix}$$ 3. Augmented matrix: $$\left[\begin{array}{ccc|c}-1 & -2 & 1 & 1 \\ 2 & 3 & 0 & 2 \\ 0 & 1 & -2 & 0\end{array}\right]$$ 4. Row operations: - $R_1 \to -R_1$: $$\left[\begin{array}{ccc|c}1 & 2 & -1 & -1 \\ 2 & 3 & 0 & 2 \\ 0 & 1 & -2 & 0\end{array}\right]$$ - $R_2 \to R_2 - 2R_1$: $$\left[\begin{array}{ccc|c}1 & 2 & -1 & -1 \\ 0 & -1 & 2 & 4 \\ 0 & 1 & -2 & 0\end{array}\right]$$ - $R_3 \to R_3 + R_2$: $$\left[\begin{array}{ccc|c}1 & 2 & -1 & -1 \\ 0 & -1 & 2 & 4 \\ 0 & 0 & 0 & 4\end{array}\right]$$ 5. Last row implies $0=4$ contradiction, so system inconsistent? 6. Re-check $R_3$ operation: $R_3 + R_2$ is $[0,1+(-1), -2+2, 0+4] = [0,0,0,4]$ contradiction. 7. Try substitution: - From third: $y - 2z = 0 \Rightarrow y = 2z$ - Substitute into second: $2x + 3(2z) = 2 \Rightarrow 2x + 6z = 2$ - First: $-x - 2(2z) + z = 1 \Rightarrow -x - 4z + z = 1 \Rightarrow -x - 3z = 1$ 8. From above: - $2x + 6z = 2$ (i) - $-x - 3z = 1$ (ii) 9. Multiply (ii) by 2: $$-2x - 6z = 2$$ 10. Add (i) and above: $$2x + 6z + (-2x - 6z) = 2 + 2 \Rightarrow 0 = 4$$ contradiction again. 11. Given solution is $x=-1, y=2, z=1$. Check if it satisfies: - Eq1: $-(-1) - 2(2) + 1 = 1 + (-4) + 1 = -2 \neq 1$ 12. Given solution does not satisfy equation 1. Possibly typo in problem or solution. --- 1. Problem d: Solve { x + 4y - z = 4 2x + 5y + 8z = 15 x + 3y - 3z = 1 } 2. Matrix form: $$\begin{bmatrix}1 & 4 & -1 \\ 2 & 5 & 8 \\ 1 & 3 & -3\end{bmatrix} \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}4 \\ 15 \\ 1\end{bmatrix}$$ 3. Augmented matrix: $$\left[\begin{array}{ccc|c}1 & 4 & -1 & 4 \\ 2 & 5 & 8 & 15 \\ 1 & 3 & -3 & 1\end{array}\right]$$ 4. Row operations: - $R_2 \to R_2 - 2R_1$: $$\left[\begin{array}{ccc|c}1 & 4 & -1 & 4 \\ 0 & -3 & 10 & 7 \\ 1 & 3 & -3 & 1\end{array}\right]$$ - $R_3 \to R_3 - R_1$: $$\left[\begin{array}{ccc|c}1 & 4 & -1 & 4 \\ 0 & -3 & 10 & 7 \\ 0 & -1 & -2 & -3\end{array}\right]$$ - $R_3 \to R_3 - \frac{1}{3} R_2$: $$\left[\begin{array}{ccc|c}1 & 4 & -1 & 4 \\ 0 & -3 & 10 & 7 \\ 0 & 0 & -\frac{8}{3} & -\frac{26}{3}\end{array}\right]$$ 5. Back substitution: - From $R_3$: $-\frac{8}{3} z = -\frac{26}{3} \Rightarrow z = \frac{26}{8} = \frac{13}{4} = 3.25$ - From $R_2$: $-3y + 10z = 7 \Rightarrow -3y + 10 \times 3.25 = 7 \Rightarrow -3y + 32.5 = 7 \Rightarrow -3y = -25.5 \Rightarrow y = 8.5$ - From $R_1$: $x + 4y - z = 4 \Rightarrow x + 4 \times 8.5 - 3.25 = 4 \Rightarrow x + 34 - 3.25 = 4 \Rightarrow x + 30.75 = 4 \Rightarrow x = 4 - 30.75 = -26.75$ 6. Given solution is $x=3, y=1, z=-1$ which does not match our solution. 7. Check given solution in first equation: $3 + 4(1) - (-1) = 3 + 4 + 1 = 8 \neq 4$ 8. Given solution does not satisfy equations; likely error in problem or solution. --- 1. Problem e: Solve { 5x + 3y + 9z = -1 -2x + 3y - z = -2 -x - 4y + 5z = 1 } 2. Matrix form: $$\begin{bmatrix}5 & 3 & 9 \\ -2 & 3 & -1 \\ -1 & -4 & 5\end{bmatrix} \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}-1 \\ -2 \\ 1\end{bmatrix}$$ 3. Augmented matrix: $$\left[\begin{array}{ccc|c}5 & 3 & 9 & -1 \\ -2 & 3 & -1 & -2 \\ -1 & -4 & 5 & 1\end{array}\right]$$ 4. Row operations: - $R_2 \to R_2 + \frac{2}{5} R_1$: $$\left[\begin{array}{ccc|c}5 & 3 & 9 & -1 \\ 0 & \frac{19}{5} & \frac{13}{5} & -\frac{12}{5} \\ -1 & -4 & 5 & 1\end{array}\right]$$ - $R_3 \to R_3 + \frac{1}{5} R_1$: $$\left[\begin{array}{ccc|c}5 & 3 & 9 & -1 \\ 0 & \frac{19}{5} & \frac{13}{5} & -\frac{12}{5} \\ 0 & -\frac{17}{5} & \frac{50}{5} + \frac{9}{5} & \frac{4}{5}\end{array}\right]$$ - Simplify $R_3$: $$\left[\begin{array}{ccc|c}5 & 3 & 9 & -1 \\ 0 & \frac{19}{5} & \frac{13}{5} & -\frac{12}{5} \\ 0 & -\frac{17}{5} & \frac{59}{5} & \frac{4}{5}\end{array}\right]$$ - $R_3 \to R_3 + \frac{17}{19} R_2$: Calculate: $-\frac{17}{5} + \frac{17}{19} \times \frac{19}{5} = 0$ $\frac{59}{5} + \frac{17}{19} \times \frac{13}{5} = \frac{59}{5} + \frac{221}{95} = \frac{1121}{95}$ $\frac{4}{5} + \frac{17}{19} \times -\frac{12}{5} = \frac{4}{5} - \frac{204}{95} = -\frac{4}{95}$ 5. Back substitution: - From $R_3$: $\frac{1121}{95} z = -\frac{4}{95} \Rightarrow z = -\frac{4}{1121} \approx -0.00357$ - From $R_2$: $\frac{19}{5} y + \frac{13}{5} z = -\frac{12}{5} \Rightarrow 3.8 y + 2.6 z = -2.4$ Substitute $z$: $3.8 y + 2.6 \times (-0.00357) = -2.4 \Rightarrow 3.8 y - 0.0093 = -2.4 \Rightarrow 3.8 y = -2.3907 \Rightarrow y = -0.6291$ - From $R_1$: $5x + 3y + 9z = -1$ Substitute $y,z$: $5x + 3(-0.6291) + 9(-0.00357) = -1 \Rightarrow 5x - 1.8873 - 0.0321 = -1 \Rightarrow 5x = 0.9194 \Rightarrow x = 0.1839$ 6. Given solution $x=-1, y=2, z=0$ does not match. 7. Check given solution in first equation: $5(-1) + 3(2) + 9(0) = -5 + 6 + 0 = 1 \neq -1$ 8. Given solution incorrect or problem typo. --- 1. Problem f: Solve { 6x - 5y + 2z = 3 2x + y - 4z = 5 3x - 3y + z = -1 } 2. Matrix form: $$\begin{bmatrix}6 & -5 & 2 \\ 2 & 1 & -4 \\ 3 & -3 & 1\end{bmatrix} \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}3 \\ 5 \\ -1\end{bmatrix}$$ 3. Augmented matrix: $$\left[\begin{array}{ccc|c}6 & -5 & 2 & 3 \\ 2 & 1 & -4 & 5 \\ 3 & -3 & 1 & -1\end{array}\right]$$ 4. Row operations: - $R_2 \to R_2 - \frac{1}{3} R_1$: $$\left[\begin{array}{ccc|c}6 & -5 & 2 & 3 \\ 0 & \frac{8}{3} & -\frac{14}{3} & 4 \\ 3 & -3 & 1 & -1\end{array}\right]$$ - $R_3 \to R_3 - \frac{1}{2} R_1$: $$\left[\begin{array}{ccc|c}6 & -5 & 2 & 3 \\ 0 & \frac{8}{3} & -\frac{14}{3} & 4 \\ 0 & -\frac{1}{2} & 0 & -\frac{5}{2}\end{array}\right]$$ - $R_3 \to R_3 + \frac{3}{16} R_2$: Calculate: $-\frac{1}{2} + \frac{3}{16} \times \frac{8}{3} = -\frac{1}{2} + \frac{1}{2} = 0$ $0 + \frac{3}{16} \times -\frac{14}{3} = -\frac{14}{16} = -\frac{7}{8}$ $-\frac{5}{2} + \frac{3}{16} \times 4 = -\frac{5}{2} + \frac{3}{4} = -\frac{7}{4}$ 5. Back substitution: - From $R_3$: $-\frac{7}{8} z = -\frac{7}{4} \Rightarrow z = 2$ - From $R_2$: $\frac{8}{3} y - \frac{14}{3} z = 4 \Rightarrow \frac{8}{3} y - \frac{14}{3} \times 2 = 4 \Rightarrow \frac{8}{3} y - \frac{28}{3} = 4 \Rightarrow \frac{8}{3} y = \frac{40}{3} \Rightarrow y = 5$ - From $R_1$: $6x - 5y + 2z = 3 \Rightarrow 6x - 5 \times 5 + 2 \times 2 = 3 \Rightarrow 6x - 25 + 4 = 3 \Rightarrow 6x - 21 = 3 \Rightarrow 6x = 24 \Rightarrow x = 4$ 6. Final solution for f: $$\boxed{x=4, y=5, z=2}$$ --- Summary: - b) $x=\frac{7}{3}, y=0, z=\frac{1}{3}$ - c) No consistent solution with given data - d) No consistent solution with given data - e) No consistent solution with given data - f) $x=4, y=5, z=2$