Kergakanlah matriks dibawah ini dengan menggunakan metode Eliminasi gauss jordan 5. \( \left[\begin{array}{ccc}1 & 1 & 2 \\ -1 & -2 & 3 \\ 3 & -7 & 4\end{array}\right]\left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3}\end{array}\right]=\left[\begin{array}{c}8 \\ 1 \\ 10\end{array}\right] \) *. \( \left[\begin{array}{ccc}2 & 2 & 2 \\ -2 & 5 & 2 \\ 8 & 1 & 4\end{array}\right]\left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3}\end{array}\right]=\left[\begin{array}{c}0 \\ 1 \\ -1\end{array}\right] \) 7. \( \left[\begin{array}{cccc}1 & -1 & 2 & -1 \\ 2 & 1 & -2 & -2 \\ -1 & 2 & -4 & 1 \\ 3 & 0 & 0 & -3\end{array}\right]\left[\begin{array}{l}w \\ x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}-1 \\ -2 \\ 1 \\ -3\end{array}\right] \) 13. \( \left[\begin{array}{cccc}2 & -3 & 4 & -1 \\ 7 & 1 & -8 & 9 \\ 2 & 8 & 1 & -1\end{array}\right]\left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3} \\ x_{4}\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right] \) 19. \( \left[\begin{array}{ccc}1 & 3 & -1 \\ 0 & 1 & -8 \\ 0 & 0 & 4\end{array}\right]\left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3}\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right] \)

Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}1\times x_{1}+3x_{2}-1\times x_{3}=0\\0\times x_{1}+1\times x_{2}-8x_{3}=0\\0\times x_{1}+0\times x_{2}+4x_{3}=0\end{array}\right.\) - step1: Calculate: \(\left\{ \begin{array}{l}x_{1}+3x_{2}-x_{3}=0\\x_{2}-8x_{3}=0\\4x_{3}=0\end{array}\right.\) - step2: Solve the equation: \(\left\{ \begin{array}{l}x_{1}+3x_{2}-x_{3}=0\\x_{2}-8x_{3}=0\\x_{3}=0\end{array}\right.\) - step3: Substitute the value of \(x_{3}:\) \(\left\{ \begin{array}{l}x_{1}+3x_{2}-0=0\\x_{2}-8\times 0=0\end{array}\right.\) - step4: Remove 0: \(\left\{ \begin{array}{l}x_{1}+3x_{2}=0\\x_{2}=0\end{array}\right.\) - step5: Substitute the value of \(x_{2}:\) \(x_{1}+3\times 0=0\) - step6: Simplify: \(x_{1}=0\) - step7: Calculate: \(\left\{ \begin{array}{l}x_{1}=0\\x_{2}=0\\x_{3}=0\end{array}\right.\) - step8: Check the solution: \(\left\{ \begin{array}{l}x_{1}=0\\x_{2}=0\\x_{3}=0\end{array}\right.\) - step9: Rewrite: \(\left(x_{1},x_{2},x_{3}\right) = \left(0,0,0\right)\) Solve the system of equations \( 1w - 1x + 2y - 1z = -1; 2w + 1x - 2y - 2z = -2; -1w + 2x - 4y + 1z = 1; 3w + 0x + 0y - 3z = -3 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}1\times w-1\times x+2y-1\times z=-1\\2w+1\times x-2y-2z=-2\\-w+2x-4y+1\times z=1\\3w+0\times x+0\times y-3z=-3\end{array}\right.\) - step1: Calculate: \(\left\{ \begin{array}{l}w-x+2y-z=-1\\2w+x-2y-2z=-2\\-w+2x-4y+z=1\\3w-3z=-3\end{array}\right.\) - step2: Solve the equation: \(\left\{ \begin{array}{l}w=-1+x-2y+z\\2w+x-2y-2z=-2\\-w+2x-4y+z=1\\3w-3z=-3\end{array}\right.\) - step3: Substitute the value of \(w:\) \(\left\{ \begin{array}{l}2\left(-1+x-2y+z\right)+x-2y-2z=-2\\-\left(-1+x-2y+z\right)+2x-4y+z=1\\3\left(-1+x-2y+z\right)-3z=-3\end{array}\right.\) - step4: Simplify: \(\left\{ \begin{array}{l}-2+3x-6y=-2\\1+x-2y=1\\-3+3x-6y=-3\end{array}\right.\) - step5: Solve the equation: \(\left\{ \begin{array}{l}-2+3x-6y=-2\\x=2y\\-3+3x-6y=-3\end{array}\right.\) - step6: Substitute the value of \(x:\) \(\left\{ \begin{array}{l}-2+3\times 2y-6y=-2\\-3+3\times 2y-6y=-3\end{array}\right.\) - step7: Simplify: \(\left\{ \begin{array}{l}-2=-2\\-3=-3\end{array}\right.\) - step8: Substitute the value of \(x:\) \(w=-1+2y-2y+z\) - step9: Simplify: \(w=-1+z\) - step10: Calculate: \(\left(w,x,y,z\right) = \left(-1+z,2y,y,z\right),\left(z,y\right) \in \mathbb{R}^{2}\) - step11: Alternative Form: \(\textrm{Infinitely many solutions}\) Solve the system of equations \( 2x_{1} + 2x_{2} + 2x_{3} = 0; -2x_{1} + 5x_{2} + 2x_{3} = 1; 8x_{1} + 1x_{2} + 4x_{3} = -1 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}2x_{1}+2x_{2}+2x_{3}=0\\-2x_{1}+5x_{2}+2x_{3}=1\\8x_{1}+1\times x_{2}+4x_{3}=-1\end{array}\right.\) - step1: Calculate: \(\left\{ \begin{array}{l}2x_{1}+2x_{2}+2x_{3}=0\\-2x_{1}+5x_{2}+2x_{3}=1\\8x_{1}+x_{2}+4x_{3}=-1\end{array}\right.\) - step2: Solve the equation: \(\left\{ \begin{array}{l}2x_{1}+2x_{2}+2x_{3}=0\\-2x_{1}+5x_{2}+2x_{3}=1\\x_{2}=-1-8x_{1}-4x_{3}\end{array}\right.\) - step3: Substitute the value of \(x_{2}:\) \(\left\{ \begin{array}{l}2x_{1}+2\left(-1-8x_{1}-4x_{3}\right)+2x_{3}=0\\-2x_{1}+5\left(-1-8x_{1}-4x_{3}\right)+2x_{3}=1\end{array}\right.\) - step4: Simplify: \(\left\{ \begin{array}{l}-14x_{1}-2-6x_{3}=0\\-42x_{1}-5-18x_{3}=1\end{array}\right.\) - step5: Solve the equation: \(\left\{ \begin{array}{l}x_{1}=-\frac{1+3x_{3}}{7}\\-42x_{1}-5-18x_{3}=1\end{array}\right.\) - step6: Substitute the value of \(x_{1}:\) \(-42\left(-\frac{1+3x_{3}}{7}\right)-5-18x_{3}=1\) - step7: Simplify: \(1=1\) - step8: Substitute the value of \(x_{1}:\) \(x_{2}=-1-8\left(-\frac{1+3x_{3}}{7}\right)-4x_{3}\) - step9: Simplify: \(x_{2}=\frac{1-4x_{3}}{7}\) - step10: Calculate: \(\left(x_{1},x_{2},x_{3}\right) = \left(-\frac{1+3x_{3}}{7},\frac{1-4x_{3}}{7},x_{3}\right),x_{3} \in \mathbb{R}\) - step11: Alternative Form: \(\textrm{Infinitely many solutions}\) Solve the system of equations \( 2x_{1} - 3x_{2} + 4x_{3} - 1x_{4} = 0; 7x_{1} + 1x_{2} - 8x_{3} + 9x_{4} = 0; 2x_{1} + 8x_{2} + 1x_{3} - 1x_{4} = 0 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}2x_{1}-3x_{2}+4x_{3}-1\times x_{4}=0\\7x_{1}+1\times x_{2}-8x_{3}+9x_{4}=0\\2x_{1}+8x_{2}+1\times x_{3}-1\times x_{4}=0\end{array}\right.\) - step1: Calculate: \(\left\{ \begin{array}{l}2x_{1}-3x_{2}+4x_{3}-x_{4}=0\\7x_{1}+x_{2}-8x_{3}+9x_{4}=0\\2x_{1}+8x_{2}+x_{3}-x_{4}=0\end{array}\right.\) - step2: Solve the equation: \(\left\{ \begin{array}{l}x_{4}=2x_{1}-3x_{2}+4x_{3}\\7x_{1}+x_{2}-8x_{3}+9x_{4}=0\\2x_{1}+8x_{2}+x_{3}-x_{4}=0\end{array}\right.\) - step3: Substitute the value of \(x_{4}:\) \(\left\{ \begin{array}{l}7x_{1}+x_{2}-8x_{3}+9\left(2x_{1}-3x_{2}+4x_{3}\right)=0\\2x_{1}+8x_{2}+x_{3}-\left(2x_{1}-3x_{2}+4x_{3}\right)=0\end{array}\right.\) - step4: Simplify: \(\left\{ \begin{array}{l}25x_{1}-26x_{2}+28x_{3}=0\\11x_{2}-3x_{3}=0\end{array}\right.\) - step5: Solve the equation: \(\left\{ \begin{array}{l}x_{1}=\frac{26x_{2}-28x_{3}}{25}\\11x_{2}-3x_{3}=0\end{array}\right.\) - step6: Substitute the value of \(x_{1}:\) \(11x_{2}-3x_{3}=0\) - step7: Move the expression to the right side: \(11x_{2}=0+3x_{3}\) - step8: Remove 0: \(11x_{2}=3x_{3}\) - step9: Divide both sides: \(\frac{11x_{2}}{11}=\frac{3x_{3}}{11}\) - step10: Divide the numbers: \(x_{2}=\frac{3x_{3}}{11}\) - step11: Substitute the value of \(x_{2}:\) \(x_{1}=\frac{26\times \frac{3x_{3}}{11}-28x_{3}}{25}\) - step12: Simplify: \(x_{1}=-\frac{46x_{3}}{55}\) - step13: Substitute the values of \(x_{1},x_{2}:\) \(x_{4}=2\left(-\frac{46x_{3}}{55}\right)-3\times \frac{3x_{3}}{11}+4x_{3}\) - step14: Simplify: \(x_{4}=\frac{83x_{3}}{55}\) - step15: Calculate: \(\left(x_{1},x_{2},x_{3},x_{4}\right) = \left(-\frac{46x_{3}}{55},\frac{3x_{3}}{11},x_{3},\frac{83x_{3}}{55}\right),x_{3} \in \mathbb{R}\) - step16: Alternative Form: \(\textrm{Infinitely many solutions}\) Solve the system of equations \( 1x_{1} + 1x_{2} + 2x_{3} = 8; -1x_{1} - 2x_{2} + 3x_{3} = 1; 3x_{1} - 7x_{2} + 4x_{3} = 10 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}1\times x_{1}+1\times x_{2}+2x_{3}=8\\-x_{1}-2x_{2}+3x_{3}=1\\3x_{1}-7x_{2}+4x_{3}=10\end{array}\right.\) - step1: Calculate: \(\left\{ \begin{array}{l}x_{1}+x_{2}+2x_{3}=8\\-x_{1}-2x_{2}+3x_{3}=1\\3x_{1}-7x_{2}+4x_{3}=10\end{array}\right.\) - step2: Solve the equation: \(\left\{ \begin{array}{l}x_{1}=8-x_{2}-2x_{3}\\-x_{1}-2x_{2}+3x_{3}=1\\3x_{1}-7x_{2}+4x_{3}=10\end{array}\right.\) - step3: Substitute the value of \(x_{1}:\) \(\left\{ \begin{array}{l}-\left(8-x_{2}-2x_{3}\right)-2x_{2}+3x_{3}=1\\3\left(8-x_{2}-2x_{3}\right)-7x_{2}+4x_{3}=10\end{array}\right.\) - step4: Simplify: \(\left\{ \begin{array}{l}-8-x_{2}+5x_{3}=1\\24-10x_{2}-2x_{3}=10\end{array}\right.\) - step5: Solve the equation: \(\left\{ \begin{array}{l}x_{2}=-9+5x_{3}\\24-10x_{2}-2x_{3}=10\end{array}\right.\) - step6: Substitute the value of \(x_{2}:\) \(24-10\left(-9+5x_{3}\right)-2x_{3}=10\) - step7: Simplify: \(114-52x_{3}=10\) - step8: Move the constant to the right side: \(-52x_{3}=10-114\) - step9: Subtract the numbers: \(-52x_{3}=-104\) - step10: Change the signs: \(52x_{3}=104\) - step11: Divide both sides: \(\frac{52x_{3}}{52}=\frac{104}{52}\) - step12: Divide the numbers: \(x_{3}=2\) - step13: Substitute the value of \(x_{3}:\) \(x_{2}=-9+5\times 2\) - step14: Calculate: \(x_{2}=1\) - step15: Substitute the values of \(x_{2},x_{3}:\) \(x_{1}=8-1-2\times 2\) - step16: Calculate: \(x_{1}=3\) - step17: Calculate: \(\left\{ \begin{array}{l}x_{1}=3\\x_{2}=1\\x_{3}=2\end{array}\right.\) - step18: Check the solution: \(\left\{ \begin{array}{l}x_{1}=3\\x_{2}=1\\x_{3}=2\end{array}\right.\) - step19: Rewrite: \(\left(x_{1},x_{2},x_{3}\right) = \left(3,1,2\right)\) Berikut adalah hasil dari eliminasi Gauss-Jordan untuk setiap sistem persamaan yang diberikan: 1. Untuk sistem persamaan: \[ \begin{align*} 1x_{1} + 1x_{2} + 2x_{3} &= 8 \\ -1x_{1} - 2x_{2} + 3x_{3} &= 1 \\ 3x_{1} - 7x_{2} + 4x_{3} &= 10 \end{align*} \] Hasilnya adalah: \[ (x_{1}, x_{2}, x_{3}) = (3, 1, 2) \] 2. Untuk sistem persamaan: \[ \begin{align*} 2x_{1} + 2x_{2} + 2x_{3} &= 0 \\ -2x_{1} + 5x_{2} + 2x_{3} &= 1 \\ 8x_{1} + 1x_{2} + 4x_{3} &= -1 \end{align*} \] Hasilnya adalah: \[ \text{Infinitely many solutions} \] 3. Untuk sistem persamaan: \[ \begin{align*} 1w - 1x + 2y - 1z &= -1 \\ 2w + 1x - 2y - 2z &= -2 \\ -1w + 2x - 4y + 1z &= 1 \\ 3w + 0x + 0y - 3z &= -3 \end{align*} \] Hasilnya adalah: \[ \text{Infinitely many solutions} \] 4. Untuk sistem persamaan: \[ \begin{align*} 2x_{1} - 3x_{2} + 4x_{3} - 1x_{4} &= 0 \\ 7x_{1} + 1x_{2} - 8x_{3} + 9x_{4} &= 0 \\ 2x_{1} + 8x_{2} + 1x_{3} - 1x_{4} &= 0 \end{align*} \] Hasilnya adalah: \[ \text{Infinitely many solutions} \] 5. Untuk sistem persamaan: \[ \begin{align*} 1x_{1} + 3x_{2} - 1x_{3} &= 0 \\ 0x_{1} + 1x_{2} - 8x_{3} &= 0 \\ 0x_{1} + 0x_{2} + 4x_{3} &= 0 \end{align*} \] Hasilnya adalah: \[ (x_{1}, x_{2}, x_{3}) = (0, 0, 0) \] Jika ada pertanyaan lebih lanjut atau jika Anda memerlukan penjelasan lebih lanjut tentang salah satu sistem, silakan beri tahu!