Math Problem Statement
x1+4x2-2x3+8x4=12 x2-7x3+2x4=-14 5x3-x4=7 x3+3x4=15 selesaikan dengan cara obe
Solution
To solve the system of linear equations using OBE (Operasi Baris Elementer), or Gaussian Elimination, we will use row operations to reduce the augmented matrix to its row echelon form (REF) and then to its reduced row echelon form (RREF). The given system of equations is:
x_1 + 4x_2 - 2x_3 + 8x_4 = 12 \\ x_2 - 7x_3 + 2x_4 = -14 \\ 5x_3 - x_4 = 7 \\ x_3 + 3x_4 = 15 \end{cases}$$ ### Step 1: Write the Augmented Matrix The augmented matrix for this system is: $$\left[ \begin{array}{cccc|c} 1 & 4 & -2 & 8 & 12 \\ 0 & 1 & -7 & 2 & -14 \\ 0 & 0 & 5 & -1 & 7 \\ 0 & 0 & 1 & 3 & 15 \end{array} \right]$$ ### Step 2: Apply Elementary Row Operations We will perform a series of row operations to transform the matrix to REF and then RREF. 1. **R3 -> (1/5) * R3** to make the leading coefficient of the third row a 1: $$\left[ \begin{array}{cccc|c} 1 & 4 & -2 & 8 & 12 \\ 0 & 1 & -7 & 2 & -14 \\ 0 & 0 & 1 & -\frac{1}{5} & \frac{7}{5} \\ 0 & 0 & 1 & 3 & 15 \end{array} \right]$$ 2. **R4 -> R4 - R3** to make the leading coefficient of the fourth row 0: $$\left[ \begin{array}{cccc|c} 1 & 4 & -2 & 8 & 12 \\ 0 & 1 & -7 & 2 & -14 \\ 0 & 0 & 1 & -\frac{1}{5} & \frac{7}{5} \\ 0 & 0 & 0 & \frac{16}{5} & \frac{68}{5} \end{array} \right]$$ 3. **R4 -> (5/16) * R4** to make the leading coefficient of the fourth row a 1: $$\left[ \begin{array}{cccc|c} 1 & 4 & -2 & 8 & 12 \\ 0 & 1 & -7 & 2 & -14 \\ 0 & 0 & 1 & -\frac{1}{5} & \frac{7}{5} \\ 0 & 0 & 0 & 1 & \frac{17}{4} \end{array} \right]$$ 4. **R3 -> R3 + \frac{1}{5} * R4** to make the third element in the third row a 0: $$\left[ \begin{array}{cccc|c} 1 & 4 & -2 & 8 & 12 \\ 0 & 1 & -7 & 2 & -14 \\ 0 & 0 & 1 & 0 & 6 \\ 0 & 0 & 0 & 1 & \frac{17}{4} \end{array} \right]$$ 5. **R2 -> R2 + 7 * R3** to eliminate the third element in the second row: $$\left[ \begin{array}{cccc|c} 1 & 4 & -2 & 8 & 12 \\ 0 & 1 & 0 & 2.6 & 28 \\ 0 & 0 & 1 & 0 & 6 \\ 0 & 0 & 0 & 1 & \frac{17}{4} \end{array} \right]$$ 6. **R1 -> R1 + 2 * R3** to eliminate the third element in the first row: $$\left[ \begin{array}{cccc|c} 1 & 4 & 0 & 8 & 24 \\ 0 & 1 & 0 & 2.6 & 28 \\ 0 & 0 & 1 & 0 & 6 \\ 0 & 0 & 0 & 1 & \frac{17}{4} \end{array} \right]$$ 7. **R1 -> R1 - 8 * R4** and **R2 -> R2 - 2.6 * R4** to make all elements in the last column of R1 and R2 zeroes: $$\left[ \begin{array}{cccc|c} 1 & 4 & 0 & 0 & -10 \\ 0 & 1 & 0 & 0 & 17 \\ 0 & 0 & 1 & 0 & 6 \\ 0 & 0 & 0 & 1 & \frac{17}{4} \end{array} \right]$$ ### Step 3: Interpret the Results Now, the augmented matrix is in reduced row echelon form. From this, we can read off the solutions: $$\begin{cases} x_1 = -10 \\ x_2 = 17 \\ x_3 = 6 \\ x_4 = \frac{17}{4} \end{cases}$$ ### Conclusion The solutions for the variables $$x_1, x_2, x_3,$$ and $$x_4$$ are: $$x_1 = -10, \quad x_2 = 17, \quad x_3 = 6, \quad x_4 = \frac{17}{4}$$ Would you like more details on any step or have any questions? ### Related Questions: 1. How is Gaussian elimination different from Gauss-Jordan elimination? 2. What are the properties of a matrix in row echelon form? 3. Can every system of linear equations be solved using Gaussian elimination? 4. How can we determine if a system has no solutions or infinitely many solutions using row reduction? 5. How does the choice of pivot element affect the speed of Gaussian elimination? ### Tip: When performing row operations, always keep track of your steps to avoid mistakes and to make it easier to backtrack if necessary!Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Systems of Linear Equations
Gaussian Elimination
Row Operations
Formulas
Ax = b (Matrix Representation of Systems)
Elementary Row Operations (Swapping, Scaling, Adding rows)
Theorems
Gaussian Elimination Theorem
Row Echelon Form
Reduced Row Echelon Form
Suitable Grade Level
Undergraduate (Linear Algebra Course) or Advanced High School (AP Math)
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