Math Problem Statement
TAE-2 - Div- II.pdf
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CONVERT A MATRIX FROM QUESTION 5 TO UPPER TRIANGULAR MATRIX
Solution
To convert the matrix from Question 5 into an upper triangular matrix, we first extract the linear system:
3x + 2y - z &= 1, \\ x + 2y + 2z &= 0, \\ 2x + y + 3z &= -1. \end{aligned}$$ This can be written in matrix form as $$AX = B$$, where: $$A = \begin{bmatrix} 3 & 2 & -1 \\ 1 & 2 & 2 \\ 2 & 1 & 3 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}.$$ ### Steps to Convert to Upper Triangular Matrix 1. **Initial Matrix:** Form the augmented matrix $$[A|B]$$: $$\left[ \begin{array}{ccc|c} 3 & 2 & -1 & 1 \\ 1 & 2 & 2 & 0 \\ 2 & 1 & 3 & -1 \end{array} \right].$$ 2. **Row Operations:** We will use row reduction (Gaussian elimination) to convert this matrix to upper triangular form. - **Step 1:** Make the element at the first row, first column (pivot) equal to 1. Swap $$R_1$$ with $$R_2$$: $$\left[ \begin{array}{ccc|c} 1 & 2 & 2 & 0 \\ 3 & 2 & -1 & 1 \\ 2 & 1 & 3 & -1 \end{array} \right].$$ - **Step 2:** Eliminate the elements below the pivot in the first column: - For $$R_2$$: $$R_2 = R_2 - 3R_1$$: $$\left[ \begin{array}{ccc|c} 1 & 2 & 2 & 0 \\ 0 & -4 & -7 & 1 \\ 2 & 1 & 3 & -1 \end{array} \right].$$ - For $$R_3$$: $$R_3 = R_3 - 2R_1$$: $$\left[ \begin{array}{ccc|c} 1 & 2 & 2 & 0 \\ 0 & -4 & -7 & 1 \\ 0 & -3 & -1 & -1 \end{array} \right].$$ - **Step 3:** Move to the second pivot (second row, second column) and make it 1. Multiply $$R_2$$ by $$-\frac{1}{4}$$: $$\left[ \begin{array}{ccc|c} 1 & 2 & 2 & 0 \\ 0 & 1 & \frac{7}{4} & -\frac{1}{4} \\ 0 & -3 & -1 & -1 \end{array} \right].$$ - **Step 4:** Eliminate the element below the pivot in the second column: - For $$R_3$$: $$R_3 = R_3 + 3R_2$$: $$\left[ \begin{array}{ccc|c} 1 & 2 & 2 & 0 \\ 0 & 1 & \frac{7}{4} & -\frac{1}{4} \\ 0 & 0 & \frac{19}{4} & -\frac{7}{4} \end{array} \right].$$ 3. **Final Upper Triangular Matrix:** The augmented matrix is now in upper triangular form: $$\left[ \begin{array}{ccc|c} 1 & 2 & 2 & 0 \\ 0 & 1 & \frac{7}{4} & -\frac{1}{4} \\ 0 & 0 & \frac{19}{4} & -\frac{7}{4} \end{array} \right].$$ ### Summary The matrix has been successfully converted into an upper triangular form using Gaussian elimination. Would you like a step-by-step explanation of how to solve this system using back-substitution or any further details? ### 5 Related Questions 1. What is the next step after obtaining the upper triangular form of a matrix? 2. How can you find the determinant of a matrix using its triangular form? 3. What is the difference between row echelon form and reduced row echelon form? 4. How do row operations affect the solutions of a linear system? 5. Can Gaussian elimination be used on non-square matrices? ### Tip Always start Gaussian elimination by choosing the row with the largest absolute value in the pivot column to minimize numerical errors during computations.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Gaussian Elimination
Upper Triangular Matrix
Formulas
Row operations for Gaussian elimination
Upper triangular matrix form
Theorems
Gaussian Elimination
Suitable Grade Level
Undergraduate level (First-year university)
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