Math Problem Statement
Solve
Solution
We are tasked with solving the system of linear equations using the inverse matrix method, where the matrix inverse is calculated using the Gaussian elimination method. The system of equations is as follows:
Step 1: Represent the system in matrix form
The system can be written as , where:
X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 9 \\ -6 \\ 17 \end{bmatrix}.$$ ### Step 2: Find the inverse of $$A$$ using Gaussian elimination We augment $$A$$ with the identity matrix $$I$$ to form $$[A|I]$$, then perform row operations to transform $$[A|I]$$ into $$[I|A^{-1}]$$. $$[A|I] = \begin{bmatrix} 1 & -2 & 3 & 1 & 0 & 0 \\ -1 & 3 & -1 & 0 & 1 & 0 \\ 2 & -5 & 5 & 0 & 0 & 1 \end{bmatrix}.$$ Performing Gaussian elimination: 1. Make the pivot element in row 1, column 1 equal to 1 (already done). 2. Eliminate column 1 below the pivot: - Add row 1 to row 2: $$R_2 \to R_2 + R_1$$. - Subtract 2 × row 1 from row 3: $$R_3 \to R_3 - 2R_1$$. $$\begin{bmatrix} 1 & -2 & 3 & 1 & 0 & 0 \\ 0 & 1 & 2 & 1 & 1 & 0 \\ 0 & -1 & -1 & -2 & 0 & 1 \end{bmatrix}.$$ 3. Scale row 2 to make the pivot in column 2 equal to 1: - $$R_2 \to R_2 \div 1$$ (no change needed). 4. Eliminate column 2 above and below the pivot: - Add 2 × row 2 to row 1: $$R_1 \to R_1 + 2R_2$$. - Add row 2 to row 3: $$R_3 \to R_3 + R_2$$. $$\begin{bmatrix} 1 & 0 & 7 & 3 & 2 & 0 \\ 0 & 1 & 2 & 1 & 1 & 0 \\ 0 & 0 & 1 & -1 & -1 & 1 \end{bmatrix}.$$ 5. Scale row 3 to make the pivot in column 3 equal to 1: - $$R_3 \to R_3$$. 6. Eliminate column 3 above: - Subtract 7 × row 3 from row 1: $$R_1 \to R_1 - 7R_3$$. - Subtract 2 × row 3 from row 2: $$R_2 \to R_2 - 2R_3$$. $$[A|I] = \begin{bmatrix} 1 & 0 & 0 & 1 & 1 & -7 \\ 0 & 1 & 0 & 3 & 3 & -2 \\ 0 & 0 & 1 & -1 & -1 & 1 \end{bmatrix}.$$ Thus, the inverse of $$A$$ is: $$A^{-1} = \begin{bmatrix} 1 & 1 & -7 \\ 3 & 3 & -2 \\ -1 & -1 & 1 \end{bmatrix}.$$ ### Step 3: Multiply $$A^{-1}$$ by $$B$$ Finally, solve $$X = A^{-1}B$$: $$X = \begin{bmatrix} 1 & 1 & -7 \\ 3 & 3 & -2 \\ -1 & -1 & 1 \end{bmatrix} \begin{bmatrix} 9 \\ -6 \\ 17 \end{bmatrix}.$$ Performing the matrix multiplication: $$X = \begin{bmatrix} 1(9) + 1(-6) + (-7)(17) \\ 3(9) + 3(-6) + (-2)(17) \\ -1(9) + -1(-6) + 1(17) \end{bmatrix} = \begin{bmatrix} 9 - 6 - 119 \\ 27 - 18 - 34 \\ -9 + 6 + 17 \end{bmatrix}.$$ $$X = \begin{bmatrix} -116 \\ -25 \\ 14 \end{bmatrix}.$$ ### Final Answer: $$x = -116, \quad y = -25, \quad z = 14.$$ Let me know if you have questions or need further clarification! Here are related questions for practice: 1. How do you check if a matrix is invertible? 2. What are other methods to solve a system of linear equations besides the inverse matrix method? 3. What are the advantages of Gaussian elimination over other techniques? 4. How would the process change if the coefficient matrix was non-square? 5. What is the determinant's role in verifying a matrix's invertibility? **Tip**: Always verify the determinant of a matrix before attempting to calculate its inverse!Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Inverse Matrices
Gaussian Elimination
Systems of Linear Equations
Formulas
AX = B
Matrix Inversion using Gaussian Elimination
Theorems
Matrix Inverse Theorem
Gaussian Elimination Procedure
Suitable Grade Level
Undergraduate Level (or advanced high school)
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