Math Problem Statement
Consider the matrix A = [[2, 6, 6], [4, 14, 13], [2, 7, 7]]. (a) Use the inversion algorithm to find A^(-1). (b) Express Ax = b as a linear system where x = (x1, x2, x3) and b = (1, -3, 2). (c) Use A^(-1) to solve the linear system in part (b).
Solution
Let's break down the problem and solve it step by step:
We are given the matrix and asked to perform three tasks:
Matrix A:
2 & 6 & 6 \\ 4 & 14 & 13 \\ 2 & 7 & 7 \end{pmatrix}$$ ### Part (a): Use the inversion algorithm to find $$A^{-1}$$ To find $$A^{-1}$$, we will use the formula for the inverse of a matrix: $$A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A)$$ 1. **Step 1: Find $$\text{det}(A)$$** 2. **Step 2: Find the adjugate matrix $$\text{adj}(A)$$** 3. **Step 3: Calculate $$A^{-1}$$ using the formula** ### Part (b): Express $$A\mathbf{x} = \mathbf{b}$$ as a linear system Given: $$\mathbf{x} = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 1 \\ -3 \\ 2 \end{pmatrix}$$ This gives us the system of linear equations: $$\begin{aligned} 2x_1 + 6x_2 + 6x_3 &= 1 \\ 4x_1 + 14x_2 + 13x_3 &= -3 \\ 2x_1 + 7x_2 + 7x_3 &= 2 \end{aligned}$$ ### Part (c): Use $$A^{-1}$$ to solve the system Once we have $$A^{-1}$$, we will solve for $$\mathbf{x}$$ using the relation: $$\mathbf{x} = A^{-1} \mathbf{b}$$ Now, I will calculate the determinant of $$A$$, find the adjugate, and determine $$A^{-1}$$. Let's proceed with that calculation.### Part (a): Inverse of Matrix $$A$$ The determinant of $$A$$ is $$\text{det}(A) = 2$$, which is non-zero, so the inverse exists. The inverse matrix $$A^{-1}$$ is: $$A^{-1} = \begin{pmatrix} 3.5 & 0 & -3 \\ -1 & 1 & -1 \\ 0 & -1 & 2 \end{pmatrix}$$ ### Part (b): Linear System The system of equations $$A\mathbf{x} = \mathbf{b}$$ is expressed as: $$\begin{aligned} 2x_1 + 6x_2 + 6x_3 &= 1 \\ 4x_1 + 14x_2 + 13x_3 &= -3 \\ 2x_1 + 7x_2 + 7x_3 &= 2 \end{aligned}$$ ### Part (c): Solving the System Using $$A^{-1}$$ Using the inverse matrix $$A^{-1}$$, we solve for $$\mathbf{x}$$ by calculating $$\mathbf{x} = A^{-1} \mathbf{b}$$. The solution is: $$\mathbf{x} = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} -2.5 \\ -6 \\ 7 \end{pmatrix}$$ This is the solution to the system. --- Would you like further details or have any questions? Here are some related questions you could explore: 1. How is the adjugate matrix $$\text{adj}(A)$$ calculated? 2. What happens if the determinant of a matrix is zero? 3. How does matrix inversion differ from solving systems with Gaussian elimination? 4. Can you solve a system with dependent or inconsistent equations using inverses? 5. How does matrix inversion behave computationally for larger matrices? **Tip:** The determinant gives you insight into whether a matrix is invertible, but it also reveals important geometric properties like the volume scaling factor of linear transformations.Ask a new question for Free
By Image
Drop file here or Click Here to upload
Math Problem Analysis
Mathematical Concepts
Matrix Algebra
Linear Systems
Inverse of a Matrix
Formulas
A^(-1) = (1/det(A)) * adj(A)
Ax = b
x = A^(-1) * b
Theorems
Matrix Inversion Theorem
Determinant of a Matrix
Suitable Grade Level
Undergraduate