Math Problem Statement
Solve by using matrix inverses: open curly brackets table attributes columnalign left end attributes row cell 2 x plus y minus z equals 4 end cell row cell 2 x plus 2 y minus z equals short dash 5 end cell row cell short dash x short dash y plus z equals 3 end cell end table close
A.) left parenthesis 7 comma space short dash 9 comma space 1 right parenthesis
B.) left parenthesis 1 comma space short dash 9 comma space 7 right parenthesis
C.) left parenthesis 9 comma space short dash 2 comma space 10 right parenthesis
D.) left parenthesis 0 comma space short dash 5 comma space 4 right parenthesis
Solution
To solve the system of linear equations using matrix inverses, let's follow these steps:
The system of equations:
We rewrite this in matrix form:
2 & 1 & -1 \\ 2 & 2 & -1 \\ -1 & -1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 4 \\ -5 \\ 3 \end{bmatrix}$$ ### Step 1: Represent as $$AX = B$$ Where: $$A = \begin{bmatrix} 2 & 1 & -1 \\ 2 & 2 & -1 \\ -1 & -1 & 1 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 4 \\ -5 \\ 3 \end{bmatrix}.$$ ### Step 2: Find the inverse of $$A$$, denoted $$A^{-1}$$. Using matrix inverse properties, $$X = A^{-1}B$$. #### Calculate the determinant of $$A$$: $$\text{det}(A) = 2 \begin{vmatrix} 2 & -1 \\ -1 & 1 \end{vmatrix} - 1 \begin{vmatrix} 2 & -1 \\ -1 & 1 \end{vmatrix} - 1 \begin{vmatrix} 2 & 2 \\ -1 & -1 \end{vmatrix}.$$ Compute minor determinants: $$\begin{vmatrix} 2 & -1 \\ -1 & 1 \end{vmatrix} = (2)(1) - (-1)(-1) = 2 - 1 = 1.$$ $$\begin{vmatrix} 2 & 2 \\ -1 & -1 \end{vmatrix} = (2)(-1) - (2)(-1) = -2 - (-2) = 0.$$ Substitute: $$\text{det}(A) = 2(1) - 1(1) - 1(0) = 2 - 1 = 1.$$ The determinant is nonzero, so $$A$$ is invertible. #### Compute $$A^{-1}$$ (Adjugate/Determinant method): $$A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A),$$ where $$\text{adj}(A)$$ is the adjugate matrix (transpose of cofactor matrix). 1. Compute cofactors: - C_{11}: $$\begin{vmatrix} 2 & -1 \\ -1 & 1 \end{vmatrix} = 1 \quad \Rightarrow \quad (-1)^{1+1}(1) = 1.$$ - Similarly compute all cofactors... Finish computingimport numpy as np # Define the matrix A and the vector B A = np.array([[2, 1, -1], [2, 2, -1], [-1, -1, 1]]) B = np.array([4, -5, 3]) # Compute the inverse of A A_inverse = np.linalg.inv(A) # Solve for X using A_inverse * B X = np.dot(A_inverse, B) Xarray([ 7., -9., 1.])array([ 7., -9., 1.])successThe solution to the system of equations is: $$x = 7, \, y = -9, \, z = 1$$ This corresponds to **Option A: $$(7, -9, 1)$$**. Let me know if you'd like further explanation or have additional questions! ### Related Questions: 1. How do you calculate the determinant of a 3x3 matrix step by step? 2. What are the properties of matrix inverses, and how are they useful in solving systems of equations? 3. How can you verify if a matrix is invertible without calculating its determinant? 4. What are the differences between solving systems using inverses versus other methods like Gaussian elimination? 5. How can matrix solutions be extended to larger systems (e.g., 4x4 or higher)? ### Tip: For matrices with large dimensions, numerical methods or software like Python, MATLAB, or calculators are often more efficient for finding inverses and solutions.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Systems of Linear Equations
Matrix Inverses
Formulas
AX = B, where A is the coefficient matrix, X is the variable vector, and B is the constant vector
Matrix inverse formula: A⁻¹ = adj(A) / det(A)
Determinant of a 3x3 matrix
Theorems
Properties of Inverse Matrices
Cramer's Rule
Determinants and Matrix Inverses
Suitable Grade Level
Grades 10-12
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