Math Problem Statement
Cramer matrix show all calculation steps
Solution
To solve the system of linear equations using Cramer's rule, we'll go through each step carefully. Here are the equations extracted from the image:
6x + y - 3z &= 5 \quad \text{(Equation 1)} \\ x + 3y - 2z &= 5 \quad \text{(Equation 2)} \\ 2x + y + 4z &= 8 \quad \text{(Equation 3)} \end{aligned}$$ ### Step 1: Write the system in matrix form The system of equations can be written as: $$\mathbf{A} \cdot \mathbf{x} = \mathbf{B}$$ Where: $$\mathbf{A} = \begin{pmatrix} 6 & 1 & -3 \\ 1 & 3 & -2 \\ 2 & 1 & 4 \end{pmatrix}, \quad \mathbf{x} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad \mathbf{B} = \begin{pmatrix} 5 \\ 5 \\ 8 \end{pmatrix}$$ ### Step 2: Calculate the determinant of matrix $$\mathbf{A}$$ $$\text{det}(\mathbf{A}) = \begin{vmatrix} 6 & 1 & -3 \\ 1 & 3 & -2 \\ 2 & 1 & 4 \end{vmatrix}$$ This determinant can be expanded as: $$\text{det}(\mathbf{A}) = 6 \begin{vmatrix} 3 & -2 \\ 1 & 4 \end{vmatrix} - 1 \begin{vmatrix} 1 & -2 \\ 2 & 4 \end{vmatrix} + (-3) \begin{vmatrix} 1 & 3 \\ 2 & 1 \end{vmatrix}$$ Calculating each 2x2 determinant: $$\begin{vmatrix} 3 & -2 \\ 1 & 4 \end{vmatrix} = (3)(4) - (-2)(1) = 12 + 2 = 14$$ $$\begin{vmatrix} 1 & -2 \\ 2 & 4 \end{vmatrix} = (1)(4) - (-2)(2) = 4 + 4 = 8$$ $$\begin{vmatrix} 1 & 3 \\ 2 & 1 \end{vmatrix} = (1)(1) - (3)(2) = 1 - 6 = -5$$ Now, substitute back: $$\text{det}(\mathbf{A}) = 6(14) - 1(8) - 3(-5) = 84 - 8 + 15 = 91$$ ### Step 3: Calculate determinants of matrices $$\mathbf{A}_x$$, $$\mathbf{A}_y$$, and $$\mathbf{A}_z$$ #### Determinant of $$\mathbf{A}_x$$ Replace the first column of $$\mathbf{A}$$ with $$\mathbf{B}$$: $$\mathbf{A}_x = \begin{pmatrix} 5 & 1 & -3 \\ 5 & 3 & -2 \\ 8 & 1 & 4 \end{pmatrix}$$ $$\text{det}(\mathbf{A}_x) = 5 \begin{vmatrix} 3 & -2 \\ 1 & 4 \end{vmatrix} - 1 \begin{vmatrix} 5 & -2 \\ 8 & 4 \end{vmatrix} + (-3) \begin{vmatrix} 5 & 3 \\ 8 & 1 \end{vmatrix}$$ Calculate the 2x2 determinants: $$\begin{vmatrix} 3 & -2 \\ 1 & 4 \end{vmatrix} = 14 \quad \text{(as calculated earlier)}$$ $$\begin{vmatrix} 5 & -2 \\ 8 & 4 \end{vmatrix} = (5)(4) - (-2)(8) = 20 + 16 = 36$$ $$\begin{vmatrix} 5 & 3 \\ 8 & 1 \end{vmatrix} = (5)(1) - (3)(8) = 5 - 24 = -19$$ Substitute back: \[ \text{det}(\mathbf{A}_x) = 5(14) - 1(36) - 3(-19) = 70 - 36 +Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Cramer's Rule
Determinants
Matrix Operations
Formulas
Cramer's Rule for solving linear systems
Theorems
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Suitable Grade Level
Advanced High School or College
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