Math Problem Statement
Solve the following system of equations by Cramer’s rule x + y + z = 5, 3x − 2y + 2z = 3, 2x + 2y − z = 4
Solution
To solve the system of equations using Cramer's Rule, we follow these steps:
Given system of equations:
1. & \quad x + y + z = 5 \\ 2. & \quad 3x - 2y + 2z = 3 \\ 3. & \quad 2x + 2y - z = 4 \end{align*}$$ ### Step 1: Write the system in matrix form This system can be written in matrix form as: $$\mathbf{A} \mathbf{x} = \mathbf{b}$$ where: $$\mathbf{A} = \begin{pmatrix} 1 & 1 & 1 \\ 3 & -2 & 2 \\ 2 & 2 & -1 \end{pmatrix} ,\quad \mathbf{x} = \begin{pmatrix} x \\ y \\ z \end{pmatrix} ,\quad \mathbf{b} = \begin{pmatrix} 5 \\ 3 \\ 4 \end{pmatrix}$$ ### Step 2: Find the determinant of matrix $$A$$ We first calculate the determinant of the coefficient matrix $$\mathbf{A}$$: $$\det(\mathbf{A}) = \begin{vmatrix} 1 & 1 & 1 \\ 3 & -2 & 2 \\ 2 & 2 & -1 \end{vmatrix}$$ Using cofactor expansion along the first row: $$\det(\mathbf{A}) = 1 \cdot \begin{vmatrix} -2 & 2 \\ 2 & -1 \end{vmatrix} - 1 \cdot \begin{vmatrix} 3 & 2 \\ 2 & -1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 3 & -2 \\ 2 & 2 \end{vmatrix}$$ Now, calculate the 2x2 determinants: $$\det \begin{vmatrix} -2 & 2 \\ 2 & -1 \end{vmatrix} = (-2)(-1) - (2)(2) = 2 - 4 = -2$$ $$\det \begin{vmatrix} 3 & 2 \\ 2 & -1 \end{vmatrix} = (3)(-1) - (2)(2) = -3 - 4 = -7$$ $$\det \begin{vmatrix} 3 & -2 \\ 2 & 2 \end{vmatrix} = (3)(2) - (-2)(2) = 6 + 4 = 10$$ Now substitute back: $$\det(\mathbf{A}) = 1(-2) - 1(-7) + 1(10) = -2 + 7 + 10 = 15$$ ### Step 3: Find determinants for $$x$$, $$y$$, and $$z$$ Next, we replace the corresponding columns of $$\mathbf{A}$$ with $$\mathbf{b}$$ and compute the determinants to find $$x$$, $$y$$, and $$z$$. - **For $$x$$:** $$\mathbf{A}_x = \begin{pmatrix} 5 & 1 & 1 \\ 3 & -2 & 2 \\ 4 & 2 & -1 \end{pmatrix}$$ $$\det(\mathbf{A}_x) = 5 \cdot \begin{vmatrix} -2 & 2 \\ 2 & -1 \end{vmatrix} - 1 \cdot \begin{vmatrix} 3 & 2 \\ 4 & -1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 3 & -2 \\ 4 & 2 \end{vmatrix}$$ We already know $$\begin{vmatrix} -2 & 2 \\ 2 & -1 \end{vmatrix} = -2$$. Now calculate the other two: $$\det \begin{vmatrix} 3 & 2 \\ 4 & -1 \end{vmatrix} = (3)(-1) - (2)(4) = -3 - 8 = -11$$ $$\det \begin{vmatrix} 3 & -2 \\ 4 & 2 \end{vmatrix} = (3)(2) - (-2)(4) = 6 + 8 = 14$$ Now substitute: $$\det(\mathbf{A}_x) = 5(-2) - 1(-11) + 1(14) = -10 + 11 + 14 = 15$$ - **For $$y$$:** $$\mathbf{A}_y = \begin{pmatrix} 1 & 5 & 1 \\ 3 & 3 & 2 \\ 2 & 4 & -1 \end{pmatrix}$$ \[ \det(\mathbf{A}_y) = 1 \cdot \begin{vmatrix} 3 & 2 \\ 2 & -1 \end{vmatrix} - 5 \cdot \begin{vmatrix} 3 & 2 \\ 2 & -1 \end{vmatrix}Ask a new question for Free
By Image
Drop file here or Click Here to upload
Math Problem Analysis
Mathematical Concepts
Linear Algebra
Systems of Equations
Determinants
Formulas
Cramer's Rule
Determinant of a 3x3 Matrix
Theorems
Cramer's Rule Theorem
Suitable Grade Level
Grades 10-12
Related Recommendation
Solve System of Equations Using Cramer's Rule: x + y + z = 5, 3x − 2y + 2z = 3, 2x + 2y − z = 4
Solve the System of Equations Using Cramer's Rule: 3x + y + 2z = 3
Solve a 3x3 System Using Cramer's Rule - Example Problem
Solve System of Equations Using Cramer's Rule: 4x + 5y = 2, 11x + y + 2z = 3, x + 5y + 2z = 1
Solve System of Equations with Cramer's Rule: 2x + 8y + 5z = -5, x + y + z = -2, x + 2y - z = 2