Math Problem Statement

Each group should solve at least two systems of linear equations using matrices. Discuss the consistency of the systems.

Solution

To solve systems of linear equations using matrices, you'll follow these steps:

Step 1: Write the System of Equations in Matrix Form

For a system of linear equations, you can express it in the form AX=BAX = B, where:

  • AA is the coefficient matrix.
  • XX is the column matrix (vector) of variables.
  • BB is the column matrix (vector) of constants.

Example 1:

Consider the following system of linear equations:

2x + 3y &= 5 \\ 4x - y &= 11 \end{align*}$$ This can be written in matrix form as: $$\begin{pmatrix} 2 & 3 \\ 4 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 5 \\ 11 \end{pmatrix}$$ Here, $$A = \begin{pmatrix} 2 & 3 \\ 4 & -1 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \end{pmatrix}, \quad B = \begin{pmatrix} 5 \\ 11 \end{pmatrix}$$ ### Step 2: Solve the Matrix Equation To solve for $$X$$, multiply both sides of the equation by the inverse of $$A$$: $$X = A^{-1}B$$ Compute $$A^{-1}$$ and then multiply it by $$B$$ to find the values of $$x$$ and $$y$$. ### Example 2: Consider another system: $$\begin{align*} x - 2y + 3z &= 9 \\ 2x + y - z &= 1 \\ 3x + y + 2z &= 7 \end{align*}$$ This can be written in matrix form as: $$\begin{pmatrix} 1 & -2 & 3 \\ 2 & 1 & -1 \\ 3 & 1 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 9 \\ 1 \\ 7 \end{pmatrix}$$ Here, $$A = \begin{pmatrix} 1 & -2 & 3 \\ 2 & 1 & -1 \\ 3 & 1 & 2 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad B = \begin{pmatrix} 9 \\ 1 \\ 7 \end{pmatrix}$$ Similarly, solve for $$X$$ by finding $$A^{-1}$$ and multiplying by $$B$$. ### Step 3: Discuss the Consistency of the Systems A system is **consistent** if it has at least one solution. There are two types of consistent systems: 1. **Exactly one solution**: This occurs when the determinant of $$A$$ is non-zero ($$\text{det}(A) \neq 0$$). 2. **Infinitely many solutions**: This occurs when the determinant is zero, but the augmented matrix (which includes $$B$$) does not lead to a contradiction. A system is **inconsistent** if it has no solution. This occurs when the determinant of $$A$$ is zero and the augmented matrix leads to a contradiction (e.g., a row reduces to something like $$0 = 1$$). ### Consistency of Example 1: Calculate the determinant of $$A$$: $$\text{det}(A) = (2)(-1) - (3)(4) = -2 - 12 = -14$$ Since $$\text{det}(A) \neq 0$$, the system has a unique solution, making it consistent. ### Consistency of Example 2: Calculate the determinant of $$A$$: $$\text{det}(A) = 1(1 \times 2 - (-1) \times 1) - (-2)(2 \times 2 - (-1) \times 3) + 3(2 \times 1 - 1 \times 3)$$ $$= 1(2 + 1) - (-2)(4 + 3) + 3(2 - 3) = 3 + 14 - 3 = 14$$ Since $$\text{det}(A) \neq 0$$, the system has a unique solution, making it consistent. Would you like more details on the solution process or matrix operations? Here are 5 related questions: 1. How do you calculate the inverse of a matrix? 2. What is the significance of the determinant in determining the consistency of a system? 3. How do you solve a system of linear equations if the determinant is zero? 4. Can a system with more equations than variables still have a unique solution? 5. How would you handle systems with infinitely many solutions in practice? **Tip**: When solving systems of equations, always check the determinant first to quickly determine if a unique solution exists.

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Math Problem Analysis

Mathematical Concepts

Linear Equations
Matrices
Determinants

Formulas

Matrix multiplication
Inverse matrix formula
Determinant calculation

Theorems

Consistency of systems of linear equations

Suitable Grade Level

Grades 10-12