Math Problem Statement

Solve the system of equations using augmented matrix methods: 3x_1 + 4x_2 = 1 x_1 - 2x_2 = 7

Solution

To solve this system of linear equations using the augmented matrix method, we need to start by setting up the augmented matrix for the system:

Given equations: 3x1+4x2=13x_1 + 4x_2 = 1 x12x2=7x_1 - 2x_2 = 7

Step 1: Write the Augmented Matrix

The augmented matrix for this system is: [341127]\begin{bmatrix} 3 & 4 & | & 1 \\ 1 & -2 & | & 7 \end{bmatrix}

Step 2: Use Row Operations to Achieve Row Echelon Form

  1. First, make the first element in the first row (pivot in the first column) a 1 by dividing the entire row by 3: R1=[334313]=[14313127]R_1 = \left[\frac{3}{3} \quad \frac{4}{3} \quad | \quad \frac{1}{3}\right] = \begin{bmatrix} 1 & \frac{4}{3} & | & \frac{1}{3} \\ 1 & -2 & | & 7 \end{bmatrix}

  2. Next, eliminate the 1 below this pivot in the second row by performing the operation R2=R2R1R_2 = R_2 - R_1: R2=[127][14313]=[0103203]R_2 = \begin{bmatrix} 1 & -2 & | & 7 \end{bmatrix} - \begin{bmatrix} 1 & \frac{4}{3} & | & \frac{1}{3} \end{bmatrix} = \begin{bmatrix} 0 & -\frac{10}{3} & | & \frac{20}{3} \end{bmatrix} So, our matrix now looks like: [143130103203]\begin{bmatrix} 1 & \frac{4}{3} & | & \frac{1}{3} \\ 0 & -\frac{10}{3} & | & \frac{20}{3} \end{bmatrix}

  3. Simplify the second row by dividing by 103-\frac{10}{3}: R2=[012]R_2 = \left[0 \quad 1 \quad | \quad -2\right] The matrix becomes: [14313012]\begin{bmatrix} 1 & \frac{4}{3} & | & \frac{1}{3} \\ 0 & 1 & | & -2 \end{bmatrix}

  4. Finally, use this result to eliminate the 43\frac{4}{3} in the first row. Perform the operation R1=R143R2R_1 = R_1 - \frac{4}{3} R_2: R1=[14313]43×[012]=[103]R_1 = \begin{bmatrix} 1 & \frac{4}{3} & | & \frac{1}{3} \end{bmatrix} - \frac{4}{3} \times \begin{bmatrix} 0 & 1 & | & -2 \end{bmatrix} = \begin{bmatrix} 1 & 0 & | & 3 \end{bmatrix} Now, the augmented matrix is: [103012]\begin{bmatrix} 1 & 0 & | & 3 \\ 0 & 1 & | & -2 \end{bmatrix}

Step 3: Write the Solution

From this final matrix, we can see that: x1=3andx2=2x_1 = 3 \quad \text{and} \quad x_2 = -2

Solution

The solution to the system is: (x1,x2)=(3,2)(x_1, x_2) = (3, -2)

Would you like more details on any of these steps?


Here are 5 related questions:

  1. How do you interpret row operations in the context of solving a matrix?
  2. What are other methods for solving systems of linear equations besides the augmented matrix method?
  3. How can you verify the solution of a system of equations?
  4. What is the difference between row echelon form and reduced row echelon form?
  5. How does the augmented matrix method handle inconsistent systems with no solution?

Tip: When solving systems of equations, always check if the system is consistent (has solutions) or inconsistent (has no solutions) before applying row reduction.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Systems of Equations
Augmented Matrices
Row Operations

Formulas

Row operations to achieve row echelon form in an augmented matrix

Theorems

Row Echelon Form
Reduced Row Echelon Form

Suitable Grade Level

Grades 10-12