Math Problem Statement

Given the matrix A and vector b, solve the system A * x = b and write the solution as a vector.

Solution

This image contains a set of exercises and problems related to linear algebra, including systems of equations, vector spaces, matrix transformations, and concepts like span, linear combinations, and pivot positions. Let's address one of the questions.

Let's start by tackling Exercise 11:

Exercise 11:

Given the matrix AA and vector bb:

A=[10452143],b=[29]A = \begin{bmatrix} 1 & 0 & 4 & 5 \\ -2 & -1 & -4 & -3 \end{bmatrix}, \quad b = \begin{bmatrix} -2 \\ 9 \end{bmatrix}

Task:

Write the augmented matrix for the system Ax=bA \mathbf{x} = b, solve the system, and write the solution as a vector.

Step 1: Form the augmented matrix

The augmented matrix for Ax=bA \mathbf{x} = b is:

[Ab]=[1045221439]\left[ A | b \right] = \begin{bmatrix} 1 & 0 & 4 & 5 & -2 \\ -2 & -1 & -4 & -3 & 9 \end{bmatrix}

Step 2: Use row reduction (Gaussian elimination)

We will perform row operations to bring this matrix to row echelon form:

  1. Multiply Row 1 by 2 and add it to Row 2: R2=2R1+R2R_2 = 2R_1 + R_2

    This gives:

    [1045201475]\begin{bmatrix} 1 & 0 & 4 & 5 & -2 \\ 0 & -1 & 4 & 7 & 5 \end{bmatrix}

  2. Multiply Row 2 by 1-1 to simplify: R2=1×R2R_2 = -1 \times R_2

    The result is: [1045201475]\begin{bmatrix} 1 & 0 & 4 & 5 & -2 \\ 0 & 1 & -4 & -7 & -5 \end{bmatrix}

  3. Eliminate the second entry in Row 1 by subtracting 4 times Row 2 from Row 1: R1=R14R2R_1 = R_1 - 4R_2

    The final matrix is:

    [10011801475]\begin{bmatrix} 1 & 0 & 0 & -1 & 18 \\ 0 & 1 & -4 & -7 & -5 \end{bmatrix}

Step 3: Back-substitute to find the solution

We now interpret this as the system of equations:

x1x4=18andx24x37x4=5x_1 - x_4 = 18 \quad \text{and} \quad x_2 - 4x_3 - 7x_4 = -5

From here, we can solve for x1x_1, x2x_2, x3x_3, and x4x_4, expressing x1x_1 and x2x_2 in terms of free variables x3x_3 and x4x_4. The solution is:

x1=x4+18,x2=4x3+7x45x_1 = x_4 + 18, \quad x_2 = 4x_3 + 7x_4 - 5

Thus, the general solution can be written as:

x=[x1x2x3x4]=[18500]+x3[0410]+x4[1701]\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} 18 \\ -5 \\ 0 \\ 0 \end{bmatrix} + x_3 \begin{bmatrix} 0 \\ 4 \\ 1 \\ 0 \end{bmatrix} + x_4 \begin{bmatrix} 1 \\ 7 \\ 0 \\ 1 \end{bmatrix}

This describes the solution space.


Would you like more details or explanations on any specific part? Let me know if you'd like me to continue with other exercises!

Here are 5 related questions to expand on this:

  1. What is the geometric interpretation of the solution space in Exercise 11?
  2. How does Gaussian elimination help in solving linear systems efficiently?
  3. What does the presence of free variables indicate about the nature of the solutions to a system of equations?
  4. How can we determine if a matrix is consistent or inconsistent from its augmented form?
  5. What role do pivot positions play in the solvability of a system Ax=bA\mathbf{x} = b?

Tip: In linear algebra, understanding the row-reduced echelon form (RREF) of a matrix is crucial, as it reveals whether a system is consistent and how many solutions it has.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Matrix Equations
Gaussian Elimination
Vector Spaces

Formulas

Augmented matrix form: [A | b]
Row reduction via Gaussian elimination
Back-substitution for solving systems

Theorems

Gaussian Elimination
Row Echelon Form (REF)

Suitable Grade Level

Undergraduate (Linear Algebra Course)