Ax = b: Solving Systems of Linear Equations and Describing Solution Sets

Math Problem Statement

Let Aequalsleft bracket Start 3 By 3 Matrix 1st Row 1st Column 1 2nd Column negative 4 3rd Column negative 3 2nd Row 1st Column negative 2 2nd Column 2 3rd Column 0 3rd Row 1st Column 3 2nd Column 0 3rd Column 3 EndMatrix right bracket and bequalsleft bracket Start 3 By 1 Matrix 1st Row 1st Column b 1 2nd Row 1st Column b 2 3rd Row 1st Column b 3 EndMatrix right bracket . Show that the equation Axequalsb does not have a solution for all possible b, and describe the set of all b for which Axequalsb does have a solution. Question content area bottom Part 1 How can it be shown that the equation Axequalsb does not have a solution for all possible b? Choose the correct answer below. A. Find a vector x for which Axequalsb is the zero vector. B. Row reduce the matrix A to demonstrate that A does not have a pivot position in every row. Your answer is correct.C. Row reduce the augmented matrix left bracket Start 1 By 2 Matrix 1st Row 1st Column Upper A 2nd Column Bold b EndMatrix right bracket to demonstrate that left bracket Start 1 By 2 Matrix 1st Row 1st Column Upper A 2nd Column Bold b EndMatrix right bracket has a pivot position in every row. D. Find a vector b for which the solution to Axequalsb is the zero vector. E. Row reduce the matrix A to demonstrate that A has a pivot position in every row. Part 2 Describe the set of all b for which Axequalsb does have a solution. 0equals enter your response here (Type an expression using b 1, b 2, and b 3 as the variables and 1 as the coefficient of b 3.)

Solution

Part 1: How can it be shown that the equation $A \mathbf{x} = \mathbf{b}$ does not have a solution for all possible $\mathbf{b}$ ?

To show that the equation does not have a solution for all possible $\mathbf{b}$ , the correct approach is:

B. Row reduce the matrix $A$ to demonstrate that $A$ does not have a pivot position in every row.

This is because if $A$ does not have a pivot in every row, the system $A \mathbf{x} = \mathbf{b}$ will be inconsistent for some choices of $\mathbf{b}$ . Specifically, if the row-reduced form of $A$ shows that there is at least one row of zeros in $A$ , there will be cases where the augmented system cannot be consistent, and hence, no solution will exist for all $\mathbf{b}$ .

Part 2: Describe the set of all $\mathbf{b}$ for which $A \mathbf{x} = \mathbf{b}$ does have a solution.

To determine the conditions on $\mathbf{b}$ for which the equation $A \mathbf{x} = \mathbf{b}$ has a solution, we need to row reduce the augmented matrix $\left[ A | \mathbf{b} \right]$ .

The matrix $A$ is:

1 & -4 & -3 \\ -2 & 2 & 0 \\ 3 & 0 & 3 \end{pmatrix}$$ Let $$\mathbf{b} = \begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix}$$. So, the augmented matrix is: $$\left[ A | \mathbf{b} \right] = \begin{pmatrix} 1 & -4 & -3 & | & b_1 \\ -2 & 2 & 0 & | & b_2 \\ 3 & 0 & 3 & | & b_3 \end{pmatrix}$$ Now, perform the row reduction: 1. **First, row reduce the first column**: $$R_2 \leftarrow R_2 + 2R_1$$, and $$R_3 \leftarrow R_3 - 3R_1$$: $$\left[ A | \mathbf{b} \right] = \begin{pmatrix} 1 & -4 & -3 & | & b_1 \\ 0 & -6 & -6 & | & b_2 + 2b_1 \\ 0 & 12 & 12 & | & b_3 - 3b_1 \end{pmatrix}$$ 2. **Next, simplify the second column**: $$R_3 \leftarrow R_3 + 2R_2$$: $$\left[ A | \mathbf{b} \right] = \begin{pmatrix} 1 & -4 & -3 & | & b_1 \\ 0 & -6 & -6 & | & b_2 + 2b_1 \\ 0 & 0 & 0 & | & b_3 - 3b_1 + 2(b_2 + 2b_1) \end{pmatrix}$$ 3. **Simplify further**: $$b_3 - 3b_1 + 2(b_2 + 2b_1) = b_3 - 3b_1 + 2b_2 + 4b_1 = b_3 + 2b_2 + b_1$$. Now the row-reduced matrix looks like: $$\left[ A | \mathbf{b} \right] = \begin{pmatrix} 1 & -4 & -3 & | & b_1 \\ 0 & -6 & -6 & | & b_2 + 2b_1 \\ 0 & 0 & 0 & | & b_3 + 2b_2 + b_1 \end{pmatrix}$$ From this, the system is consistent if and only if the last row does not lead to a contradiction. Specifically, for the system to be consistent, we require: $$b_3 + 2b_2 + b_1 = 0$$ Therefore, the set of all $$\mathbf{b}$$ for which $$A \mathbf{x} = \mathbf{b}$$ has a solution is: $$b_3 + 2b_2 + b_1 = 0$$ This equation describes the set of all $$\mathbf{b}$$ for which the system $$A \mathbf{x} = \mathbf{b}$$ has a solution. --- Would you like further details or clarifications on any part of the solution? Here are some related questions to explore further: 1. What does it mean for a matrix to have a pivot in every row? 2. How do we interpret the row-reduced form of a matrix in terms of the solution to a system? 3. What happens if $$b_3 + 2b_2 + b_1 \neq 0$$? 4. How can we use this row reduction method for larger matrices? 5. What is the geometric interpretation of having no solution in a system like this? **Tip**: When row reducing augmented matrices, always check the last row for inconsistency—it can reveal whether there are no solutions for certain values of $$b$$.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Systems of Linear Equations
Row Reduction
Pivot Positions
Inconsistent Systems

Formulas

Row reduction algorithm
Conditions for a system to have a solution (e.g., consistency of augmented matrix)

Theorems

Theorem of consistency of linear systems
Rank and pivot position in matrix analysis

Suitable Grade Level

Grades 11-12

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