Math Problem Statement

Consider a linear system whose augmented matrix is of the form [[1, 1, 0, 2], [1, 2, 0, 1], [3, 5, a, b]]. For what values of a will the system have a unique solution? For what values of a and b will the system have infinitely many solutions? For what values of a and b will the system be inconsistent?

Solution

We are given an augmented matrix representing a linear system, and the questions are related to the values of aa and bb that determine whether the system has:

  1. A unique solution.
  2. Infinitely many solutions.
  3. No solution (inconsistent system).

The augmented matrix is:

1 & 1 & 0 & 2 \\ 1 & 2 & 0 & 1 \\ 3 & 5 & a & b \end{bmatrix}$$ Let's analyze the system for each question. --- ### (a) **Unique Solution** To find when the system has a unique solution, we need to ensure that the matrix has full rank, i.e., no row is a linear combination of the others. This typically involves performing Gaussian elimination and checking for any contradictions or dependencies between the rows. Let’s start by subtracting Row 1 from Row 2 to eliminate the first column from Row 2: $$R_2 \rightarrow R_2 - R_1: \quad \begin{bmatrix} 1 & 1 & 0 & 2 \\ 0 & 1 & 0 & -1 \\ 3 & 5 & a & b \end{bmatrix}$$ Next, we subtract $$3R_1$$ from Row 3 to eliminate the first column from Row 3: $$R_3 \rightarrow R_3 - 3R_1: \quad \begin{bmatrix} 1 & 1 & 0 & 2 \\ 0 & 1 & 0 & -1 \\ 0 & 2 & a & (b - 6) \end{bmatrix}$$ Now, subtract $$2R_2$$ from Row 3 to simplify the second column: $$R_3 \rightarrow R_3 - 2R_2: \quad \begin{bmatrix} 1 & 1 & 0 & 2 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & a & (b - 4) \end{bmatrix}$$ For the system to have a unique solution, the matrix must be of full rank, meaning $$a \neq 0$$. When $$a \neq 0$$, we can perform back substitution to find the unique solution. Thus, for a unique solution, **$$a \neq 0$$**. The solution depends on the values of $$b$$, and it can be determined through back substitution if required. --- ### (b) **Infinitely Many Solutions** The system will have infinitely many solutions when one of the rows becomes dependent, leading to a free variable. This occurs if $$a = 0$$. Substituting $$a = 0$$ into the matrix: $$\begin{bmatrix} 1 & 1 & 0 & 2 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 0 & (b - 4) \end{bmatrix}$$ For infinitely many solutions, the last row must correspond to the equation $$0 = 0$$, meaning $$b = 4$$. Therefore, the system has infinitely many solutions when **$$a = 0$$ and $$b = 4$$**. --- ### (c) **Inconsistent System** The system is inconsistent if it leads to a contradictory equation like $$0 = c$$, where $$c$$ is a non-zero constant. This occurs if the last row is of the form: $$0 \ 0 \ 0 \ | \ (b - 4)$$ which gives the equation $$0 = b - 4$$. For inconsistency, $$b \neq 4$$ when $$a = 0$$. Thus, the system is inconsistent when **$$a = 0$$ and $$b \neq 4$$**. --- ### Summary: - (a) Unique solution: $$a \neq 0$$. - (b) Infinitely many solutions: $$a = 0$$ and $$b = 4$$. - (c) Inconsistent: $$a = 0$$ and $$b \neq 4$$. Would you like a detailed step-by-step explanation or further clarification? Here are 5 related questions to extend your understanding: 1. What is the significance of matrix rank in determining the type of solutions? 2. How does Gaussian elimination help identify the solution types? 3. What happens geometrically when a system has infinitely many solutions? 4. How do dependent equations affect the structure of the matrix? 5. How does back substitution work in solving linear systems? **Tip:** Always check for both zero rows and contradictions (like $$0 = 1$$) when determining inconsistency in a system.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Gaussian Elimination
Rank of Matrix

Formulas

Row reduction
System of linear equations

Theorems

Unique solution theorem
Consistency of linear systems

Suitable Grade Level

University level

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