Math Problem Statement
We are given a matrix equation Ax=b where A=⎡⎣⎢101210311⎤⎦⎥b=⎡⎣⎢6+k5−kk⎤⎦⎥ Determine for which values of k this equation has no solutions, for which it has exactly one solution, and for which it has infinitely many solutions.
Note. Answers must be entered in one of the following forms:
What you can enter: What it means: Example k=a The condition is satisfied only if k is equal to a. k=4 k!=a The condition is satisfied for any value of k different than a. k!=4 k>a The condition is satisfied for any value of k greater than a. k>4 k>=a The condition is satisfied for any value of k greater or equal to a. k>=4 k<a The condition is satisfied for any value of k smaller than a. k<4 k<=a The condition is satisfied for any value of k smaller or equal to a. k<=4 Any The condition is satisfied when k is an arbitrary real number. Any None There is no value of k that would satisfy the condition. None
Values of k for which give no solutions: Any incorrect Any
Values of k which give exactly one solution: None correct None
Values of k which infinitely many solutions: K = 11 incorrect K=11
Solution
We are given the matrix equation , where:
1 & 0 & 1 \\ 2 & 1 & 0 \\ 3 & 1 & 1 \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} 6 + k \\ 5 - k \\ k \end{bmatrix}.$$ We need to determine the values of $$k$$ for which the system has no solutions, exactly one solution, or infinitely many solutions. ### Step 1: Write the augmented matrix for the system We can represent the system as the augmented matrix $$[A | b]$$: $$\left[\begin{array}{ccc|c} 1 & 0 & 1 & 6 + k \\ 2 & 1 & 0 & 5 - k \\ 3 & 1 & 1 & k \end{array}\right]$$ ### Step 2: Perform Gaussian elimination We perform row operations to reduce the augmented matrix to row echelon form. 1. Subtract $$2 \times$$ row 1 from row 2: $$R_2 \rightarrow R_2 - 2R_1$$ $$\left[\begin{array}{ccc|c} 1 & 0 & 1 & 6 + k \\ 0 & 1 & -2 & -7 - k \\ 3 & 1 & 1 & k \end{array}\right]$$ 2. Subtract $$3 \times$$ row 1 from row 3: $$R_3 \rightarrow R_3 - 3R_1$$ $$\left[\begin{array}{ccc|c} 1 & 0 & 1 & 6 + k \\ 0 & 1 & -2 & -7 - k \\ 0 & 1 & -2 & k - 3(6 + k) \end{array}\right]$$ Simplifying the right-hand side of row 3: $$k - 3(6 + k) = k - 18 - 3k = -2k - 18$$ So the augmented matrix becomes: $$\left[\begin{array}{ccc|c} 1 & 0 & 1 & 6 + k \\ 0 & 1 & -2 & -7 - k \\ 0 & 1 & -2 & -2k - 18 \end{array}\right]$$ 3. Subtract row 2 from row 3: $$R_3 \rightarrow R_3 - R_2$$ $$\left[\begin{array}{ccc|c} 1 & 0 & 1 & 6 + k \\ 0 & 1 & -2 & -7 - k \\ 0 & 0 & 0 & -2k - 18 - (-7 - k) \end{array}\right]$$ Simplifying the right-hand side of row 3: $$-2k - 18 - (-7 - k) = -2k - 18 + 7 + k = -k - 11$$ Now the augmented matrix becomes: $$\left[\begin{array}{ccc|c} 1 & 0 & 1 & 6 + k \\ 0 & 1 & -2 & -7 - k \\ 0 & 0 & 0 & -k - 11 \end{array}\right]$$ ### Step 3: Analyze the system The third row of the augmented matrix is: $$0x + 0y + 0z = -k - 11$$ This represents the equation: $$0 = -k - 11$$ This equation holds only when $$k = -11$$. Thus, the system has a contradiction (i.e., no solution) when $$k = -11$$. ### Step 4: Determine the types of solutions 1. **No solution**: The system has no solution when $$k = -11$$ because the last row represents the inconsistent equation $$0 = -k - 11$$, which is false. 2. **Exactly one solution**: The system has exactly one solution when $$k \neq -11$$. In this case, the rank of the augmented matrix will be 3, which means the system is consistent and has a unique solution. 3. **Infinitely many solutions**: There are no values of $$k$$ that would lead to infinitely many solutions. This would only occur if the rank of the augmented matrix was less than the number of variables, which does not happen here. ### Final Answer: - **No solutions**: $$k = -11$$ - **Exactly one solution**: $$k \neq -11$$ - **Infinitely many solutions**: None Let me know if you need further details! ### Related questions to explore: 1. How does Gaussian elimination help identify the number of solutions in a system of linear equations? 2. What does it mean for a system to be consistent or inconsistent? 3. How can we apply the rank of a matrix to determine the number of solutions to a system? 4. What would happen if the matrix $$A$$ was singular (i.e., its determinant was zero)? 5. How do the values of $$k$$ affect the solution set of the system? **Tip**: Always check for possible inconsistencies, like zero rows with non-zero constants, when solving systems of linear equations.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Matrix Equations
Gaussian Elimination
Systems of Linear Equations
Formulas
Gaussian Elimination
Matrix Operations
Row Echelon Form
Theorems
Rank of a Matrix
Consistency of Linear Systems
Suitable Grade Level
Grades 11-12
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