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
Let's break down the problem systematically. We are given a matrix equation:
where
\quad \text{and} \quad b = \begin{bmatrix} 6+k \\ 5-k \\ k \end{bmatrix}$$ We are tasked with determining the values of $$k$$ for which this system has: - No solutions, - Exactly one solution, - Infinitely many solutions. ### Step 1: The augmented matrix of the system We can set up the augmented matrix for the system: $$\left[\begin{array}{ccc|c} 1 & 0 & 1 & 6+k \\ 2 & 1 & 0 & 5-k \\ 3 & 1 & 1 & k \\ \end{array}\right]$$ We will use **row reduction** to analyze the system of equations. ### Step 2: Row Reduction Start with the augmented matrix: $$\left[\begin{array}{ccc|c} 1 & 0 & 1 & 6+k \\ 2 & 1 & 0 & 5-k \\ 3 & 1 & 1 & k \\ \end{array}\right]$$ **First**, perform row operations to eliminate entries below the pivot (the $$1$$ in the first row, first column): - Subtract 2 times row 1 from row 2 to eliminate the $$2$$ in the first column of row 2. - Subtract 3 times row 1 from row 3 to eliminate the $$3$$ in the first column of row 3. $$R_2 \to R_2 - 2R_1, \quad R_3 \to R_3 - 3R_1$$ This gives: $$\left[\begin{array}{ccc|c} 1 & 0 & 1 & 6+k \\ 0 & 1 & -2 & -7+3k \\ 0 & 1 & -2 & -18+3k \\ \end{array}\right]$$ **Second**, subtract row 2 from row 3 to eliminate the $$1$$ in the second column of row 3: $$R_3 \to R_3 - R_2$$ This gives: $$\left[\begin{array}{ccc|c} 1 & 0 & 1 & 6+k \\ 0 & 1 & -2 & -7+3k \\ 0 & 0 & 0 & -11+3k \\ \end{array}\right]$$ ### Step 3: Analyzing the system Now, let's examine the last row of the reduced matrix: $$0x + 0y + 0z = -11 + 3k$$ For the system to have: - **No solution**: The last row represents an inconsistent equation, i.e., $$0 = -11 + 3k$$, which is impossible if $$-11 + 3k \neq 0$$. This happens when $$k \neq \frac{11}{3}$$. - **Exactly one solution**: This case occurs when the rank of the matrix is 3 (i.e., there are three nonzero rows in the reduced matrix). This happens when $$k \neq \frac{11}{3}$$, because the third row becomes inconsistent when $$k \neq \frac{11}{3}$$. - **Infinitely many solutions**: This occurs when the last row becomes a row of zeros and doesn't introduce any inconsistency. This happens when $$k = \frac{11}{3}$$, because the equation $$0 = 0$$ holds, and the system has fewer independent equations than unknowns, leading to infinitely many solutions. ### Step 4: Conclusion - **No solutions**: This occurs for $$k \neq \frac{11}{3}$$. - **Exactly one solution**: This occurs for $$k \neq \frac{11}{3}$$. - **Infinitely many solutions**: This occurs for $$k = \frac{11}{3}$$. ### Final Answer: - **No solutions**: $$k \neq \frac{11}{3}$$ - **Exactly one solution**: $$k \neq \frac{11}{3}$$ - **Infinitely many solutions**: $$k = \frac{11}{3}$$ Let me know if you'd like further clarification on any part of the solution!Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Matrix Equations
Row Reduction
Systems of Linear Equations
Formulas
Matrix equation Ax = b
Row reduction techniques
Rank of a matrix
Theorems
Rank-Nullity Theorem
Conditions for consistency of a system of linear equations
Suitable Grade Level
Grades 11-12
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