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Hello!
I have a simple question about solutions, better said number of solutions for this system of equations.
\[ \begin{cases} x_{1 } − x_{2 } + 3x_{3 } − 2x_{4 } = 1\\ −2x_{1 } + 2cx_{2 } − 4x_{3 } + 2x_{4 } = −7\\ − 2x_{3 } + (−c + 6)x_{4 } = 2c + 15\\ − 2x_{3 } + c^{2 }x_{4 } = c^{2 }\end{cases} \]
I know it's only possible that this system has either 0, 1 or \( \infty \) number of solutions, for different values of c:
\[ c = -3 \rightarrow \infty\\ c = 1 \rightarrow \infty\\ c = 2 \rightarrow 0 \\ c \in ℝ \setminus \{-3, 1, 2\} \rightarrow 1 \]
My question is: for which c has this system at the utmost 2 solutions? Should it be only for when the whole system has only one solution or also when it has none? Thank you for your help!
I have a simple question about solutions, better said number of solutions for this system of equations.
\[ \begin{cases} x_{1 } − x_{2 } + 3x_{3 } − 2x_{4 } = 1\\ −2x_{1 } + 2cx_{2 } − 4x_{3 } + 2x_{4 } = −7\\ − 2x_{3 } + (−c + 6)x_{4 } = 2c + 15\\ − 2x_{3 } + c^{2 }x_{4 } = c^{2 }\end{cases} \]
I know it's only possible that this system has either 0, 1 or \( \infty \) number of solutions, for different values of c:
\[ c = -3 \rightarrow \infty\\ c = 1 \rightarrow \infty\\ c = 2 \rightarrow 0 \\ c \in ℝ \setminus \{-3, 1, 2\} \rightarrow 1 \]
My question is: for which c has this system at the utmost 2 solutions? Should it be only for when the whole system has only one solution or also when it has none? Thank you for your help!