# Number of solutions for system of equations

• MHB
• bargaj

#### bargaj

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!

YOU just said that such a system has either 0 or 1 or infinitely many solutions. So you know it is impossible to have 2 solutions. "At the utmost 2 solutions" must mean no solutions or 1 solution. There will be exactly one solution if c is such that the determinant of coefficients is NOT 0. There will be no solution if c is such that the determinant of coefficients is 0 but the right hand side is not.