Answer: β ≠ ±1
Step-by-step explanation: For a system of equations to have an unique solution, its determinant must be different from 0: det |A| ≠ 0. So,
det [tex]\left[\begin{array}{ccc}1&\beta&1-\beta\\2&2&0\\2-2\beta&4&0\end{array}\right][/tex] ≠ 0
Determinant of a 3x3 matrix is calculated by:
det [tex]\left[\begin{array}{ccc}1&\beta&1-\beta\\2&2&0\\2-2\beta&4&0\end{array}\right]\left[\begin{array}{ccc}1&\beta\\2&2\\2-2\beta&4\end{array}\right][/tex]
[tex]8(1-\beta)-[2(2-2\beta)(1-\beta)][/tex]
[tex]8-8\beta-4+8\beta-4\beta^{2}[/tex]
[tex]-4\beta^{2}+4\neq 0[/tex]
[tex]\beta^{2}\neq 1[/tex]
[tex]\beta \neq \sqrt{1}[/tex]
β ≠ ±1
For the system to have only one solution, β ≠ 1 or β ≠ -1.
