Question

For which values of the parameter B the following system admits a unique solution [ 1 b 1-b 2 2 0 2–2B 4 0] [x y z] = [1 2 0]

Answer 1

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 $\left[\begin{array}{ccc}1&\beta&1-\beta\\2&2&0\\2-2\beta&4&0\end{array}\right]$ ≠ 0

Determinant of a 3x3 matrix is calculated by:

det $\left[\begin{array}{ccc}1&\beta&1-\beta\\2&2&0\\2-2\beta&4&0\end{array}\right]\left[\begin{array}{ccc}1&\beta\\2&2\\2-2\beta&4\end{array}\right]$

$8(1-\beta)-[2(2-2\beta)(1-\beta)]$

$8-8\beta-4+8\beta-4\beta^{2}$

$-4\beta^{2}+4\neq 0$

$\beta^{2}\neq 1$

$\beta \neq \sqrt{1}$

β ≠ ±1

For the system to have only one solution, β ≠ 1 or β ≠ -1.