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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]

User Selvamani
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1 Answer

2 votes

Answer: β ≠ ±1

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\\eq 0


\beta^(2)\\eq 1


\beta \\eq √(1)

β ≠ ±1

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

User Erix
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