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
≠ 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]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26%5Cbeta%261-%5Cbeta%5C%5C2%262%260%5C%5C2-2%5Cbeta%264%260%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26%5Cbeta%5C%5C2%262%5C%5C2-2%5Cbeta%264%5Cend%7Barray%7D%5Cright%5D)
![8(1-\beta)-[2(2-2\beta)(1-\beta)]](https://tex.z-dn.net/?f=8%281-%5Cbeta%29-%5B2%282-2%5Cbeta%29%281-%5Cbeta%29%5D)




β ≠ ±1
For the system to have only one solution, β ≠ 1 or β ≠ -1.
x=8 is the answer fellow user
Answer:
−
3
x
=
15 Divide each term in −
3
x
=
15 by -
3
.
−
3
x
−
3
+15
=−
3
Cancel the common factor of −
3
.
Cancel the common factor.
−
3
x
−
3
=
15
−
3
Divide x
by 1
.
x
=
15
−
3 by −
3
.
x
=
−
5