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]](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)
≠ 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.
Answer:
d
Step-by-step explanation:
Answer:
Let coordinates of vertex D be (x,y)
In parallelogram diagonals are bisect each other.
∴ Mid-point of AC= Mid-point of BD
⇒ (
2
3+(−6)
,
2
−4+2
)=(
2
−1+x
,
2
−3+y
)
⇒ (
2
−3
,
2
−2
)=(
2
−1+x
,
2
−3+y
)
⇒ (
2
−3
,−1)=(
2
−1+x
,
2
−3+y
)
Now,
⇒
2
−3
=
2
−1+x
⇒ −6=−2+2x
⇒ −4=2x
∴ x=−2
⇒ −1=
2
−3+y
⇒ −2=−3+y
⇒ 1=y
∴ y=1
∴ Coordinates of vertex D is (−2,1)
Answer:
$4.8
Step-by-step explanation:
$24×20/100=$4.8
Undefined is the answer for me. Lol
