Answer with explanation:
For, a Matrix A , having eigenvector 'v' has eigenvalue =2
The order of matrix is not given.
It has one eigenvalue it means it is of order , 1×1.
→A=[a]
Determinant [a-k I]=0, where k is eigenvalue of the given matrix.
It is given that,
k=2
For, k=2, the matrix [a-2 I] will become singular,that is
→ Determinant |a-2 I|=0
→I=[1]
→a=2
Let , v be the corresponding eigenvector of the given eigenvalue.
→[a-I] v=0
→[2-1] v=[0]
→[v]=[0]
→v=0
Now, corresponding eigenvector(v), when eigenvalue is 2 =0
We have to find solution of the system
→Ax=v
→[2] x=0
→[2 x] =[0]
→x=0, is one solution of the system.
Answer:
The inverse is ±sqrt(100-x)
Step-by-step explanation:
To find the inverse, exchange x and y
x = 100 -y^2
Solve for y
Subtract 100 from each side
x - 100 = -y^2
Divide by -1
-x +100 = y^2
Take the square root of each side
±sqrt(100-x) = sqrt(y^2)
±sqrt(100-x) =y
The inverse is ±sqrt(100-x)
O would be the answer, but I’m not sure why you have 3
(360.00 to be exact)
