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:
D. y = -5/7x + 5
Step-by-step explanation:
5x + 7y = 35
-5x -5x
7y = 35 - 5x
/7 /7 /7
y = 35/7 - 5/7 x
y = 5 - 5/7x
Answer:
30 centimeters
Step-by-step explanation:
12 times 2.5 = 30
Answer:
(10-8) + n
Step-by-step explanation:
Good luck with your work!
Answer:
C
Step-by-step explanation:
The center of inscribed circle into triangle is point of intersection of all interior angles of triangle.
The center of circumscribed circle over triabgle is point of intersection of perpendicular bisectors to the sides.
Circumscribed circle always passes through the vertices of the triangle.
Inscribed circle is always tangent to the triangle's sides.
In your case angles' bisectors and perpendicular bisectors intesect at one point, so point A is the center of inscribed circle and the center of corcumsribed circle. Thus, these circles pass through the points X, Y, Z and G, E, F, respectively.