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.
Take the y and x and 4 as a common factor and you will get 4xy(2yz+3x) the second one is the answer
X² + x - 20
= (x+5)(x-4)
x + 5 = 0
x = -5
x - 4 = 0
x = 4
hence the answer is C
First divide by 2
|x+1/4|=3
Then you pick the possible answers
|x+1/4|=3
|x+1/4|=-3
2 3/4 or -3 1/4
