1)
Let (2−i)=a+bi where a,b∈R
2−i=a2−b2+2 abi
a2−b2=2 and 2ab=−1 ⟹b=2a−1
a2−(2a−1)2=2⟹4a4−8a2−1=0
a2=88−+64+16 But a∈R
∴a2=2.118
a=−+1.4553
b=2a−1=−+0.3436
2−i=−+(1.4553−0.3436i)
2)
A=(5−6−32)
Characteristic polynomial
P(t)=det(A−tI)=(5−t−6−32−t)
=(5−t)(2−t)−18
=t2−7t−8
Solving t2−7t−8=0,t=8or−1
Eigenvalues are λ=8λ=−1
For (A−8)x=0
A−8=(−3−6−3−6) →then3−1R1R2−2R1 (1010)
x1=−x2
Eigen vector =(−11)
For (A+1)x=0
A+1=(6−6−33)→Then31R1R2+R1(20−10)
2x1=x2
Eigenvector = (12)
P=(−1112)
3)
The zero of R is defined as
0+0+a(0)(0)=0
This shows that S is a subspace of R2 if and only b=0
Also S is a subspace of R2 if it is closed under addition and scalar multiplication
Let x1y1∈R2 and x2y2∈R2
(x1+y1),(x2+y2):(x1+y1)+(x2+y2)+a(x1+y1)(x2+y2)=0
(x1y1)+(x2y2)=x1+y1+ax1y1+x2+y2+ax2y2
=(x1+y)+(y1+y2)+a(x1y1+x2y2)
S is not a subspace of R2 unless a=0 b=0
Let α∈R
α(x1y)=α(x+y+axy)
=αx+xy+aαxy
α(x)x(y)=αx+αy+aα2xy
S is not a subspace of R2 Unless a=0,b=0
4)
U=x⎝⎛10100⎠⎞+y⎝⎛01102⎠⎞+z⎝⎛00011⎠⎞
U is a subspace spanned by linearly independent set S={(1,0,1,0,0)+(0,1,1,0,2)+(0,0,0,1,1)}
By adjoining elements of F5 that will not affect linear independence of
(0,0,0,1,1) and (0,0,0,0,1)
Span W
= { ⎣⎡00011⎦⎤ ,⎣⎡00001⎦⎤ }
Or W={(0,0,0,x,x+y)}
Such that F5 =u⨁w
5)
The subset S of the vector space V is not a basis if
(i) S is linearly dependent
(ii) S is not a spanning set
Row echelon form of S is ⎝⎛2000270007−5t7+15⎠⎞
⟹7−5t+715=0
∴t=3
The vectors are not basis of R3 if t=3
Comments
Leave a comment