5 advanced math questions

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Answered Same Day Aug 07, 2021

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Rajeswari answered on Aug 08 2021

140 Votes

63070 assignment
6a)
Is the linear combination where a,b,c are any real scalars
) B is a basis of C because B is linearly independent since determinant value not equals 0. Also C^3 has rank of 3, and this basis has dimension 3.
c)
This is linear combination of using above.
Then we have a+ib+(1+i)c = 0
+c+i(a+b) = -1+i
a+b+i(a+c) =0
We find no solution for a,b and c.
Qno.6d
Hence adj A is transpose of cofactor matrix
=
Q.no.7
Qno.9
a) Substitute for a,b,c to get
=
) Matrix for T = coefficients of x,y,z =
c) Trace of A = -2+4-4 = -2 (sum of diagonal elements)
d) |A| = -2(-16+7)-5(4-7)+5(1-4) = 18+10+15 =43
e) Av = kV
Hence Av = (-2, 2, -2)T=k(1,-1,1)
Solving k = -2
f) A-kI =0 gives
(k+2)(k-3)(k+3) =0
Other eigen values are 3, -3
Eigen vectors are:
A-3I =0 gives vector as (1 1 0)
X1+5x2+5x3 =0: -x1+x2+7x3 =0 and x1-x2-7x3 =0
So we get vector as (1 1 0)
and
A+3I = 0 gives vector as (0 -1 0)
g) TB is a diagonal matrix means T is adj basis
Hence adj would be =
Qno.7
a)
Plane given is 2x-8y+5z =0
The reflection point would be at equidistant from the plane as the...

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