Math Problem: Linear Algebra Assignment Paper #4 (Math Problem Sample)

Linear Algebra: Assignment 4
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Introduction
Assignment 4 of the linear algebra module deals with elementary vector operations with applications of algebra methods to solve linear systems of equations. Methods, properties and theorems are adopted from Howard Anton & Chris Rorres book, Elementary Linear Algebra (11th Edition). The solutions are herein indicated stepwise and in order of questions as they appear in the original assignment file.
Question 1(a)
u=(1,2,-3,5,0)
v=(0,4,-1,1,2)
w=(7,1,-4,-2,3)
* v+w=0,4,-1,1,2+7,1,-4,-2,3
=0+7,4+,1,-1-4,1-2,2+3
=7,5,-5,-1.5
* 3(2u-v=3(21,2,-3,5,0-(0,4,-1,1,2))
=3(2,4,-6,10,0-(0,4,-1,1,2))
=3(2-0,4-4,-6+1,10-1,0-2)
=3(2,0,-5,9,-2))
=(6,0,-15,27,-6)
* 3u-v)-(2u+4w)=
=31,2,-3,5,0-0,4,-1,1,2-21,2,-3,5,0+7,1,-4,-2,3
=(3,6,-9,15,0-(0,4,-1,1,2))-(2,4,-6,10,0+(28,4,-16,-8,12))
=3-0,6-4,-9+1,15-1,0-2-2+28,4+4,-6-16,10-8,0+12
=3,2,-8,14,-2-30,8,-22,2,12
=(-28,-6,-30,12,-14)
* 12(w-5v+2u)
=127,1,-4,-2,3-50,4,-1,1,2+21,2,-3,5,0+0,4,-1,1,2
=127,1,-4,-2,3-0,20, -5,5,10+2,4,-6,10,0+(0,4-1,1,2)
=127-0+2,1-20+4,-4+5-6,-2-5+10,3-10+0+0,4,-1,1,12
=129,-15,-5,3-7+(0,4,-1,1,2)
=92,-152,-52,32,-72+0,4-1,1,2
=(92,-72,-72,52,-32)
Question 1(b)
Given:
c1-1,0,2+c12,2,-2+c31,-2,1=(-6,12,4)
-c1+2c2+c3=-6 …i
2c2-2c3=12 …ii
2c1-2c2+c3=4 …(iii)
Solve eqn i and eqn (iii) to eliminate the variable c1
-2c1+4c2+2c3=-12
2c1-2c2+c3=4
Adding the two equations, the following resultant equation is obtained:
2c2+3c3=-8 …(iv)
Solving equation (ii) and equation (iv):
2c2-2c3=12
2c2+3c3=-8
⟹5c3=-20
c3=-4
∴ c2-c3=6 ⟹ c2=2
Also:
c1=6+2c2+c3
c1=6+4-4
c1=6
Question 2(a)
* Given:
u=(1,2-3,0)
v=5,1,2,-2
∴d=1-52+2-12+-3-22+(0--21)2
=16+1+25+4
46
and
cosθ=u.vuv=1*5+2*1+-3*2+0*212+22+(-3)2+02*52+12+22+(-2)2
=5+2-6+01434=11434
θ=cos-1(11434)=87.37
Hence θ<900
* u=(0,1,1,1,2)
v=2,1,0,-1,3
∴d=4+0+1+4+1=10
cosθ=0+1+0-1+61+1+1+44+1+1+9=6715
θ=cos-167=54.150
θ is an acute angle since θ<900
Question 2(b)
According to Cauchy-Schwarz inequality
u.v≤uv
* u=(4,1,1)
v=1,2,3
u.v=4*1+1*2+1*3
=9
u.v=9
u=16+1+1=18
v=1+4+9=15
u.v=18*15=16.43
9<16.43 ∴Cauchy-Schwarz inequality holds.
*
u=(1,2,1,2,3)
v=0,1,1,5,-2
u.v=0+2+1+10+-6=7
u.v=7=7
u=1+4+1+4+9=19
v=0+1+25+4=31
u.v=19*31=24.26
7<24.26∴Cauchy-Schwarz inequality holds
Question 3(a)
The given planes are:
1 2x-y+z-1=0
2 2x-y+z+1=0
Here:
a=2, b=-1, c=1, d1=-1, d2=1
As the planes are parallel; therefore the distance between the plane is:
d=d1-d2a2+b2+c2
= -1-122+(-1)2+12
=-24+1+1
=26=63≈0.816496
distance between parrallel planes=d=0.816496
Question 3(b)
We know that if a and b are orthogonal vectors, then:
a.b=0
* Here v.w
=a,-b.(-b,a)
=a*-b+b*a
=-ab+ab
=0
Since v.w=0
∴ v & w are orthogonal vectors
* Let another unit vector be:
u=(cosθ,sinθ)
u=cos2θ+sin2θ=1
u.v=cosθ,sinθ.-3,4=1
⟹-3cosθ+4sinθ=0
4sinθ=3cosθ
tanθ=34
θ=tan-134
=Ï€+tan-134
when θ= tan-134
u=(costan-134, sin⁡(tan-134))
=(45,35)
when θ=π+tan-134
u=(cosπ+tan-134, sin⁡(π+tan-134))
=(-45,-35)
Question 4(a)
x1+3x2-4x3=0
x1+2x2+3x3=0
13-4123 x1x2x3=00
The augmented matrix is
13-412300
Implementing row operations on the augmented matrix
R2-R1→13-40-1700
R1+3.R2→10170-1700
-1.R2→101701-700
x1+17x3=0 ⟹ x1=-17x3
x2-7x3=0 ⟹ x2=7x3
Let x3=t
⟹ x1=-17t x2=7t t∈R
The solution is:
=-17t,7t,tt∈R }
Which are orthogonal to:
1,3,-4 & (1,2,3)
Question 4(b)
*
a=-3,2,-1
b=(0,-2,-2)<...