LINEAR ALGEBRA 1 Linear Algebra Name Institution

Running head: LINEAR ALGEBRA 1
Linear Algebra
Name
Institution

LINEAR ALGEBRA 2
Linear Algebra 1
Question 1
Question 2.
Part a.
~c can be written in terms of ~
d, ~e and ~
f as follows,
~c =
~
−f + ~e−~d
Part b.
~g can be written in terms of ~c, ~
d, ~e and ~
k as follows,
~g=− ~
k + ~c+ ~
d− ~e
Part c.
~x can be found by solving the equation below,
~x + ~b= ~
f
From figure 2, ~
f is represented by the equation ~
f = ~a+ ~b.
~
f = ~a+ ~b
~
f = ~x + ~b
Therefore,

LINEAR ALGEBRA 3
~x + ~
b=¿ ~a+ ~
b
~x= ~a+ ~
b− ~
b
Hence,
~x= ~a
Part d
~x can be found by solving the equation below,
~x + ~
h=¿ ~
d− ~e
~
h=¿ ~
d− ~e− ~x
From figure 2, ~
h is represented by the equation ~
h=¿ ~
d− ~e− ~g
Therefore,
~
h=¿ ~
d− ~e− ~x
~
h=¿ ~
d− ~e− ~g
~
d− ~e− ~x= ~
d − ~e− ~g
− ~x=− ~g
Hence.
~x= ~g
Part e.
~x can be found by solving the equation below,
~x= ~e−~a− ~
b− ~
d− ~
k
Making ~
b the subject of the formula, ~
b= ~e− ~a− ~
d − ~
k −~x
From figure 2, ~
b is represented by the equation,
~
b=− ~a+ ~e− ~
d− ~c

LINEAR ALGEBRA 4
Therefore,
− ~a+ ~e− ~
d− ~c= ~e− ~a− ~
d− ~
k− ~x
~c =
~
k− ~x
From figure 2, ~c is represented by the equation,
~c =
~
k− ~
h
hence,
~
k − ~x= ~
k − ~
h
− ~x=− ~
h
~x= ~
h
Question 3.
Part a.
For the function ~a= ~
i+2 ~
j− ~
k
||~a||= √12+22 +12
||~a||= √ 6
Part b.
The unit vector withthe same direction as ~a is given by : ~a
||~a ||
¿
~
i+2 ~
j− ~
k
√ 6
¿ 1
√ 6
~
i+ 2
√ 6
~
j− 1
√ 6
~
k
Part c
The vector will be given by multiplying the unit vector by 5

LINEAR ALGEBRA 5
¿ 5 ( 1
√6
~i+ 2
√6
~j− 1
√6
~k )= 5
√6
~i+ 10
√6
~j− 5
√6
~k
Part d
The opposite direction for a vector ~a= ~
i+2 ~
j− ~
k is given as − ~a
The unit vector withthe same direction as ~a is 1
√6
~i+ 2
√6
~j− 1
√6
~k
The unit vector withopposite direction as ~a is−( 1
√6
~i + 2
√6
~j− 1
√6
~k )
The vector of length 2∈opposite direction=−2 ( 1
√6
~i+ 2
√ 6
~j− 1
√ 6
~k )
¿ −2
√ 6
~i− 4
√ 6
~j+ 2
√ 6
~k
Question 4
Part a
Considering the points P= (1,0,2 ) and Q= (−1,2,5 )
i.⃗ OP= ~
i+0 ~
j+ 2 ~
k= ~
i+2 ~
k⃗
OQ=− ~
i+ 2 ~
j+5 ~
k
ii. To find⃗ PQ, subtract⃗ OPfrom⃗ OQ⃗
PQ =⃗ OQ−⃗ OP⃗
PQ = (−1−1,2−0,5−2 ) = (−2,2,3 )⃗
PQ = (−2,2,3 )
¿−2 ~i+2 ~j+3 ~
k
iii. ¿⃗ PQ= √ (−22 ) + ( 22 )+ ( 32 )= √17=4.1231
Part b
The value of t can be calculated from the two points R= ( 2, t ,3 ) and S= ( 1,1, 11 ) as follows;
First we find ⃗ RS, given by,⃗
RS= ( 1−2,1−t , 11−3 )

LINEAR ALGEBRA 6⃗
RS= (−1 ,1−t , 8 )
d (⃗ RS ) =9
Hence the distance 9 is given by finding the magnitude,
9= √ (−12 ) + ( 1−t )2+ ( 82 )
92 =1+ ( 1−t )2+ 64
81−1−64= ( 1−t ) 2
16= ( 1−t )2
16=1 ( 1−t ) −t ( 1−t )
t2−2 t−15=0
Solving the quadratic equation by the use of quadratic formula,
−b ± √ b2−4 ac
2a
a=1 , b=−2 , c=−15
−2± √ −22 + ( 60 )
2
−2± 8
2
Hence t=3∨t=−5
Part c
the distance from the starting point can be calculated by the use of vectors,
let ~arepresent a vector direction through which the crawl travelled
hence ~a= ( 4 ~i−12 ~j+3 ~k ) in component notation
~a= ( 4 ,−12 , 3 )

LINEAR ALGEBRA 7
Distance is given by the magnitude of ~a
||~a||= √42±122+ 32= √169=13
Therefore, the distance from the starting point is 13 cm.
Part d
Whent=1 , r ( t ) =3 ~
i +12 ~
j−3 ( 1 ) ~
k=3 ~
i+ ~
j−3 ~
k
When t=3 , r ( t ) =3 ~
i+32 ~
j −3 ( 3 ) ~
k=3 ~
i+9 ~
j−9 ~
k
Displacement= ( 3 ~
i+9 ~
j−9 ~
k ) − ( 3 ~
i+ ~
j−3 ~
k )=8 ~
j−6 ~
k
scalar displacement = √ 82 +(−6)2=10
Question 5
Part a
P= (1,2,1 ) , Q= ( 3 , 3 , 6 )∧R (−3 , 0 ,−9 )⃗
PQ =Q−P= (3−1 , 3−2 , 6−1 )= ( 2 ,1 , 5 )⃗
QR =R−Q= (−3−3 , 0−3 ,−9−6 )= (−6 ,−3 ,−15 )⃗
PR=R−P= (−3−1 , 0−2 ,−9−1 )= (−4 ,−2,−10 )
Part b
The three points are collinear since:⃗
QR =−3⃗ PQ ∧⃗ PR=−2⃗ PQ
Question 6
Part a
For the following points in space
P= (1,2,1 ) , Q= ( 3 , 3 , 4 ) , R ( 4 ,3 , 5 ) and S ( 2 ,2 , 2 )
The direction ⃗ PQ ,⃗ QR , ,⃗ SR∧⃗ PS are given as follows;

LINEAR ALGEBRA 8⃗
PQ =Q−P= (3−1 , 3−2 , 4−1 )= (2 , 1 ,3 )⃗
QR =R−Q= ( 4−3 , 3−3 , 5−4 )= ( 1 , 0 ,1 )⃗
SR=R−S= ( 4−2 , 3−2 , 5−2 ) = ( 2 ,1 , 3 )⃗
PS=S−P= ( 2−1 , 2−2 , 2−1 ) = ( 1 , 0 , 1 )
Part b
The quadrilateral is a parallelogram since two opposite sides are equal as we can see from the
vectors calculated above.
Part c ⃗
PQ =Q−P= (3−1 , 3−2 , 4−1 )= (2 , 1 ,3 )⃗
QS=S−Q= ( 2−3 , 2−3 , 2−4 )= (−1 ,−1 ,−2 ) ,
1
2⃗ QS= 1
2 ( −1 ,−1 ,−2 ) = ( −1
2 ,− 1
2 ,−1 )⃗
PM =⃗ PQ+ 1
2⃗ QS⃗
PM = ( 2 ,1 , 3 ) +(−1
2 ,− 1
2 ,−1 )= (2− 1
2 ,1− 1
2 ,3−1)=( 3
2 , 1
2 , 2 )
Part d.
To the direction of ⃗ PN can be determined using the routes,⃗
PN =⃗ PS+ 1
2⃗ SQ and ⃗ PN =⃗ PQ+ 1
2⃗ QS⃗
PS=S−P= ( 2−1 , 2−2 , 2−1 ) = ( 1 , 0 , 1 )
1
2⃗ SQ=1
2 ( Q−S ) =1
2 ( 3−2 , 3−2 , 4−2 ) =1
2 ( 1 , 1 ,2 ) =( 1
2 , 1
2 , 1 )⃗
PN =⃗ PS+ 1
2⃗ SQ

LINEAR ALGEBRA 9⃗
PN = ( 1 ,0 , 1 ) +( 1
2 , 1
2 , 1 )=(1+ 1
2 ,0+ 1
2 ,1+1 )= (3
2 , 1
2 ,2 )
The mid points M=N this is because ⃗
PS+ 1
2⃗ SQ=⃗ PQ + 1
2⃗ QS= (3
2 , 1
2 ,2 )
Since ⃗ PM =⃗ PN then M=N.
Question 7
x , y ∈ R3 having ||~x + ~y||=||~x||+||~y || cab be proved as a false statement as follows
Let ~x be (−1 , 0 ,−2 ) and ~y be (2 , 1 ,3 )
Therefore ~x + ~y= (−1, 0 ,−2 )+( 2 , 1, 3 ) = ( 2−1 ,1+ 0 ,3−2 ) = ( 1 , 1, 1 )
||~x + ~y||= √ 12+12+12= √ 3=1.7321
||~x||= √−12+02 +−22= √5=2.2361
||~y||= √ 22 +12+32= √ 15=3.8730
Hence ||~x||+||~y||=2.2361+3.8730=6.1091
From the above analysis, ||~x + ~y||≠ ||~x||+||~y||
This justifies that the statement is false
LINEAR ALGEBRA 2
QUESTION 1
Part a
~u . ~v =¿ ||u||.||v||. cos θ
¿ 2 ×3 ×cos ( π
3 )=6 × 0.5=3
Part b

LINEAR ALGEBRA 10
Since ~u . ~v =¿ ||u||.||v ||. cos θ
||v||= ~u . ~v
||u||. cos θ
Since ~u=3∧ ~u . ~v =¿ 6, having given θ= π
4
||v||= 6
3 cos ( π
4 )=2.8293
Part c
When ~u . ~v =0 , this indicates that
cos θ=0 ,
Therefore, θ=cos−1 0=90 °
θ=90 °= π
2
Part d
When ~u . ~v =6 ,||v||=3∧||u||=2 , θwould be calculated as;
Making cos θ the subject of the formular
~u . ~v
||u||.||v|| =¿ cos θ
Substituting 6
2× 3 =¿ cos θ
cos θ=1
θ=cos−1 0=0 °=π
Question 2
Part a
~a= ~
i−2 ~
j∧
~
b=3 ~
i+6 ~
j
~a . ~
b= ( ~i−2 ~j ) .(3 ~i+6 ~j)
¿ ( 1 ×3 ) + (−2 ×6 )=−9
Part b
~a=3 ~
i−5 ~
j+4 ~
k

LINEAR ALGEBRA 11
~
b= ~i+ ~j+2 ~
k
~a . ~
b=(3 ~
i−5 ~
j+4 ~
k).( ~
i+ ~
j +2 ~
k )
~a . ~
b= ( 3 ×1 ) + (−5 ×1 ) + ( 4 ×2 ) =6
Part c
~a= ~
i− ~
j+2 ~
k
~
b= ~
i− ~
j− ~
k
~a . ~
b= ( ~
i− ~
j+ 2 ~
k ) .( ~
i− ~
j−
~
k )
~a . ~
b= ( 1× 1 ) + ( −1×−1 ) + ( 2×−1 ) =0
Question 3
Part a
~a= ~
i−2 ~
j
||~a||= √12+−22 = √5=2.2361
||~a||=2.2361
~
b=3 ~i+6 ~j
||~
b||= √ 32+62= √ 25=5
~a . ~
b=¿ ||~a||. ||~
b||. cos θ
From question above part a, ~a . ~
b=−9
Substituting each component to the formula,
−9=¿ 2.2239 ×5 cos θ
cos θ=¿ −9
11.1195 =−0.8094 ¿
θ=cos−1−0.8094=144.04 °
Part b
~a= ( 3 ,−5 , 4 )
||~a||= √32+−52 + 42 = √50=7.0711
||~a||=7.0711
~
b= ( 1 ,1 , 2 )

LINEAR ALGEBRA 12
||~b||= √12 +12 +22= √6=2.4495
||~
b||=2.4495
~a . ~
b=¿ ||~a||. ||~
b||. cos θ
From question above part b, ~a . ~
b=6
Substituting each component to the formula,
6=¿ 7.0711× 2.4495 cos θ
cos θ=¿ 6
17.3207 =0.3464 ¿
θ=cos−1 0.3464=69.73 °
Part c
~a= ( 1 ,−1 , 2 )
||~a||= √12+12 +22= √6=2.4495
||~a||=2.4495
~
b= ( 1 ,−1 ,−1 )
||~
b||= √12 +−12 +12= √3=1.7321
||~b||=1.7321
~a . ~
b=¿ ||~a||. ||~b||. cos θ
From question above part c, ~a . ~
b=0
Substituting each component to the formula,
0=¿ 2.4495 ×1.7321 cos θ
cos θ=¿ 0
4.2428 =0 ¿
θ=cos−1 0=90 °
Question 4
Part a
Finding ~u . ( 3 ~v +5 ~w ) if ~u . ~v =4∧ ~u . ~w=−2

LINEAR ALGEBRA 13
~u . ( 3 ~v +5 ~w )=3 ( ~u . ~v ) +5 ( ~u .~w )
¿ 3 ( 4 ) +5 (−2 )=2
Part b
~u . ( a ~v +2 ~w ) =4, ~u . ~v =2, and ~u . ~w=−3
~u . ( a ~v +2 ~w )=a ( ~u . ~v ) +2 ( ~u . ~w )
¿ a ( 2 ) +2 (−3 )=2a−6=4
a= 4+ 6
2 =5
Part c
Finding ~u . ( 4 ~v +6 ~w ) if ~u . ~v =3∧ ~uis perpendicular to ~w
For a non-zero vectors ~uand ~w to be perpendicular to each
~u . ~w=0
~u . ( 4 ~v +6 ~w )=4 ( ~u . ~v )+ 6 ( ~u . ~w )
¿ 4 ( 3 ) +6 ( 0 ) =12
Part c
~u . ( 5 ~u−3 ~v ) if ~u . ~v =7and ~u as a length of 2
~u . ( 5 ~u−3 ~v )=5 ( ~u . ~u )−3 ( ~u . ~v )
¿ 5 ( 2× 2 )−3 ( 7 )=−1
Part e
finding out ~a . ( ~
b ∥+ ~
b ⊥ ) given that ~
b ∥=2 ~a
length of ¿|~a|∨¿ 5
implying that ¿|~a|∨¿ √ 52
simplify a2=52

LINEAR ALGEBRA 14
~a=5
~
b ∥=2 ~a
~
b ⊥meaning its perpendicular to ~a
Therefore ~
b . ~a=0
~b ⊥=0
~a
Substituting 0
~a and 10 to the function
~a . ( ~b ∥+ ~b ⊥ )= ~a× ( 0
~a +2 ~a )=2 a2 =50
Question 5
Part a

LINEAR ALGEBRA 15
Part b
~
b∨¿ ¿
~
b × ~a
~a × ~a ~a
From part a above,
~
b=4 ~i+3 ~j
~a=4 ~
i−2 ~
j
~
b . ~a= ( 4 ,3 ) × ( 4 ,−2 )= ( 4 × 4 )+ (3 ×−2 )=10
~
b . ~a=10
~a . ~a= ( 4 ,−2 ) × ( 4 ,−2 )= ( 4 × 4 ) + (−2 ×−2 )=20
~a . ~a=20
Hence
~
b ∥= 10
20 ( 4 ,−2 )= ( 2 ,−1 )
~
b ∥= ( 2 ,−1 )
The function is drawn on the graph in part a.
Part c
Calculating the vector
~
b ⊥=
~
b− ~
b ∥
~
b=4 ~
i+3 ~
j
~
i=4∧ ~
j=3
~
b= ( 4 , 3 ) and ~
b ∥= ( 2 ,−1 )
Therefore
~
b− ~
b ∥= ( 4 ,3 )− ( 2 ,−1 ) = ( 4−2 , 3+1 )= ( 2 , 4 )

LINEAR ALGEBRA 16
~
b ⊥= ( 2 , 4 ) and ~a= ( 4 ,−2 )
Verifying that, ~
b ⊥ and ~a are perpendicular
~
b ⊥ . ~a=0 implying that
( 2 , 4 ) × ( 4 ,−2 )= ( 4 ×2 ) + (−2× 4 )=0
Hence this proves that ~
b ⊥ and ~a are perpendicular.
Question 6
Part a
~a=4 ~
i− ~
j+ ~
k , ~
b=−2 ~
i− ~
j−2 ~
k
we take ~
b¿∨¿= −1
2
~i+ 1
4
~j− 1
2
~
k∧ ~
b¿=−3
2
~i− 5
4
~j− 1
2
~
k ¿
Part b
~
b ∥=λ ~a
In the above case λ is the Scalar constant that gives the multiplication ratios.
λ< 0 this implies the scalar is a negative value which is less than zero implying that its
multiplication will switch the direction of the vector point in an opposite direction making it gain
a stretch or shrink on the magnitude of the vector. This depends on the size of the constant.
Question 7
Part a
For the quadrilateral ABCD with vertices A= (−1 ,2 , 3 ), B= ( 0 , 4 , 1 ) , C= ( 3 , 4 ,1 ) and
d= ( 2, 2 , 3 )
About the origin O= ( 0 ,0 , 0 )⃗
OA = A−O= ( −1−0,2−0 , 3−0 ) = ( −1 , 2 ,3 )⃗
OA = (−1 , 2, 3 )

LINEAR ALGEBRA 17⃗
OB =B−O= ( 0−0,4−0 , 1−0 )= ( 0 , 4 ,1 )⃗
OB = ( 0 , 4 , 1 )⃗
OC=C−O= (3−0,4−0 , 1−0 )= ( 3 , 4 ,1 )⃗
OC= ( 3 , 4 ,1 )⃗
OD=D−O= ( 2−0,2−0 ,3−0 ) = ( 2 ,2 , 3 )
Part b
Finding⃗ AC and⃗ BD⃗
AC=⃗ OC−⃗ OA = ( 3 , 4 ,1 ) − (−1 ,2 , 3 )= ( 3+1 , 4−2 , 1−3 )= ( 4 ,2 ,−2 )⃗
AC= ( 4 , 2,−2 )⃗
BD=⃗ OD−⃗ OB= ( 2, 2 , 3 )− ( 0 , 4 ,1 )= ( 2−0,2−4 ,3−1 )= ( 2 ,−2 ,2 )⃗
BD= ( 2 ,−2 ,2 )
Part c⃗
AC and⃗ BD are the diagonals of the quadrilateral
If⃗ AC and⃗ BD are perpendicular to each other, the⃗
AC .⃗ BD=0⃗
AC= ( 4 , 2,−2 )⃗
BD= ( 2 ,−2 ,2 )
Hence⃗ AC .⃗ BD=0
This proves that the diagonals are
perpendicular
QUESTION 8

LINEAR ALGEBRA 18
The parallelogram above ABCD have the vectors ⃗
AB= ~a, ⃗ BC = ~
b ,⃗ AD= ~
b, ⃗ DC= ~a
´BD, and ´AC are the diagonals
´AC=⃗ AB+⃗ BC= ~a+ ~
b
´BD=−⃗ AB+,⃗ AD= ~
b−¿ ~a
´AC . ´BD= ( ~a+ ~
b ) . ( ~
b− ~a )
¿ ¿ ~
b∨¿2−|~a|2=0 ¿
This implies that ´AC is perpendicular ¿ ´BD , hence it is a rhombus