j.n. ridley engineering mathematics 2, guide to solving problems unofficial soln manual

MATH280 Algebra CHAPTER 3 (Linear Algebra) by Mo Val

1

Tutorial questions-‐Linear spaces and subspaces Page 78 Question 1 Required To Prove: The linear combination !!!!!

!!! is equivalent to the matrix product CX. Given: C is m x n matrix with columns !! ,!!,… ,!! that are n x 1 column vectors

i.e. ! = !! !! …!! ; !! =

!!!!!"!!"⋮

!!!

; !! =

!!"!!!!!"⋮

!!!

; !! =

!!!!!!!!!⋮

!!"

X is a n x 1 column vector with entries !! , !!,… , !!

I.e. ! =

!!!!!"⋮!!!

Step 1: !!!!!

!!! = !! !! + !! !! +⋯+ !! !!

Step 2: !" = !! !! …!!

!!!!!"⋮!!!

= !! !! + !!!! +⋯+ !! !!

But !! !! = !! !! since constant x vector is commutative

∴ !!!!

!

!!!

≡ !"


2

Question 2 To find if the vector set ! [with 2 complex entries (ℂ!)] whose entries satisfy a given equation is a complex linear space. [Note a vector set is a set of vectors satisfying the given equation. ! is a vector set and not a vector] Q 2 a) ! = !,! ! ℂ!: ! + !" = 0 Step 1: Need to verify ! is a complex linear space Step 2: By definition: ! is a linear space if every linear combination of elements of the set is again in the set. Are the elements of a vector set vectors?

!"# !!, !! !" !"#$%& !"!#!$%& !" !ℎ! !"#$%& !"# ! !! = !!,!! ; !! = !!,!!

!"#$": !! + !!! = 0 & !! + !!! = 0 Step 3: !"# !! !" ! !"#$%& !"#$%&'(%"& !" !!, !!

∴ !! = !!!! + !!!! !! = !! !!,!! + !! !!,!!

!! = (!!!! + !!!!,!!!! + !!!!) Step 4: Test whether !! is in the set, i.e. its entries !!,!! satisfy the equation ! + !" = 0 LHS: !! + !!! = (!!!! + !!!!)+ !(!!!! + !!,!!) = !!(!! + !!!)+ !!(!! + !!!)

= !!(0)+ !!(0) = 0 = !"# ∴ !ℎ! !"#$%& !"# ! !" ! !"#$%&' !"#$%& !"#$% !" the linear combination !! of two general vectors of the set is again in the set


3

Q 2 b) ! = !,! ! ℂ!: ! + !" = 1+ ! Step 1: Need to verify ! is a complex linear space Step 2: By definition: ! is a linear space if every linear combination of elements of the set is again in the set.

!"# !!, !! !" !"#$!" !"!#!$%& !" !ℎ! !"#$%& !"# ! !"#$": !! + !!! = 1+ ! & !! + !!! = 1+ !

Step 3: !"# !! !" ! !"#$%& !"#$%&'(%"& !" !!, !!

∴ !! = !!!! + !!!! !! = (!!!! + !!!!,!!!! + !!!!)

Step 4: Test whether !! is in the set, i.e. its entries !!,!! satisfy the equation ! + !" = 1+ ! LHS: !! + !!! = (!!!! + !!!!)+ !(!!!! + !!,!!) = !!(!! + !!!)+ !!(!! + !!!)

= !!(1+ !)+ !!(1+ !) = (!! + !!)(1+ !) ≠ 1+ !

∴ !ℎ! !"#$%& !"# ! !" !"# ! !"#$%&' !"#$%& !"#$%#! !" !ℎ! !"#$%& !"#$%&'(%"& !! is not again in the set ! Q 2 c) ! = !,! ! ℂ!: !! − !!! = 1+ ! Step 1: Need to verify ! is a complex linear space Step 2: By definition: ! is a linear space if every linear combination of elements of the set is again in the set.

!"# !!, !! !" !"#$%& !"!#!$%& !" !ℎ! !"#$%& !"# ! !"#$": !!! − !!!! = 1+ ! & !!! − !!!! = 1+ !

Step 3: !"# !! !" ! !"#$%& !"#$%&'(%"& !" !!, !!

∴ !! = !!!! + !!!! !! = (!!!! + !!!!,!!!! + !!!!)

Step 4: Test whether !! is in the set!, i.e. its entries !!,!! satisfy the equation !! − !!! = 1+ ! LHS:

!! + !!! = (!!!! + !!!!)! − ! !!!! + !!!! ! !! + !!! = !!!!!! + 2 !!!!!!!! + !!!!!! − ! !!!!!! + 2 !!!!!!!! + !!!!!!

!! + !!! = !!! !!! − !!!! + !!! !!! − !!!! + 2!!!! !!!! − !!!!! = !!! 1+ ! +!!! 1+ ! + 2!!!! !!!! − !!!!! = !!! + !!! 1+ ! + 2!!!! !!!! − !!!!! ≠1+ ! ∴ !ℎ! !"#$%& !"# ! !" !"# ! !"#$%&' !"#$%& !"#$% !" !ℎ! !"#$%& !"#$%&'(%"& !! is not again in the set !


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Q 2 d) ! = !,! ! ℂ!: !! − !!! = 0 Step 1: Need to verify ! is a complex linear space Step 2: By definition: ! is a linear space if every linear combination of elements of the set is again in the set.

!"# !!, !! !" !"#$%& !"!#!$%& !" !ℎ! !"#$%& !"# ! !"#$": !!! − !!!! = 0+ ! & !!! − !!!! = 0

Step 3: !"# !! !" ! !"#$%& !"#$%&'(%"& !" !!, !!

∴ !! = !!!! + !!!! !! = (!!!! + !!!!,!!!! + !!!!)

Step 4: Test whether !! is in the set !, i.e. its entries !!,!! satisfy the equation !! − !!! = 0 LHS:

!! + !!! = (!!!! + !!!!)! − ! !!!! + !!!! ! !! + !!! = !!!!!! + 2 !!!!!!!! + !!!!!! − ! !!!!!! + 2 !!!!!!!! + !!!!!!

!! + !!! = !!! !!! − !!!! + !!! !!! − !!!! + 2!!!! !!!! − !!!!! = !!! 0 +!!! 0 + 2!!!! !!!! − !!!!! = 2!!!! !!!! − !!!!! ≠ 0 ∴ !ℎ! !"#$%& !"# ! !" !"# ! !"#$%&' !"#$%& !"#$% !" !ℎ! !"#$%& !"#$%&'(%"& !! is not again in the set !


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Question 3 To solve the systems of equations AZ=B using Gauss-‐Jordan elimination Q 3 a)

1 2+ ! : 1+ !1− ! 3 : !

Step 1: make 1− ! = 0; ∴ !! − (1− !)!!

!!! = 3− 1− ! 2+ ! = 3− 2− ! + 1 = ! !!" = −! − 1− ! 1+ ! = ! − 2 = −2+ !

~ 1 2+ ! : 1+ !

0 ! : −2+ ! Step 2 make ! = 1; !!×−!

!!" = 0 !!! = 1

!!" = −! −2+ ! = 1+ !2

~ 1 2+ ! : 1+ !0 1 : 1+ !2

Step 3: make 2+ ! = 0; !! − 2+ ! !!

!!" = 1+ ! − 2+ ! 1+ !2 = 1+ ! − 5! = 1− !4

~ 1 0 : 1− !40 1 : 1+ !2

! = !!+!!

Let Z be a complex No. with 2 components (!,!) i.e. ! =

!!

! = 1− !4 ! = 1+ 2!

! =!! = 1− 4!

1+ 2! = !!


6

Q 3 b) 1 −1+ ! : 1+ !1− ! 2! : 2

Step 1: make 1− ! = 0; ∴ !! − (1− !)!!

!!! = 2! − 1− ! −1+ ! = 2! − 2! = 0 !!" = 2− 1− ! 1+ ! = 2− 2 = 0

~ 1 −1+ ! : 1+ !

0 0 : 0 Step 2: 1 equation 2 unknowns therefore 1 parameter.

!"# ! = ! 1! + −1+ ! ! = 1+ ! ! = 1+ ! − ! −1+ !

! = !!+!! Let Z be a complex No. with 2 components (!,!) i.e. ! =

!!

! = 1+ ! + ! 1− ! ! = 0+ !

! =!! =

1+ !0!!

+ !1− !1!!

Q 3 c) 1 ! 1− ! : 1+ !

1+ ! −1+ ! 2 : 2! Step 1: make 1+ ! = 0; ∴ !! − (1+ !) !!

!!! = −1+ ! − 1+ ! ! = −1+ ! − ! − 1 = 0 !!" = 2− 1+ ! 1− ! = 2− 2 = 0

!!" = 2! − 1+ ! 1+ ! = 2! − 2! = 2− 2!

∼ 1 ! 1− ! : 1+ !0 0 0 : 0

Step 2: 1 equations 3 unknowns therefore 2 parameters

!"# ! = ! !"# ! = !

1! + !" + 1− ! ! = 1+ ! ! = 1+ ! − !" − 1− ! !

! = !!+!!

Let Z be a complex No. with 3 components !,!, ! i.e. ! =!!!

! =!!!

=

1+ !00!!

+ !−!10

+ !−1+ !01

!!


7

Q 3 d) 1 2+ ! : 1− !

1− ! 3− ! : −2!! −1+ 2! : 1+ !

Step 1: make 1− ! = 0; ∴ !! − (1− !) !!

!!! = 3− ! − 1− ! 2+ ! = 3− ! − (3− !) = 0 !!" = −2! − 1− ! 1− ! = −2! − −2! = 0

Step 2: make ! = 0; ∴ !! − ! !!

!!" = −1+ 2! − ! 2+ ! = −1+ 2! − (−1+ 2!) = 0 !!! = 1+ ! − ! 1− ! = 1+ ! − 1+ ! = 0

~1 2+ ! : 1− !0 0 : 00 0 : 0

Step 3: 1 equation 2 unknowns therefore 1 parameter

!"# ! = ! 1! + 2+ ! ! = 1− ! ! = 1− ! − 2+ ! !

! = !!+!! Let Z be a complex No. with 2 components (!,!) i.e. ! =

!!

! = 1− ! − 2+ ! ! ! = 0+ !

! =!! =

1− !0!!

+ !— 2− !

1!!


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Question 4 Q 4 i) To find the inverse of the matrix T using Gauss-‐Jordan Elimination

Q 4 i a) ! = 1 1/3 1/2 1/2

Step 1: Augment ! !"#ℎ !

1 1/3 : 1 01/2 1/2 : 0 1

Step 2: Get the Augmented Matrix in the form !:!!! Step 1: make !

!= 0; ∴ !! − (

!!) !!

!!! =12−

12

13 =

12−

16 =

26

!!" = 0−12 1 = −

12 = −

12

!!" = 1−12 0 = 1− 0 = 1

~ 1 1/3 : 1 00 1/3 : −1/2 1

Step 2: make !!= 1; ∴ !!× 3

!!" = 0 !!! = 1

!!" = −32

!!" = 3

~ 1 1/3 : 1 00 1 : −3/2 3

Step 1: make !

!= 0; ∴ !! − (

!!) !!

!!! = 1

!!" = 1−13 −

32 = 1− −

12 =

32

!!" = 0−13 3 = 0− 1 = −1

~ 1 0 : 3/2 −10 1 : −3/2 3

Therefore !!! = 3/2 −1−3/2 3


9

Q 4 i b) ! = 1 1+ ! 1− 2! 3


1 1+ ! : 1 01− 2! 3 : 0 1

Step 2: Get the Augmented Matrix in the form !:!!! Step 1: make 1− 2! = 0; ∴ !! − (1− 2!) !!

!!! = 3− 1− 2! 1+ ! = 3− (3− !) = ! !!" = 0− 1− 2! 1 = 0− 1− 2! = −1+ 2!

!!" = 1− 1− 2! 0 = 1

~ 1 1+ ! : 1 00 ! : −1+ 2! 1

Step 2: make ! = 1; ∴ !!× −!

!!" = 0 !!! = 1

!!" = −! −1+ 2! = 2+ ! !!" = −!

~ 1 1+ ! : 1 0

0 1 : −2+ ! −! Step 1: make 1+ ! = 0; ∴ !! − (1+ !) !!

!!! = 1 !!" = 1− 1+ ! −2+ ! = 1− −3− ! = 4+ ! !!" = 0− 1+ ! −! = 0− 1− ! = −1+ !

~ 1 0 : 4+ ! −1+ !

0 1 : −2+ ! −! Therefore !!! = 4+ ! −1+ !

−2+ ! −!


10

Q 4 i c) ! =1 2 32 3 43 4 6


1 2 3 : 1 0 02 3 4 : 0 1 03 4 6 : 0 0 1

Step 2: Get the Augmented Matrix in the form !:!!! Step 1: make 2 = 0; ∴ !! − (2) !!

!!! = 3− 2 2 = 3− 4 = −1 !!" = 4− 2 3 = 4− 6 = −2

!!" = 0− 2 1 = −2 !!" = 1− 2 0 = 1 !!" = 0− 2 0 = 0

Step 2: make 3 = 0; ∴ !! − (3) !!

!!" = 4− 3 2 = 4− 6 = −2 !!! = 6− 3 3 = 6− 9 = −3

!!" = 0− 3 1 = −3 !!" = 0− 3 0 = 0 !!" = 1− 3 0 = 1

~1 2 3 : 1 0 00 −1 −2 : −2 1 00 −2 −3 : −3 0 1

Step 3: make −1 = 1; ∴ !!× −1

!!" = 0 !!! = 1

!!" = −1 −2 = 2 !!" = 2 !!" = −1 !!" = 0

~1 2 3 : 1 0 00 1 2 : 2 −1 00 −2 −3 : −3 0 1

Step 4: make 2 = 0; ∴ !! − (2) !!

!!! = 1 !!" = 3− 2 2 = −1 !!" = 1− 2 2 = −3 !!" = 0− 2 −1 = 2 !!" = 0− 2 0 = 0


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Step 5: make −2 = 0; ∴ !! + 2 !!

!!! = −3+ 2 2 = 1 !!" = −3+ 2 2 = 1 !!" = 0+ 2 −1 = −2 !!" = 1+ 2 0 = 1

~1 0 −1 : −3 2 00 1 2 : 2 −1 00 0 1 : 1 −2 1

Step 6: make 2 = 0; ∴ !! − (2) !!

!!" = 2− 2 −1 = 0 !!" = −1− 2 −2 = 3 !!" = 0− 2 1 = −2

Step 7: make −1 = 0; ∴ !!+ !!

!!" = −3+ 1 = −2 !!" = 2− 2 = 0 !!" = 0+ 1 = 1

~1 0 0 : −2 0 10 1 0 : 0 3 −20 0 1 : 1 −2 1

Therefore !!! =−2 0 10 3 −2−1 −2 1


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Q 4 i d) ! =1 ! 1+ !! 0 11 −1 −1+ !


1 ! 1+ ! : 1 0 0! 0 1 : 0 1 01 −1+ ! −1+ ! : 0 0 1

Step 2: Get the Augmented Matrix in the form !:!!! Step 3: make ! = 0; ∴ !! − (!) !!

!!! = 0− ! ! = 1 !!" = 1− ! 1+ ! = 1− −1+ ! = 2− !

!!" = 0− ! 1 = −! !!" = 1− ! 0 = 1 !!" = 0− ! 0 = 0

Step 2: make 1 = 0; ∴ !! − !!

!!" = −1+ ! − ! = −1 !!! = −1+ ! − 1+ ! = −2

!!" = 0− 1 = −1 !!" = 0− 0 = 0

!!" = 1− 3 0 = 1

~1 ! 1+ ! : 1 0 00 1 2− ! : −! 1 00 −1 −2 : −1 0 1

Step 3: make ! = 0; ∴ !! − (!) !!

!!" = 1+ ! − ! 2− ! = 1+ ! − −1+ 2! = 2− ! !!" = 1− ! −! = 1− 1 = 0

!!" = 0− ! 1 = −! !!" = 0− ! 0 = 0

Step 4: make −1 = 0; ∴ !! + !!

!!! = −2+ 2− ! = −! !!" = −1+ ! !!" = 0+ 1 = 1

!!" = 1

~1 0 2− ! : 0 −! 00 1 2− ! : −! 1 00 0 −! : −1+ ! 1 1

Step 4: make −! = 1; ∴ !!×!


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~1 0 2− ! : 0 −! 00 1 2− ! : −! 1 00 0 1 : −1− ! ! !

Step 5: make 2− ! = 0; ∴ !! − 2− ! !!

!!" = 0− 2− ! 1− ! = − 1− 3! = −1+ 3! !!" = −! − 2− ! ! = −! − 1+ 2! = −1+ ! !!" = 0− 2− ! ! = − 1+ 2! = −1− 2!

Step 6: make2− ! = 0; ∴ !! − 2− ! !!

!!" = −− ! − 2− ! 1− ! = −! − 1− 3! = 2! − 1 !!" = 1− 2− ! ! = 1− 1+ 2! = −2! !!" = 0− 2− ! ! = − 1+ 2! = −1− 2!

~1 0 0 : −1+ 3! −1+ ! −1− 2!0 1 0 : −1+ 2! −2! −1− 2!0 0 1 : −1− ! ! !

Therefore !!! =−1+ 3! −1+ ! −1− 2!−1+ 2! −2! −1− 2!−1− ! ! !

Tutorial questions-‐Bases and dimension Page 80 Question 5 To find a basis of the null space of the coefficient matrix A and state its dimension. [Note: The general solution of !" = ! is of the form ! = !! + !! where !! is a particular solution and !! is a general element of the null space. The number of arbitrary constants or parameters in !! is equal to the number of basis elements.] [Note: The dimension of the null space of a linear operator is equal to the number of arbitrary constants in the general solution] Q 5 a) ! = 1 2+ !

1− ! 3

! =!! = 1− 4!

1+ 2! = !! Step 1:

!! = 0, ℎ!"#! !" !"#$%"!"& !"#$%&#%$ !" !ℎ! !"## !"#$%. ∴ !" !"#$# !"!#!$%&

Step 2: !"#$%&"'% = 0


14

Q 5 b) ! = 1 −1+ !1− ! 2!

! =!! =

1+ !0!!

+ !1− !1!!

Basis: 1− !1 ; !"#$%&"'% 1 Q 5 c) ! = 1 ! 1− !

1+ ! −1+ ! 2

! =!!!

=

1+ !00!!

+ !−!10

+ !−1+ !01

!!

Basis: −!10

!"# −1+ !01

; !"#$%&"'% 2

Q 5 d) ! =1 2+ !

1− ! 3− !! −1+ 2!

! =!! =

1− !0!!

+ !— 2− !

1!!

Basis: — 2− !1

; !"#$%&"'% 1

Question 6 To find a basis for the null space of the differential operator noting that dimension of the null space should be equal to the degree of the operator Q 6 a) !! − 4! + 3 Step 1: To find the null space i.e. when !! − 4! + 3 !! = 0 Step 2: Make the equation linear in terms of D i.e. !!

! − 3 ! − 1 !! = 0 ! = 3 !" ! = 1

Step 3: Write the equation in the form of !! = !!∝! ! = ! + !";!,! !"# !"#$%"!"& !"#$%&#%$

!! = !!!! + !!!! Basis: !!! , !!! ∴ !"#$%&"'% !" 2 Q 6 b) !! + 2! + 2 Step 1: To find the null space i.e. when !! + 2! + 2 !! = 0 Step 2: Make the equation linear in terms of D i.e. !!


15

Using the quadratic equation

! =−2± 4!!

2(1)

! = −1± ! ! = −1± !

Step 3: Write the equation in the form of !! = !!∝!

! = ! + !";!,! !"# !!"#$!%!& !"#$%&#%$ !! = !!(!!!!)! + !!(!!!!)!

Basis: ! !!!! ! , ! !!!! ! ∴ !"#$%&"'% !" 2 Q 6 c) !! Step 1: To find the null space i.e. when (!!)!! = 0 Step 2: Make the equation linear in terms of D i.e. !!

! + 0 !!! = 0

! = 0 Step 3: Write the equation in the form of !! = !!∝!

! = ! + !";!,! !"# !"#$%"!"& !"#$%&#%$ [Note: For repeated factors we add similar expressions that are each multiplied by a power of t]

!! = ! + !" + !!! !∝! !! = !!(!)! + !"!(!)! + !!!! ! !

Basis: ! ! ! = 1 !" 0 ! = ! !!! ! ! = !!

∴ !"#$%&"'% !" 3

Q 6 d)!! Step 1: To find the null space i.e. when (!!)!! = 0 Step 2: Make the equation linear in terms of D i.e. !!

! + 0 !!! = 0 ! = 0


! = ! + !";!,! !"# !"#$%"!"& !"#$%&#%$


16

[Note: For repeated factors we add similar expressions that are each multiplied by a power of t]

!! = ! + !" + !!!+. . .+!!!!! !∝! !! = ! + !" + !!!+. . .+!!!!! !!!

Basis: 1, !, !!,… , !!!!

∴ !"#$%&"'% !" ! Q 6 e) !! + 4 Step 1: To find the null space i.e. when (!! + 4)!! = 0 Step 2: Make the equation linear in terms of D i.e. !!

!! − 4!! !! = 0 !! + 2! !! − 2! !! = 0 !! + 2! !! − 2! !! = 0

STOP Start over Step 3: In order to reduce the equation we need to complete the square Add and subtract 4!!

[!! + 4+ 4!! − 4!!]!! = 0

!! + 2 ! − 4!! !! = 0 !! + 2 − 2! !! + 2 + 2! !! = 0 !! − 2! + 2 !! + 2! + 2 !! = 0

Using the quadratic equation

! =2± 4!!

2(1)

! = 1± ! ! = 1± !

Or

! =−2± 4!!

2(1)

! = −1± ! ! = −1± !


! = ! + !";!,! !"# !"#$%"!"& !"#$%&#%$ !! = !!(!!!)! + !!(!!!)! + !!(!!!!)! + !!(!!!!)!

Basis: !±!±! ∴ !"#$%&"'% !" 4


17

Question 7 Tutorial questions-‐ Independence and rank Page 82 Question 8 To determine if a matrix is independent. Using coefficient matrices from Q 3. [Note: !" = !, !ℎ! !"#$%& !"#$!"# !" ! !"# !"#$%$"#$"& !" !" =0; !. !.!ℎ! !"## !"#$% !"#$%$&$ !" !"#$ ! !"#$ !"#$%& !"# ℎ!"#! !ℎ! !"#$%&"' ! !" !"#$!%. If A is a square matrix the columns are independent if and only if det A≠ 0 i.e !!!!"#$%$.]

!! = 0 !"# !3 !) Question 9 To test the sets of functions for independence using Wronskians. [Note: Functions are independent if their Wronskian≠ 0. The Wronskian is the determinant of a (square) matrix with entries of functions and their respective derivatives of successive orders. Where the No. of Derivatives !"# !"#$%& = 1−!". !" !"#$%&'#(. ] Q 9 a) !! , !!!

! !! , !!! = !! !!!!! −!!!

! = !! −!!! − !!! !! = −!! − !! = −2 ≠ 0 ∴ !"#$%&'#( !! , !!! !"# !"#$%$"#&"'

Q 9 b) cos !, sin !, 1

!(cos !, sin !, 1) =cos ! sin ! 1− sin ! cos ! 0− cos ! − sin ! 0

!"#$%& !"#$% !"#$%& 3

! = −1 !!!!!" !!"

! = −1 !!! 1 − sin ! cos !−cos ! −sin ! + −1 !!! 0 cos ! sin !

− cos ! − sin !+ −1 !!! 0 cos ! sin !

− sin ! − cos ! ! = (− sin !)(− sin !)− (cos !)(−cos !) = sin! ! + cos! ! = 1 ≠ 0

∴ !"#$%&'#( cos !, sin !, 1 !"# !"#$%$"#&"' Q 9 c) cos! !, sin! !, 1

! cos! !, sin! !, 1


18

=cos! ! sin! ! 1

2!"#$ – sin ! 2 sin ! cos ! 0−2 sin ! − sin ! + 2 cos ! – cos ! 2 cos ! cos ! + 2 sin ! −!"#$ 0

!"#$%& !3

! = 1[ 1 −2 sin ! cos ! 2 sin ! cos !

2 sin! ! − 2 cos! ! 2 cos! ! − 2 sin! ! ] + 0+ 0 ! = (− sin 2!)(2)(cos! ! − sin! !)− (sin 2!)(2)(sin! ! − cos! !) = 0

∴ !"#$%&'#( cos! !, sin! !, 1 !"# !"# !"#$%$"#$"&

Q 9 d) 1, sec !, tan ! ! 1, sec !, tan !

=1 sec ! tan !0 sec ! tan ! sec! !0 tan !(sec ! tan !)+ sec !(sec! !) 2 sec !(sec ! tan !)

!"#$%& !"#$% !1

! = 1[ 1 sec ! tan ! sec! !

!"# ! tan! ! + sec! ! 2 !"!! ! !"#$]+ 0+ 0

! = sec ! tan ! 2 !"!! ! !"#$ − sec! ! !"# ! tan! ! + sec! !

= 2 sec! ! !"!!! − sec! ! − sec! ! tan! ! = sec! ! tan! ! − sec! != sec! !(tan! ! − sec! !) = sec! ! 1 ≠ 0

∴ !"#$%&'#( 1, sec !, tan ! !"# !"#$%$"#&"' Question 10) Show that the basis functions i.e. functions in the null space are independent using Wronskians. Q 10 a) Basis: !!! , !!!

! !!! , !!! = !!! !!!3!!! !!

! = (!!!)(!!)− !! 3!!! = !!! − 3!!! = −2!!! ≠ 0

∴ !"#$%&'#( !!! , !!! !"# !"#$%$"#&"' Q 10 b) Basis: ! !!!! ! , ! !!!! !

! ! !!!! ! , ! !!!! ! = ! !!!! ! ! !!!! !

−1+ ! ! !!!! ! −1− ! ! !!!! !

! = ! !!!! !) −1− ! ! !!!! ! − ! !!!! ! −1+ ! ! !!!! !

= −1− ! ! !! ! − −1+ ! ! !! ! = !!!!(−2!) ≠ 0 ∴ !"#!"#$%& ! !!!! ! , ! !!!! ! !"# !"#$%$"#&"'


19

Q 10 c) Basis: 1, !, !!

! 1, !, !! = 1 ! !!0 1 2!0 0 2

!"#$%& !"#$% !1

! = 1 1 1 2!0 2 + 0+ 0 = 2− 2! 0 = 2 ≠ 0

∴ !"#$%&'#( 1, !, !! !"# !"#$%$"#&"' Question 12 To find the rank of the coefficient matrix and note that the dimension of the null space is equal to !". !" !"#$%&' !" ! − !"#$ Q 12 a) ! = 1 2+ !

1− ! 3

! =!! = 1− 4!

1+ 2! = !! Step 1:

!! ℎ!" 2 !"! − !"#$ !"#$ ∴ !"#$ !" 2

!"#$%&"'% = ! − ! Step 2: Check !"#$%&"'% = 2− 2 = 0 Note: ! = !: !ℎ! !"#$%&"' !" !"# !"#$%&'( !" = ! !"## !"#!$% !"#$%

! = !:!ℎ! !"#$!"#$ !" !"# !"#$%&'( !" = ! !" !"#$!% ! = !"#$%& !" !"#"$%"& !. !. !,!

Q 12 b) ! = 1 −1+ !1− ! 2!

! =!! =

1+ !0!!

+ !1− !1!!

Step 1: !! ℎ!" 1− !"! − !"#$ !"#$ ∴ !"!" !" 1

Step 2: Check !"#$%&"'% = 2− 1 = 1 Q 5 c) ! = 1 ! 1− !

1+ ! −1+ ! 2

! =!!!

=

1+ !00!!

+ !−!10

+ !−1+ !01

!!

Step 1: !! ℎ!" 1− !"! − !"#$ !"#$ ∴ !"#$ !" 1

Step 2: Check !"#$!"#$! = 3− 1 = 2


20

Q 5 d) ! =1 2+ !

1− ! 3− !! −1+ 2!

! =!! =

1− !0!!

+ !— 2− !

1!!

Step 1: !! ℎ!" 1− !"! − !"#$ !"#$ ∴ !"#$ !" 1

Step 2: Check !"#$%&"'% = 2− 1 = 1 Q 12 b) In a square matrix ! = !. If ! = ! ! !"#$%&"' !"#$%$ !ℎ!" ! =! !ℎ! !"#$%&"' !"## !" !"#$!% !" !ℎ! !"#$%&"'% !" !"#$


21

Tutorial questions –Eigenvalues and eigenvectors Page 83 Question 13 To show the matrix product AX is a scalar multiple of X Q 13 a i) ! = 0 1

1 0 ; ! = 11

!" = 0 1 + 1 11 1 + 0 1 = 1

1

!" = 1! Q 13 a ii) ! = 0 1

1 0 ; ! = 1−1

!" = 0 1 + 1 −11 1 + 0 −1 = −1

−1 = − 11

!" = −! Eigenvalues of A are 1& -‐1 Eigenvectors are X Q 13 b i) ! = 1 2

3 2 ; ! = 23

!" = 1 2 + 2 33 2 + 2 3 = 8

12 = 4 23

!" = 4! Q 13 b ii) ! = 1 2

3 2 ; ! = 1−1

!" = 1 1 + 2 −13 1 + 2 −1 = −1

1 = − 1−1

!! = −! Eigenvalues of A are 4& -‐1 Eigenvectors are X Q 13 c i) ! = 1 −1

−2 2 ; ! = 11

!" = 1 1 + −1 1−2 1 + 2 1 = 0

0 = 11

!" = 0! Q 13 c ii) ! = 1 −1

−2 2 ; ! = −12

!" = 1 −1 + −1 2−2 −1 + 2 2 = −3

6 = 3 −12

!" = 3! Eigenvalues of A are 0& 3 Eigenvectors are X Q 13 d i) ! = 1 −2

1 3 ; ! = 1+ !−1


22

!" = 1 1+ ! + −2 −11 1+ ! + 3 −1 = 3+ !

−2+ !

Make −2+ ! = −1; !"#"!$ !" − −2+ ! = 2− !

3+ !2− !×

2+ !2+ ! =

5+ 5! 5 = 1+ 1!

= 2− ! 1+ !−1

!" = (2− !)! Q 13 d ii) ! = 1 −2

1 3 ; ! = 1− !−1

!" = 1 1− ! + −2 −11 1− ! + 3 −1 = 3− !

−2− !

Make −2− ! = −1; !"#"!$ !" − −2− ! = 2+ !

3− !2+ !×

2− !2− ! =

5− 5! 5 = 1− 1!

= 2+ ! 1− !−1

!" = 2+ !"

Eigenvalues of A are 2− ! & 2+ ! Eigenvectors are X Q13 e) ! = 0 −1

1 2 ; ! = 1−1

!" = 0 1 + −1 −11 1 + 2 −1 = 1

−1

!" = !

Eigenvalue of A is 1 Eigenvector is X


23

Question 14 To find det ! − !!! [Note: ! is an eigenvalue of A if and only if det ! − !!! = 0] Q 14 a) ! = 0 1

1 0 Step 1: Calculate !!! =

! 00 !

Step 2: Calculate det ! − !!! : det

0− ! 11 0− ! = (0− !)! − 1 = !! − 1

Q 14 b) ! = 1 2


! 00 !


1− ! 23 2− ! = 1− ! 2− ! − 2 3 =

!! − 3! − 4 Q 14 c) ! = 1 −1

−2 2 Step 1: Calculate !!! =

! 00 !


1− ! −1−2 2− ! = 1− ! 2− ! — 1 −2 =

!! − 3! Q 14 d) ! = 1 −2


! 00 !

Step 2: Calculate det ! − !!! : det1− ! −21 3− ! = 1− ! 3− ! — 2(1) =

!! − 4! + 5 Q 14 e) ! = 0 −1


! 00 !

Step 2: Calculate det ! − !!! : det−! −11 2− ! = −! 2− ! − (−1)(1) =

!! − 2! + 1 = (! − 1)!


24

Page 84 Question 15 To solve the characteristic equation of a matrix and hence find its eigenvalues and corresponding eigenvectors. [Note the equation det ! − !!! = 0 !" !"##$% !ℎ! !ℎ!"!#$%"&'$&# !"#$%&'( !" the matrix A. The characteristic polynomial det ! − !!! is found from subtracting ! !"#$ !ℎ! !"#$%&#' !"#$%!& !" !.! !"#"!$ !" !ℎ! !"#!$%&'(!) which are determined from the characteristic equation. By augmenting

! − !!! !"#ℎ !"#$% we can determine the variables of Z (AZ=0) ! =!!!

The set of solutions to ! = !" − ! ! is called the Eigenspace of A corresponding to ! . There is one eigenspace for every distinct eigenvalue. By solving this system using the calculated eigenvalues we can determine the general eigenvectors of the matrix and also determine the basis elements

!"#$!"# !"#$%& !!! = !

Q 15 a) 0 0 41 0 40 1 −1

Step 1: Calculate det ! − !!! by subtracting ! from the diagonal entries of A

!"#0− ! 0 41 0− ! 40 1 −1− !

!"#$%& !"#$% !1

= 1 −! −! −1− ! − 4 1 + −1 !!! 0 + 1 4 [ 1 1 — ! 0 ]

−! !! + ! − 4 + 4 = −!! − !! + 4! + 4 = −!! ! + 1 + 4 ! + 1= 4− !! ! + 1

Step 2: Solve for ! in det ! − !!! = 0

4− !! ! + 1 = 0 2− ! 2+ ! ! + 1 = 0

Eigenvalues: ! = ±2 or ! = −1 Step 3: Solve for the unknown variables of the eigenvector Z, AZ=0

! = 2

0− 2 0 4 : 01 0− 2 4 : 00 1 −1− 2 : 0

Step 4: Make −2 = 1;∴ − !!!!

Step 5: Make 1 = 0; ∴ !! − !!


25

~1 0 −2 : 00 −2 6 : 00 1 −3 : 0

Step 6: Make −2 = 1;∴ − !

!!!

Step 7: Make 1 = 0; ∴ !! − !!

~1 0 −2 : 00 1 −3 : 00 0 0 : 0

!"# ! = ! ! − 3! = 0;! = 3! 1! − 2! = 0; ! = 2!

Eigenvector 1= !231

! = −2

Q 15 b) −1 −1 11 0 −20 1 2

Step 1: Find the eigenvalues by solving det ! − !" = 0

det ! − !" = det−1− ! −1 11 −! −20 1 2− !

!"#$%& !"#$% !!

= 1 −1− ! −! 2− ! + 2 + −1 1 −1 2− ! − 1 + 0

= −1− ! −2! + !! + 2 − 1 −2+ ! − 1 = 2! − !! − 2+ 2!! − !! − 2! + 2− ! + 1

= −! + !! + 1− !! = −(!! − !! + ! − 1)

Solve det !" − ! = 0

!! − !! + ! − 1 = 0 ! − 1 !! + 1 = 0 ! = 1, ! = !, ! = −!

Step 2: Substitute the eigenvalues into the eigenspace ! − !" ! = ! and solve for Z the eigenvectors

−1− ! −1 1 : 01 −! −2 : 00 1 2− ! : 0


26

! = 1 −1− 1 −1 1 : 01 −1 −2 : 00 1 2− 1 : 0

~−2 −1 1 : 01 −1 −2 : 00 1 1 : 0

!!×−12

112

−12

: 01 −1 −2 : 00 1 1 : 0

!! − !!

112

−12

: 0

0 −32

−32

: 00 1 1 : 0

!!×−23

112

−12

: 00 1 1 : 00 1 1 : 0

!! −12!!

!! − !! 1 0 −1 : 00 1 1 : 00 0 0 : 0

!"# ! = ! ! = ! ! = −!

!! = !1−11

Eigenvector for ! = !

−1− ! −1 1 : 01 −! −2 : 00 1 2− ! : 0

!"#$ !!!"# !!

~1 −! −2 : 0

−1− ! −1 1 : 00 1 2− ! : 0

!! − −1− ! !! = !! + (1+ !)!!


27

!!! = −1+ 1+ ! −! = −1− ! + 1 = −! !!" = 1− 2− 2! = −1− 2! 1 −! −2 : 00 −! −1− 2! : 00 1 2− ! : 0

!"#$ !!!"# !!

1 −! −2 : 00 1 2− ! : 00 −! −1− 2! : 0

!! + !!! !! + !!!

!!" = −2+ 2! + 1 = −1+ 2! !!! = −1− 2! + 2! + 1 = 0

1 0 −1+ 2! : 00 1 2− ! : 00 0 0 : 0

2 !"#$%&'() !ℎ!"" !"#"$%"& !"# ! = !

! = 1− 2! ! ! = ! − 2 !

!! =!!!

= !1− 2!−2+ !1

Since ! is a real matrix !! = !; !! = −! !! = !!

Therefore !! = !! = !1+ 2!−2− !1


28

Tutorial questions-‐Diagonalization Page 85 [Note: A diagonal matrix is a square matrix where all entries not on the diagonal are zero. A matrix is non-‐singular if its invertible i.e. !"# ! ≠ 0. An n x n matrix A can be diagonalized if and only if it has ! independent eigenvectors]

Guide to finding the !!! Q 15 a)

! =0 0 41 0 40 1 −1

! =2 −2 −43 −1 01 1 1

!!! =1

det! !"# ! det! = + −4 3+ 1 + − 0 + + 1 −2+ 6 = −16+ 4 = −12

!!" =−1 3 4−1 6 4−4 12 4

!!" =−1 −3 4+1 6 −4−4 −12 4

!!"! =

−1 +1 −4−3 6 −124 −4 4

!!! =1−12

−1 +1 −4−3 6 −124 −4 4

Question 16 Find !!! and verify that !!!!" is a diagonal matrix with eigenvalues on the diagonal Q 16 a) ! = 0 1

1 0 ;! = 11 ;! = 1

−1

! = !"#!! 00 !"#!!

= 1 00 −1

! = !!, !! = 1 1

1 −1


29

!!! =1

det! !"# !

det! = −1− 1 = −2 !"# ! = !!"

! =

!!" =+1 −1−1 −1

!"# ! = −1 −1

−1 +1

!!! =1−2

−1 −1−1 +1

!!!!" =1−2

−1 −1−1 +1

0 11 0

1 11 −1

=1−2

−1 −1−1 +1

1 −11 1 = −1− 1 1− 1

−1+ 1 1+ 1 =1−2

−2 00 2 = 1 0

0 −1 = ! Question 18 Diagonalize ! = !

!1 43 2 !"# !"#$ ! lim!→! !!

Goal: To find !! = !!!!!!. We need to find ! and D Step 1: Take the fraction inside the matrix

15

45

35

25

Step 2: Find the eigenvalues by finding the characteristic polynomial and then solving it

!! ! = det ! − !" =

15− !

45

35

25− !

!! ! =15− !

25− ! −

45

35 =

225−

35 ! + !

! −1225 = −

1025−

35 ! + !

!

= !! −25 ! −

35

!! ! = 0 !ℎ!" !! −25 ! −

25 = 0

!! −35 ! −

25 = 0


30

! − 1 ! +25 = 0

!! = 1; !! = −25

Step 3: Find the eigenvectors by substituting the eigenvalues into the eigenspace ! − !" ! = !

!! = 1

! − 1! =

15− 1

45

35

25− 1

=−45

45

35

−35

!"#$% ! − !" ! = !

−45

45 : 0

35

−35

: 0

!!×−54

1 −1 : 035 −

35 : 0

!! −35!!

1 −1 : 00 0 : 0

!"# !! = ! !! = !

!! = ! 11

!! = −25

! +25 ! =

15+

25

45

35

25+25

=

35

45

35

45


31

!"#$% ! − !" ! = !

35

45 : 0

35

45

: 0

!!×53

143 : 0

35

45

: 0

!! −35!!

143

: 00 0 : 0

!"# !! = !

!! = −43

!! = ! −431

Step 3: Find ! !"! !

! =1 0

0 −25

;! = 1 −43

1 1

Step 4: Calculate !!!

!!! =1

det! !"# !

det! = 1+43 =

73

!"# ! = 143

−1 1

!!! =37

143

−1 1

Step 4: Determine !!

! = !"!!!


32

!! = !!!!!!

!! = 1 −43

1 1

1 0

0 −25

!37

143

−1 1

!! = 1 −43

1 1

1! 0

0 −25

! 37

143

−1 1

As ! → ∞

!! → 1 −43

1 1

1 00 0

37

143

−1 1=37

1 −43

1 11

43

0 0=37

143

143

=173 43 4

Question 20 Given the difference equation !!!! −

!!!! + !!!! = 0, where ! represents the

voltage. Find a constant matrix A such that

!!!!!! = !

!!!!!!

[Note: difference equations are of the form !!!! = !!! + !!!!!] Step 1: Express the difference equation in terms of !!!!

!!!! =52 !! − !!!!

Step 2:

!!!!!! =

52!! − !!!!!!

Step 3: Separate the matrix product into two factors

!!!!!! =

52

−11 0

!!!!!!

! =52

−11 0

Q 20 ii) Find an expression for !! !" !"#$% !" !! !"# !!, by diagonalizing A. To diagonalize A we need to find D which represent the eigenvalues of A


33

Step 1: Determine the eigenvalues of a by solving the characteristic polynomial !! ! = det(! − !")

!! ! =52− ! −11 −!

=52− ! −! + 1 = !! −

5!2 + 1

!! −52 ! + 1 = 0

2!! − 5! + 2 = 0 2! − 1 ! − 2 = 0

!! =12 ; !! = 2

! =12 00 2

The solution of the equation depends on getting an expression for !!

!!!!!! = !!

!!!!!!

!! = !!!!!!

Step 2: Determine ! whose entries are the eigenvectors To find the eigenvectors

!"#$% ! − !" ! = ! For !! =

!!

52−

12 −1 : 0

1 −12 : 0

~2 −1 : 0

1 −12 : 0

Swap !!!"# !!

1 −12 : 0

2 −1 : 0

!! − 2!!

1 −12 : 0

0 0 : 0


34

!"# !! = !

!! =12!

!! = !121

For !! = 2

52− 2 −1 : 01 −2 : 0

~12 −1 : 01 −2 : 0

Swap !!!"# !!

1 −2 : 012 −1 : 0

!! −12!!

1 −2 : 00 0 : 0 !"# !! = ! !! = 2!

!! = !121

+ ! 21

! =12 21 1

To get the Ridley solution swap !!!"# !! and multiply the new !! by 2

! = 2 11 2

! =2 0

012

!!! =13

2 −1−1 2

!!!!!! = !!!!!!

!!!!

!!!!!! = 2 1

1 2

2 0

012

!13

2 −1−1 2

!!!!


35

2 11 2

2! 0

012

! 13

2 −1−1 2

!!!!

=13

2 11 2

2! 00 2!!

2 −1−1 2

!!!!

=13 2 1

1 22! 00 2!!

2!! − !!−!! + 2!!

=13 2 1

1 22! 2!! − !!2!! −!! + 2!!

!!!!!! =

13 2 2! 2!! − !! + 2!! −!! + 2!!

2! 2!! − !! + 2(2!! −!! + 2!! )

!! =13 2! 2!! − !! + 2 2!! −!! + 2!!

!! =13 !!2!!! − !!2! − !!2!!! + !!2!!!!!

!! =13 2!!! − 2!!! !! − 2! − 2!!! !!

Q20 iii) Given: !! = 10, !!! = 0 find !!"

0 =13 2!!!! − 2!!!! !! − 2!! − 2!!!! (10)

10 2!! − 2!!!! = 2!!!! − 2!!!! !!

!! =10 2!! − 2!!!!

2!" − 2!!"= 10

2!! − 2!!

2!" − 2!!" = 102!! − 2!!

2 2!! − 2!!! ≈ 20

!!" =13 2!! − 2!! 20 − 2!" − 2!! (10)}

!!" =!ℎ!"#$#% I got far enough Question 21 Find an expression for !! in terms of !! and !! Given !!!! − ! + 1 !! + !!!!! = 0 Step 1: !!!! = ! + 1 !! − !!!!!

Step 2: !!!!!! = ! + 1 !! − !!!!!

!!= ! + 1 −!

1 0!!!!!!


36

Step 3: Diagonalize ! + 1 −!1 0 = !

Find the eigenvalues by finding the characteristic polynomial and then solving it

!! ! = det ! − !" = ! + 1− ! −!1 −! = !! − ! − !" + !

!! − ! − !" + ! = 0

! ! − 1 + ! 1− ! = 0 ! ! − 1 − ! ! − 1 = 0

! − ! ! − 1 = 0 !! = !; !! = 1

Find the corresponding eigenvectors by solving the eigenspace ! − !" ! = !

!"# !! = !

! + 1− ! −! : 01 −! : 0

~ 1 −! : 0

1 −! : 0

!! − !!

1 −! : 00 0 : 0

!"# !! = ! !! = !

!! = ! !

1 !"# !! = 1

! + 1− 1 −! : 0

1 −1 : 0

~ ! −! : 01 −1 : 0

Swap !!!"# !! 1 −1 : 0! −! : 0

!! − !"!

1 −1 : 00 0 : 0

!"# !! = ! !! = !


37

!! = ! 1

1 Step 4: ! = !"!!!

! = ! 11 1

!!! =1

! − 11 −1−1 !

! = ! 0

0 1 Step 6:!! = !!!!!!

!!!!!! = !!!!!!

!!!!!!

!!!!!! = !!!!!!

!!!!!!

!!!!!! = ! 1

1 1! 00 1

! 1! − 1

1 −1−1 !

!!!!!!

!!!!!! =

1! − 1

! 11 1

! 00 1

! !! − !!!!−!! + !!!!!

!!!!!! =

1! − 1

! 11 1

!! !! − !!!!−!! + !!!!!

!!!!!! =

1! − 1

!!!! !! − !!!! − !! + !!!!!!! !! − !!!! − !! + !!!!!

Substitute !! !"# !!

!! =1

! − 1 !! !! − !! − !! + !!!

!! =1

! − 1 !!!! − !!!! − !! + !!!

!! =1

! − 1 !! − 1 !! − !! − 10 !!

Kapow!!!


38

Question 22 Find the trajectories in the vector fields using Diagonalization Q 22 a) ! + !, 4! + ! [Note: Let ! be the points in a region in space. We have a vector ! = ! (!). ! forms a vector field in the region. The vector field attaches an arrow to each point in the region. If we join the arrows with curves v will always be tangent to the curve at the point r. These curves are called trajectories of the field.] The principle !" = !" Where P is a linear transformation Every number ! is an eigenvalue of the differentiation operator !

!" with

corresponding eigenfunctions/eigenvectors ! = !!!" The Aim of this is thus to find an equation for !,!, !. These equations are eigenfunctions which are in the null space of D= !

!"= !

To find the trajectory at a point

!!!" = ! ! !!!" = !"

! ! = !"

! ! = ! + !, 4! + !

! ! ! =

! + !4! + ! = 1 1

4 1!!

! =

!! ;! = 1 1

4 1 Now ! is a function of t ! = !(!) is the general trajectory For Matrices, we call !, ! In calculus this problem is solved by !"

!", !"!"

= ! + !, 4! + !

In algebra !!"! = ! + !, 4! + ! ! = !!

! = !!


39

!! = 1 1

4 1!!

Now we can’t simply stop here and solve for !,!"# ! !"#$% ! − !"#$%&!$ methods as we will have two variables i.e.

! = 1! + 1! ! − 1 ! = !

We therefore require a different approach. We know that the eigenfunctions consist of eigenvalues !!" and

! = !!!!" = !! 00 !!

= ⊿{!" !"#$% !"#$%&'"# !" !!"#$%&!$ methods are also

used. We define a new variable to separate the variables. Now X the eigenvector of A with x, y components that are unknown is equal to T(columns of eigenvectors of A )multiplied by U Define ! = !!!!

! = !!

Therefore

! = !!!!

! = !!!!! ! = !!!!"

! = !!!!"#

! = ⊿!

Step 1: Find the eigenvalues of A

! = 1 1 4 1

!! ! = det ! − !" = 1− ! 1

4 1− ! = 1− ! ! − 4 = !! − 2! − 3

!! − 2! − 3 = 0

! − 3 ! + 1 = 0 !! = 3; !! = −1

Step 2: Find the corresponding eigenvectors

!! = 3:

1− 3 1 : 0 4 1− 3 : 0 ~ −2 1 : 0

4 −2 : 0 ~

!!×−12


40

1 −12 : 0

4 −2 : 0

!! − 4!!

1 −12 : 0

0 0 : 0

!"# !! = !

!! =12!

!! = !121

= ! 12

!! = −1:

1+ 1 1 : 0 4 1+ 1 : 0 ~ 2 1 : 0

4 2 : 0 ~

!!×12

112 : 0

4 2 : 0

!! − 4!!

112 : 0

0 0 : 0

!"# !! = !

!! = −12 !

!! = ! −121

= ! −12

! = 1 −1

2 2

! = 3 00 −1

! = ⊿! !! = 3 0

0 −1!!


41

! = 3!

! − 3 ! = 0 ! = !!!!

! = −1! ! + 1 ! = 0 ! = !!!!

! = !"

!! = 1 −1

2 2!!!!!!!!

!! = 1 −1

2 2!!!!!!!!!

Q 22 b) −! − ! + !, ! − 2!,! + 2!

! = −! − ! + !, ! − 2!,! + 2!

! = !,!, !

Therefore ! =!!!

! = !′

! = !"

! =−! − ! + !! − 2!! + 2!

=−1 −1 11 0 −20 1 2

!!!

! =−1 −1 11 0 −20 1 2

Define ! = !!!! !" ! = !"

! = !" ! = !!!! ! = !!!!" ! = !!!!"#

But ! = !!!!" ! = ⊿!


42

Step 1: Find the eigenvalues and eigenvectors of A

! =−1 −1 11 0 −20 1 2

!! ! = det ! − !" =−1− ! −1 11 −! −20 1 2− !

!"#$%& !!

!! ! = + −1− ! −! 2− ! + 2 + − 1 −1 2− ! − 1 + 0

!! ! = −1− ! !! − 2! + 2 − ! − 3= −!! + 2! − 2− !! + 2!! − 2! − ! + 3 = −!! + !! − ! + 1

!! ! = −(!! − !! + ! − 1)

!! − !! + ! − 1 = 0 ! − 1 !! + 1 = 0

!! = 1; !! = !; !! = −!

!! = 1

−1− 1 −1 1 : 01 0− 1 −2 : 00 1 2− 1 : 0

~−2 −1 1 : 01 −1 −2 : 00 1 1 : 0

1 −1 −2 : 0−2 −1 1 : 00 1 1 : 0

~!! + 2!!

1 −1 −2 : 00 −3 −3 : 00 1 1 : 0

1 −1 −2 : 00 1 1 : 00 −3 −3 : 0

~ !! + !!!! + 3!!

1 0 −1 : 00 1 1 : 00 0 0 : 0

!"# !! = ! !! = −! !! = !


43

!! = !1−11

!! = !

−1− ! −1 1 : 01 0− ! −2 : 00 1 2− ! : 0

1 −! −2 : 0−1− ! −1 1 : 00 1 2− ! : 0

!! − −1− ! !! = !! + 1+ ! !!

1 −! −2 : 00 −! −1− 2! : 00 1 2− ! : 0

1 −! −2 : 00 1 2− ! : 00 ! −1− 2! : 0

!! + !!!!! − !!!

1 0 −1+ 2! : 00 1 2− ! : 00 0 0 : 0

!"# !! = ! !! = !(−2+ !) !! = ! 1− 2!

!! = !1− 2!−2+ !1

!"#$% ! !" !"#$ !! = !!

!! = !1−11

!! = !1− 2!−2+ !1


44

!! = !1+ 2!−2− !1

1 1− 2! 1+ 2!−1 −2+ ! −2− !1 1 1

! =1 0 00 ! 00 0 −!

!!!

=1 0 00 ! 00 0 −!

!!!

! − 1 ! = 0 ! = !!!!

! − ! ! = 0 ! = !!!"

! + ! ! = 0 ! = !!!!"

! = !"

!!!

=1 1− 2! 1+ 2!−1 −2+ ! −2− !1 1 1

!!!!!!!"!!!!"

Question 23 Use diagonalization to find the paths of particles in the velocity fields Q 23 a) ! + ! − !! , 4! + ! + !!

! = !" + ! !

! = !!!! ! = !!! !" + ! ! ! = !!! !"# + ! ! ! = ⊿! + !!!! !

!! =

! + !4! + ! + −!!

!!= 1 1

4 1!! + −!!

!!

! = 1 1

4 1

!! ! = det ! − !" = 1− ! 14 1− ! = !! − 2! − 3


45

!! − 2! − 3 = 0 ! − 3 ! + 1 = 0 !! = 3; !! = −1

1− 3 1 : 04 1− 3 : 0 ~ −2 1 : 0

4 −2 : 0

−2 1 : 04 −2 : 0

!!×−12

1 −12 : 0

4 −2 : 0

!! − 4!!

1 −12 : 0

0 0 : 0

!"# !! = !

!! =12!

!! = !121

= ! 12

!! = −1

1+ 1 1 : 04 1+ 1 : 0

2 1 : 04 2 : 0 ~ 1

12 : 0

4 2 : 0~ 1

12 : 0

0 0 : 0

!! = ! −12

! = 1 −12 2

! = 3 0

0 −1

!!! =14

2 −1−2 1

! = ⊿! + !!!! !


46

!! = 3 0

0 −1!! +

14

2 −1−2 1

−!!!!

! − 3 ! =14 −2!! − !!

[Note ! = !! + !! = !"#$%&#&'()*+ !"#$%&'# + !"#$%&'("# !"#$%&'(] [For ! ! !!" !"# !ℎ!"# !"#$: !!"! (!)]

! = !!!! +1

! − 3 −34 !

! = !!!! −12 !

! 11− 3

! = !!!! −12 !

! 1−2 = !!!! +

14 !

!

! + 1 ! =14 2!! + !!

! = !!!! +1

! + 134 !

! = !!!! +12 !

! 12 = !!!! +

14 !

!

! =!!!! +

14!!

!!!! +14!!

But ! = !"

!! = 1 −1

2 2

!!!! +14!!

!!!! +14!!


47

Tutorial questions-‐The characteristic polynomial Page 88 [Note: The sum of the eigenvalues of a square matrix is equal to the trace. The product of the eigenvalues is equal to the determinant. If A is a square matrix, then the trace of A [tr(A)] is the sum of the entries on the main diagonal. If A is not square then the trace is undefined.] Question 24 Find the trace of the matrices in Q 15 also find the sum of the eigenvalues these should be equal and equal to the coefficient of !! in the characteristic polynomial

Q 24 A) 0 0 41 0 40 1 −1

!! ! = −!! − !! + 4! + 4

!" ! = −1

! = ±2 or ! = −1

! = +2− 2− 1 = −1 ! = 3 [3×3 !"#$%&]

The coefficient of !! = −1 ! !! = −1

The coefficient of !!!! = −1 !!!!" !

!! = −1 ! −1 = −1 Question 25 Q 25 i) Find the characteristic polynomial of matrix ! = ! !

! !

!! ! = det ! − !" = ! − ! !! ! − ! = !! − !" − !" + !" − !"

Q 25 ii) Show that !! ! = !! − !"#$ + det!

!" ! = ! + ! det! = !" − !"

!! ! = !! − ! ! + ! + !" − !" = !! − !" − !" + !" − !"

Q 25 iii) Verify the Cayley Hamilton theorem for 2×2 matrices (Show that !!(!) is the zero matrix)

!! ! = !! − ! ! + ! + !" − !"


48

!"#$%&'( !" ! !"# !"#$%#& !"#$%&#%$ !" ! as each term is a matrix

!! ! = !! − ! ! + ! + !" − !" !

! !! !

!− ! !

! ! ! + ! + !" − !" 1 00 1 = 0 0

0 0 Question 27 Verify the Cayley-‐Hamilton Theorem using the characteristic polynomial in factorized form

A) 0 0 41 0 40 1 −1

!! ! = −!! − !! + 4! + 4 = 4− !! ! + 1 = 2− ! 2+ ! ! + 1 = 2! − ! 2! + ! ! + ! = 0

Note This is wrong as all of the matrices should be 3x3 not 2x2

2 00 2 − + = 0 0

0 0 Question 29 Use Cayley-‐Hamilton theorem to find parametric equations for streamlines in velocity fields Q 29 a) ! = !, !,−! − 3! − 3! Solution

! = !"

! =!!

−! − 3! − !=

0 1 00 0 1−1 −3 −3

!!!

! =0 1 00 0 1−1 −3 −3

!! ! = det ! − !" =−! 1 00 −! 1−1 −3 −3− !

= + −! −! −3− ! − (−3 ]+ − 1 [0+ 1]+ 0


49

!! ! = −! !! + 3! + 3 − 1 = −! !! + 3! + 3 − 1 = −!! − 3!! − 3! − 1= −(!! + 3!! + 3! + 1)

!! ! = − ! + 1 ! Evals !! = −1

! = !! ! = !!! !! = !!!

! ! ! = ! ! !

!! ! ! = !! ! !

!! ! ! = ! !" !"#$%# − !"#$%&'( !ℎ!"#!$

!! ! ! = !

!! = −1; !! = −1; !! = −1

∴ ! + 1 !! = ! [Note: For repeated factors of D we add multiples of !]

! = !!!! + !"!!! + !!!!!! Where ! !,! !"# !"#$%"!"& !"#$%&#% !"#$%&' To find these arbitrary constant vectors corresponding vector coefficients in ! and AX must be equated

∴ !" = !"!!! + !"#!!! + !!!!!!!

! = −!!!! + !!!! + !" −1 !!! + 2!"!!! + !!! −1 !!!

! = !!! −! + ! + !"!! −! + 2! − !!!!!! Equate coefficients

!" = −!+ ! !" + ! = ! ! + ! ! = !

!" = −!+ !" !" + ! = 2! ! + ! ! = 2!


50

!" = −! !" + ! = ! ! + ! ! = 0

Now we need to find the eigenvectors Solve for C in { ! + ! ! = 0}

! − !" ! = 0 Therefore ! is the eigenvector corresponding to ! = −1

! − !" ! = ! Solve the eigenspace

! =0 1 00 0 1−1 −3 −3

1 1 0 : 00 1 1 : 0−1 −3 −2 : 0

1 1 0 : 00 1 1 : 00 −2 −2 : 0

1 0 −1 : 00 1 1 : 00 0 0 : 0

!"# !! = ! !! = −! !! = !

!! = !1−11

Solve for B

! + ! ! = 2!

! + ! =0+ 1 1 00 0+ 1 1−1 −3 −3+ 1

~1 1 00 1 1−1 −3 −2

1 1 0 : 2!0 1 1 : −2!−1 −3 −2 : !

etc.

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