Year 12 - Matrices

Chapter 2A Revision

The Term Exam will test your cumulative knowledge on Matrices, that is, anything you have ever learned in Matrices could be on the test (accept Cramer’s Rule).

So do not just do 12MC Chapter 2A, but I suggest you revise thoroughly by working through the WHOLE of the Year 11 notes, and worksheets.

We will STOP working from the Text book now, and just use this workbook.

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Eigenvalues

Let the state of stress at Point P be given in (MPa): a) Find the principal stress

and the principal directions of stress; b) Deduce the transition matrix that

diagonalises the stress matrix.

Concrete is strong when compressed, but easily pulled apart. Civil engineers need to determine whether a structure will fall apart under load.

It is possible to represent a body of material in Matrix form. By analysing the effects of applying a force to the body, engineers can assess for catastrophic failure. The force can be represented also as a matrix (the force vector is simply reformatted into a matrix).

Eg. The Force Vector ⟨ x , y ⟩ , becomes Matrix [ xy ]

Engineering students will learn that the Maximum and Minimum forces within a material (the ‘principal stresses’ in Structural Analysis) occur when the deformation is in the same direction as its initial position (the ‘principal directions’ in Structural Analysis).

** A KEY task is to explore when and by how much a Matrix A ‘stretches’ another Matrix, but maintain the same direction. We find such ‘stretching factors’, called Eigenvalues, and the corresponding directions, called Eigenvectors. **

where, Av=λv

here, we say v is an Eigenvector

and, we say λ is an Eigenvalue (a scalar)

and, A is a Matrix

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What does this mean?

We take any Matrix A and apply Eigenvector v, the result is characterised by the fact that the multiplying purely ‘stretches’ vector v.

So this ‘stretching factor’ is λ. (a scalar or “number”)

If λ is positive, then we are stretching Matrix A.

If λ is negative we are compressing Matrix A.

Task:

Confirm matrix A=[1 22 −2] has the Eigenvector ⟨−1,2 ⟩, hence also find the

Eigenvalue.

We need to show Av=λv

Hint: Convert the Eigenvector to matrix form [−12 ] and complete matrix

multiplication.

Then, find a value for λ, such that λ multiplied by your Eigenvector is the same as your previously calculated matrix multiplication.

This “scale factor” is your Eigenvalue, called LAMDA λ .

Ans: Hopefully you found that λ=−3.

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Task:

Confirm matrix A=[1 22 −2] (the same matrix as the last Task) has the

Eigenvector ⟨2,1 ⟩, hence find the Eigenvalue.

We need to show Av=λv

Hint: do the same as the previous Task.

Ans: λ=2

As you can see, we can have multiple Eigenvectors for any given Matrix, and each Eigenvector has its own Eigenvalue. Lets take this one step further.

Task:

Verify that Matrix A=[ 0 1 00 0 1

−6 7 0] has Eigenvectors ⟨1,1,1 ⟩ , ⟨1 ,−3 ,9 ⟩ , ⟨1 ,2 ,4 ⟩

And, determine the Eigenvalue corresponding to each Eigenvector.

Ans: λ=1 ,−3 ,2

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Task:

Return to the Matrix from the previous task. You verified that the vector ⟨1 ,1 ,1 ⟩ was an Eigenvector, and you found the corresponding Eigenvalue to be λ=1.

For that same Matrix, check if the following vectors are also Eigenvectors. ⟨ 2,2 ,2 ⟩ , ⟨3 ,3 ,3 ⟩ , ⟨5 ,5 ,5 ⟩ .

Or what about ⟨−2 ,−2 ,−2 ⟩ , or ⟨1.75 ,1.75 ,1.75 ⟩ … etc

What is your Hypothesis about Eigenvectors related to an Eigenvalue?

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So, from our above Task, for the Eigenvalue of λ=1, we have a “Family” of Eigenvectors and they have a relationship with each other … !

How can we describe this “family”? Clearly, as our answer in an exam, we can’t list the Eigenvectors as ⟨1 ,1 ,1 ⟩, ⟨2,2 ,2 ⟩, ⟨3 ,3 ,3 ⟩ … etc etc, as there is an Infinite number of solutions.

Can you see the Eigenvectors are all scalar multiples of each other… so using factorisation, we can pull out a scale factor … but hang on, the scale factor isn’t just a single value is it? How do we label something we don’t know, or can’t put a single value on … ? Lets just call it … t …

So we can describe the family of Eigenvectors, relating to Eigenvalue λ=1 as … t ⟨1 ,1 ,1, ⟩.

When I pulled the factor of t out, why did I leave the eigenvector as ⟨1 ,1 ,1 ⟩?

I could have said t ⟨2 ,2 ,2 ⟩ … or even … t ⟨ 32 , 32 , 32 ⟩ or an infinite of other ways.

Yep, you guessed it … like any answer, you want to try and find it’s “Simplest” form …

Lets do that again with the next Eigenvector … ⟨1 ,−3 ,9 ⟩.

Can you think of another eigenvector in this family?

How about trying ⟨ 2,−6 ,18 ⟩ … how many others are there?

Here, we can say … All vectors ( t ,−3 t ,9 t ) are Eigenvectors corresponding to Eigenvalue −3.

Eigenvector … [ t−3 t9 t ] or we could say t [ 1−39 ]

So, t [ 1−39 ] is a family of Eigenvectors of Matrix A , for the Eigenvalue of λ=−3

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But what if Eigenvectors are not known?

By definition, we know;

Av=λv

juggle a bit

0=λv−Av

This doesn’t help us as we do NOT know v. As we do not know what it is and it appears numerous times in our definition, lets reduce how much we see it in the equation by factoring.

0=( λ−A ) v

But, note that λ is a scalar and we can’t subtract Matrix A from a scalar. We know the identity Matrix is like multiplying by 1, so we can add it in without changing the overall value of our equation;

0=( λI−A ) v

Clearly if Eigenvector v = zero, this satisfies the equation, however this is called a Trivial solution, as it does not even attempt to find any Eigenvectors or Eigenvalues.

For Non-Trivial solutions, it is clear we need ( λI−A ) v=0, where v≠0

Unfortunately, its not as simple as saying λI−A=0 …

In University, you will talk about the need for a null space, and about linear independence (refers to geometric space) … but we have no need to go there…

For ( λI−A ) v=0, where v≠0, we need the determinant of λI−A equal to zero, so we say:

∴|λI−A|=0

Note: Not all matrices have eigenvectors, so the test for Eigenvalues becomes:

|λI−A|=0

Sometimes we get lost in reading text books, and just go with the flow and we forget what our purpose is … Can you recall what we are trying to find … ?

We are trying to find if Matrix A has an Eigenvector. For an Eigenvector to exist, there must be an Eigenvalue!

Can you remember what label we give an Eigenvalue?

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Hopefully you recall that an Eigenvalue is λ … !

So, we are trying to find a value for λ from our Test for Eigenvectors:

|λI−A|=0

*** recall that those vertical lines means “Determinant” ***

When we expand this determinant calculation, this becomes what is known as a “Characteristic Polynomial”. Recall that not all matrices have Eigenvalues or Eigenvectors, so some characteristic Polynomials do NOT have a solution.

Please note that I am not wanting to trick you in the exam. If I ask you to find any Eigenvalues or Eigenvectors, they will be there … !

NOTE: the text book is different pge. 63

I think my way is easier, but feel free to use anything that works for you

So again, I would use

|λI−A|=0

From our statement of det ( λI−A )=0 lets see where we go…

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Example: Find the Eigenvalues (if any) for Matrix A= [1 2 30 1 −20 1 4 ].

Solution (as you would write in your exam):

By Eigenvector definition we know;

Av=λv

so,

0=λv−Av

0=λIv−Av

0=( λI−A ) v

For non-trivial solutions we need therefore;

|λI−A|=0

Find the Characteristic Polynomial:

|[ λ 0 00 λ 00 0 λ ]−[1 2 3

0 1 −20 1 4 ]|=0

|λ−1 −2 −30 λ−1 20 −1 λ−4|=0

expand column 1

( λ−1 )|λ−1 2−1 λ−4|=0

( λ−1 ) {( λ−1 ) ( λ−4 )+2 }=0

( λ−1 ) (λ2−5 λ+6 )=0

( λ−1 ) ( λ−2 ) ( λ−3 )=0

Now by the Null Factor Law, we get our Eigenvalues of, λ=1 ,2∨3

*** Look at my setting out. I am showing my algebraic facility by manually factorising, and justifying my steps with the odd sentence … these are all in the marking rubric and is what Panel is looking for. ***

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Now you try … starting out easy with 2x2 matrices … Find the Eigenvalues of:

1.

[1 22 −2] If you are stuck, I have a fully worked solution to this one on

the next page

2.

[2 25 −1]

Ans: λ=4 ,∨λ=−3

3.

[2 11 2]

Ans: λ=3∨λ=1

4.

[4 −32 −1]

Ans: λ=1∨λ=2

5.

[5 −11 3 ]

Ans: λ=4 (multiplicity 2)

6.

[ 7 0 −3−9 −2 318 0 −8 ]

Ans: λ=1∨λ=−2∨λ=−2

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7.

[−1 1 31 2 03 0 2]

Ans: λ=2∨ λ=−3∨λ=4

8.

[0 0 −21 2 11 0 3 ]

Ans: λ=1∨λ=2∨λ=2

9.

[ 5 1 −1−3 1 30 0 4 ]

Ans: λ=2∨ λ=4 ,∨λ=4

10.

[5 0 40 3 −10 0 −2]

Ans: λ=5∨ λ=3∨λ=−2

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Eigen Question Example Solution

Find the Eigenvalues for the following matrix:

1.

[1 22 −2]

Eigen definition Av=λv

∴ ( λI−A ) v=0

Establish v as non zero as this is a trivial solution, so we can say that:

|λI−A|=0

|[ λ 00 λ]−[1 2

2 −2]|=0|λ−1 −2−2 λ+2|=0

( λ−1 ) ( λ+2 )−4=0

λ2+ λ−6=0

The characteristic Equation becomes

( λ−2 ) ( λ+3 )=0

∴ λ=2∨−3

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Let us move on …

Now we need to determine the corresponding Eigenvectors for each Eigenvalue.

We still need to comply with the general definition of an Eigenvector:

( λI−A ) v=0

but now the only unknown is the vector v.

Returning to Matrix A= [1 2 30 1 −20 1 4 ], we shall determine the Eigenvector,

corresponding with Eigenvalue 2.

By substitution, we get:

{[2 0 00 2 00 0 2]−[1 2 3

0 1 −20 1 4 ]}×[ v1v2v3]=0

[1 −2 −30 1 20 −1 −2]×[v1v2v3]=0

solve by putting it into an augmented matrix and using Gaussian Elimination

(1 −2 −30 1 20 −1 −2|

000)

(1 −1 −10 1 20 0 0 |000)(1 0 10 1 20 0 0|

000)

Can you see the bottom row is All zero’s. From that we can see that v3 is Arbitrary, and we term this a Free Variable. (we now have three variables and just two equations, which means we do NOT get a single solution). This is actually expected, as recall that Eigenvectors form a ‘family’, rather than a specific solution.

*** a column of all zero’s is another indicator of which v is arbitrary ***

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So we can deduce from the augmented matrix:

From Row 2,

0 v1+1v2+2v3=0

So,

v2=−2v3;

From Row 1,

1v1+0+1v3=0

thus , v1=−v3

looks like we hit a dead end…..but;

As v3 can be arbitrary, we term this a ‘free variable’, and I choose to …

set v3=1

So from our above “dead end” equations we can now say:

v2=−2

and

v1=−1

Thus, one Eigenvector, that corresponds to Eigenvalue 2, is: [−1−21 ]

or using the correct form for the full Family of Eigenvectors … t [−1−21 ].

This is harder than it looks …

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Task. … BEFORE you do this task (a 3X3 Matrix Eigen question), try the first three questions over the page as these are 2X2 matrices … but come back here and make sure you do this 3X3 Matrix as well!

For the same Matrix as above A= [1 2 30 1 −20 1 4 ] , determine the Eigenvectors

corresponding to Eigenvalues 1 and 3.

Even if I give you the answers … t [100] and t [−12−2] , … It’s still hard!

When there are two Eigenvalues that are the same, it usually leads to having two Eigenvectors associated with the single Eigenvalue. In this situation, the augmented matrix will normally have two free variables and they both can’t be free at the same time, so by setting an arbitrary value to each individually, will lead to your two Eigenvectors. You may not really understand this until you get to a question that requires this style of solution.

Some will be straight forward, but for others you will need to be persistent!

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Eigen Questions … Find the Eigenvalues and Eigenvectors for the following matrices:

1.

[1 22 −2] If you are stuck, I have a fully worked solution to this one on

the next page

2.

[2 25 −1]

Ans: λ=4 , v=t [11]∨ λ=−3 , v=t [−25 ]3.

[2 11 2]

Ans: λ=3 , v=t [11]∨λ=1 , v=t [−11 ]4.

[4 −32 −1]

Ans: λ=1 , v=t [11]∨λ=2 , v= t [32]

5.

[5 −11 3 ]

Ans: λ=4 (multiplicity 2) , v=t [11]∨v=t [11]

6.

[ 7 0 −3−9 −2 318 0 −8 ]

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Ans: λ=1 , v=t [ 1−12 ]∨ λ=−2, v=t [010]∨t [103]

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7.

[−1 1 31 2 03 0 2]

Ans: λ=2 , v=t [ 0−31 ]∨λ=−3 , v=t [−513 ]∨λ=4 , v=t [213]8.

[0 0 −21 2 11 0 3 ]

Ans: λ=1 , v=t [−211 ]∨λ=2, v=t [010]∨t [−101 ]9.

[ 5 1 −1−3 1 30 0 4 ]

Ans: λ=2 , v=t [ 1−30 ]∨λ=4 , v=t [−110 ]∨t [101]10.

[5 0 40 3 −10 0 −2]

Ans: λ=5 , v=t [100]∨λ=3 , v=t [010]∨λ=−2 , v=t [−20735 ]

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Eigen Question Example Solution

Understand that Exam setting out is a bit more lenient than in an Assignment due to time pressures, so the below is all that would be required in an exam, whereas in an assignment you would justify and reason your steps more:

Find the Eigenvalues and Eigenvectors for the following matrices:

1.

[1 22 −2]

Eigen definition Av=λv

∴ ( λI−A ) v=0

Establish v as non zero as this is a trivial solution, we can say that:

det|λI−A|=0

|[ λ 00 λ]−[1 2

2 −2]|=0|λ−1 −2−2 λ+2|=0

( λ−1 ) ( λ+2 )−4=0

λ2+ λ−6=0

The characteristic Equation becomes

( λ−2 ) ( λ+3 )=0

∴ λ=2∨−3

For λ=2, let Eigenvector = [v1v2] From the Eigen definition,

Av=λv

( λI−A ) v=0

we can now say;

([2 00 2 ]−[1 2

2 −2])[v1v2]=0[ 1 −2−2 4 ] [v1v2]=0

In Augmented form ( 1 −2−2 4 |00)

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(1 −20 0 |00) {R2+2×R1 }

From Row 1 → v1−2v2=0

Or v1=2v2

From Row 2, v2 is arbitrary and is a free variable, so set v2=1 and hence we can

also say, v1=2 , and our solution becomes v=t [21]

For λ=−3, let Eigenvector = [v1v2] and from the Eigen definition, we can now say;

([−3 00 −3 ]−[1 2

2 −2])[v1v2]=0[−4 −2−2 −1] [v1v2]=0

In Augmented form (−4 −2−2 −1|00)

(−2 −10 0 |00) {R2−12×R1}

From Row 1 → −2v1−v2=0

Or v2=−2v1

From Row 2, v2 is a free variable, so set v2=−2 and hence we can also say, v1=1 ,

and our solution becomes v=t [ 1−2]

∴the Eigenspace for ¿[1 22 −2¿

]

for Eigenvalue λ=2 is the Eigenvector=t (21) ,∧¿

for Eigenvalue λ=−3 ,is theEigenvactor=t( 1−2) .

Once you have finished all of the above, take a look at this tricky one.

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Example: Find (if any) the Eigenvectors for Matrix A= [1 3 03 1 00 0 −2].

I have done the first solution for you, but I suggest you do the others.

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Solution (as you would write in your exam):

By Eigenvector definition we know;

Av=λv

so,

0=λv−Av

0=λIv−Av

0=( λI−A ) v

For non-trivial solutions where v≠0, we need therefore;

|λI−A|=0

Find the Characteristic Polynomial:

|[ λ 0 00 λ 00 0 λ ]−[1 3 0

3 1 00 0 −2]|=0

|λ−1 −3 0−3 λ−1 00 0 λ+2|=0

0+0+( λ+2 )|λ−1 −3−3 λ−1|=0

( λ+2 ) {( λ−1 ) ( λ−1 )−9 }=0

( λ+2 ) ( λ2−2 λ−8 )=0

( λ+2 ) ( λ−4 ) ( λ+2 )=0

Now by the Null Factor Law, we get our Eigenvalues of, λ=¿ -2, -2 and 4.

To determine the corresponding Eigenvectors for each Eigenvalue, we still need to comply with the general definition of an Eigenvector:

( λI−A ) v=0

Where vector v is the Eigenvector.

For Eigenvalue 4, by substitution, we get:

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{[4 0 00 4 00 0 4 ]−[1 3 0

3 1 00 0 −2]}×[v1v2v3]=0

[ 3 −3 0−3 3 00 0 6]×[ v1v2v3]=0

solve via augmented matrix ( 3 −3 0−3 3 00 0 6|

000)

R1/3; R1+R2; R3/6 (1 −1 00 0 00 0 1|

000)

From Row 3,

0 v1+0 v2+1 v3=0

the only solution is v3=0;

From Row 1,

1v1−1v2+0 v3=0

thus , v1=v2

We can see that v2 is Arbitrary, and is a Free Variable.

Choose to set;

v2=1

Thus, one Eigenvector, that corresponds to Eigenvalue 4, is: [110]

And the full Family of Eigenvectors become … t [110].27

Task. Over to you to find a solution for the other 2 Eigenvalues …

For the same Matrix, determine the Eigenvectors corresponding to Eigenvalue -2.

** Care, recall above that Eigenvalue -2 was TWO solutions, so consider how many Eigen vector families you will have in your answer.

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Rotational Matrix

The Rotational Matrix R=[cosθ −sinθsinθ cosθ ]

The Rotational Matrix can rotate a vector v by theta degrees in an anticlockwise direction about the “origin”. This is done through Pre matrix multiplication … Rv. Lets investigate:

Consider the Position Vector ⟨√3 ,1 ⟩ , Firstly, we can convert this to a column

matrix [√31 ].To rotate this vector by 60 degrees in an anticlockwise direction we set:

θ=60= π3

So

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Task:

Draw vector (√3 ,1 ) on a Cartesian plane.

Now rotate that vector by 60 degrees anticlockwise and confirm our above result is true.

Task:

How far is (√3 ,1 ) from the Origin?

How far is (0,2 ) from the Origin?

What is your Hypothesis about the Rotation Matrix and how it affects the length of the rotated vector?

Task:

How do you change the length of a vector?

Task:

Create two vectors. The first at an angle of 45 degrees to the positive x axis, and the second at an angle of 60 degrees to the positive x axis.

Make both vectors a different length.

Store these two vectors in a 2 x 2 matrix. (each vector is stored vertically)

Use the rotational Matrix to rotate both vectors by 90 degrees in the CLOCKwise

direction. (hence, set θ=−π2 )

Check your resultant Matrix. Have both vectors been rotated?

What is your hypothesis about using the Rotational matrix in larger Matrices that include multiple position vectors?

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Task:

Take your two vectors from your previous task. Rather than putting your vectors

⟨a ,b ⟩ and ⟨ c ,d ⟩ vertically into a matrix like [a cb d ], keep them horizontal, like

[a bc d ]. Further, in the last task you set θ=

−π2 so that you rotated CLOCKwise. In

this task I am going to ask you to set θ=π2 , however, this time you will POST-

multiply by the Rotational matrix, rather than pre-multiplying!

Create a Hypothesis about how the Rotational Matrix works.

Note: It is “common” to see position vectors turned into column matrices, and “uncommon” to keep position vectors as row matrices. Although it is important that you understand that order does matter in PRE or POST multiplying by the rotational matrix, I suggest Always using the column transpose of the position vector, and PRE-multiplying by the Rotational Matrix and know that this moves the vectors in an ANTI-clockwise direction.

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We have seen how a Rotational Matrix can rotate a vector by a certain value. In fact we saw two ways of doing this, but we shall stick to column matrix conversions of vectors, and PRE-multiplying by the rotational matrix to rotate the vectors ANTI-clockwise about the origin …

We know a Matrix can store a system of vectors. This data does not necessarily have to relate to a line, but it could also relate to the end point of the vector. That is, vector ⟨1 ,3 ⟩ could simply be the representation for the point (1,3) in a Cartesian plane. So this would be a Position Vector!

Computer graphics utilises the Rotational Matrix properties and is used extensively in CG effects in movies and animations. As the order of the matrix increases, we can move into multidimensional rotations of 3D objects, however, lets first restrict ourselves to a 2x2 rotational matrix and 2D rotations.

Task 1:

By using the Rotational Matrix, apply a 90 degree rotation to the following Matrix.

[1 4 41 1 5]

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Task 2.

Now lets look at what you have achieved, (with relative ease), in a graphical sense.

1. On the next page, label the Cartesian plane ( −5<x<5 ; −5< y<5)

2. Split the original Matrix into three column vectors.

3. On the next page, plot the POINTS that each column Position vector

represents. (do not draw the line vector, just put a dot in the appropriate

place)

4. Now, Join the three dots in Black Pen

Now;

1. Split your rotated matrix (your answer from Task 1.) into three column

vectors.

2. Again, plot the points that these position vectors represent on your

Cartesian Plan from before.

3. Now, join the dots with a Red pen.

What is your hypothesis about what a rotational Matrix can achieve?

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Task: Do these questions and provide me with your answers.

Question1.

[122] is an Eigenvector of [4 −2 12 0 12 −2 3]. What is the corresponding Eigenvalue?

Q2.

What are the Eigenvalues of [4 7 10 −3 80 0 2]?

Q3.

Given that 1 is an Eigenvalue of A=[2 5 −61 0 00 1 0 ], find the other two Eigenvalues.

Q4.

What is the Characteristic Equation for A=[4 −1 62 1 52 −1 0]

Q5.

One Eigen value of Matrix A=[ 1 2 −4−1 4 80 1 −1 ] is 5. Find the other eigenvalues and

all of their corresponding Eigenvectors.

Q6.

Apply a rotation of π6 to matrix A=[34 ].

Also, what is the length of the vector before and after this transformation?

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Q7.

A set of vectors, v1 (0,5 ) , v2 (1,√3 ) , v3 (√3 ,1 ) are representational of a system of braces on a building structure. Designers have relocated the flanges that these braces connect to. The braces now need to be one and a half times longer and

also rotated clockwise by π6 .

What are the new vectors?

Use a separate diagram to display each of the given vectors, and the revised vector. Confirm that you have indeed made the vectors the required lengths and they have been rotated correctly.

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