IBMaths: AI HLDPExam Questions1. Number & Algebra1.8 Eigenvalues & Eigenvectors

Eigenvalues & Eigenvectors (DP IB Maths: AI HL)

Exam Questions

4 hours21 questions
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1a
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Consider the 2 cross times 2 matrix bold italic A defined by 

bold italic A equals open parentheses table row cell 0.1 end cell cell space space 0.4 end cell row cell 0.9 end cell cell space space 0.6 end cell end table close parentheses

(i)
Find the characteristic polynomial of bold italic A.

(ii)
By solving an appropriate equation with the characteristic polynomial, find the eigenvalues lambda subscript 1 and lambda subscript 2 of bold italic A.

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1b
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Let bold italic x subscript bold 1 and bold italic x subscript bold 2 be the eigenvectors of bold italic A corresponding to lambda subscript 1 and lambda subscript 2 respectively.

By solving the eigenvector equations bold italic A bold italic x subscript 1 equals lambda subscript 1 bold italic x subscript 1and bold italic A bold italic x subscript 2 equals lambda subscript 2 bold italic x subscript 2 bold comma   find eigenvectors bold italic x subscript bold 1 and bold italic x subscript bold 2 .

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1c
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Show that the answers to part (b) could alternatively have been found by solving the equations open parentheses bold italic A minus lambda subscript 1 bold italic I close parentheses space bold italic x subscript 1 equals open parentheses table row 0 row 0 end table close parentheses   and  open parentheses bold italic A minus lambda subscript 2 bold italic I close parentheses space bold italic x subscript 2 equals open parentheses table row 0 row 0 end table close parentheses  ,  where bold italic I is the 2 cross times 2 identity matrix. 

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2
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Find the eigenvalues and corresponding eigenvectors for the matrix bold italic A defined as

                   bold italic A equals open parentheses table row cell negative 1 end cell cell space space 4 end cell row 1 cell space space 2 end cell end table close parentheses

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3
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Consider the matrix bold italic B defined as

               bold italic B equals open parentheses table row 4 cell space space minus 6 end cell row 1 cell space space minus 2 end cell end table close parentheses 

Find the eigenvalues and corresponding eigenvectors of bold italic B.

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4a
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Find the eigenvalues for each of the following matrices:

                  bold italic C equals open parentheses table row cell negative 2 end cell cell space space 13 end cell row cell negative 1 end cell cell space space 2 end cell end table close parentheses

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4b
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bold italic D equals open parentheses table row 6 cell space space minus 1 end cell row 17 cell space minus 2 end cell end table close parentheses

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5a
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Consider the matrix bold italic M defined as

                  bold italic M equals open parentheses table row cell negative 1 end cell k row 3 cell negative 1 end cell end table close parentheses

where k element of straight real numbers is a constant.

The eigenvalues of bold italic M are 2 and negative 4.

Find the value of k.

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5b
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Find the eigenvectors of bold italic M that correspond to the two eigenvalues.

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5c
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Hence write bold italic M in the form bold italic P bold italic D bold italic P to the power of negative 1 end exponent, where bold italic P is a matrix of eigenvectors and bold italic D is a diagonal matrix of eigenvalues.

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6a
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It is given that, for n cross times n matrices bold italic A, bold italic B and bold italic C,

                   bold italic A equals bold italic B bold italic C bold italic B to the power of negative 1 end exponent

Use the properties of matrices and matrix inverses to show that bold italic A squared equals bold italic B bold italic C squared bold italic B to the power of negative 1 end exponent.

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6b
Sme Calculator3 marks

Consider the matrix  bold italic M equals open parentheses table row 3 cell negative 2 end cell row p 1 end table close parentheses  ,  where space p element of straight real numbers  is a constant and where it is given that open parentheses table row 1 row 2 end table close parentheses is an eigenvector of bold italic M.

Find the value ofspace p.

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6c
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Hence, by first finding the eigenvalues and the other eigenvector of bold italic M, write bold italic M in the form bold italic M equals bold italic P bold italic D bold italic P to the power of negative 1 end exponent for appropriate matrices bold italic P and bold italic D.

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6d
Sme Calculator4 marks
(i)
Use the result of part (c) to show that

             bold italic M to the power of n equals 1 third open parentheses table row cell 2 open parentheses 5 to the power of n close parentheses plus open parentheses negative 1 close parentheses to the power of n end cell cell negative 5 to the power of n plus open parentheses negative 1 close parentheses to the power of n end cell row cell negative 2 open parentheses 5 to the power of n close parentheses plus 2 open parentheses negative 1 close parentheses to the power of n end cell cell 5 to the power of n plus 2 open parentheses negative 1 close parentheses to the power of n end cell end table close parentheses

(ii)

Show that the expression for bold italic M to the power of bold n in part (d)(i) gives the expected result when n equals 1.

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7a
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Exobiologists are studying two species of animals in a region of the distant planet Dirion. In the researchers’ models the population of Heliors (a predator species) is indicated by h, while the population of Sklyveths (a competing predator species) is indicated by s.

If the respective populations at a particular point in time are h subscript n and s subscript n, then the researchers’ data suggest that the populations one year later may be given by the following system of coupled equations:

 h subscript n plus 1 end subscript equals 1.06 h subscript n minus 0.16 s subscript n 

s subscript n plus 1 end subscript equals negative 0.04 h subscript n plus 0.94 s subscript n

Represent the system of equations in the matrix form bold italic x subscript n plus 1 end subscript equals bold italic M bold italic x subscript n.

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7b
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At the start of the study, there are 600 Heliors and 500 Sklyveths in the region.

Find the expected size of the respective populations after one year.

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7c
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By first finding the eigenvalues and corresponding eigenvectors of bold italic M write bold italic M in the form bold italic P bold italic D bold italic P to the power of negative 1 end exponent, where bold italic P is a matrix of eigenvectors and bold italic D is a diagonal matrix of eigenvalues.

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7d
Sme Calculator3 marks

Hence show that the respective populations after n years are predicted by the model to be h subscript n equals 520 open parentheses 0.9 to the power of n close parentheses plus 80 open parentheses 1.1 to the power of n close parentheses  and s subscript n equals 520 open parentheses 0.9 to the power of n close parentheses minus 20 open parentheses 1.1 to the power of n close parentheses .

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7e
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Describe what the model predicts in the long term for the populations of the two species, and offer one criticism of the model based on this prediction.

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1a
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Consider the 2 cross times 2 matrix A defined by

bold italic A equals open parentheses table row cell 0.35 end cell cell 0.15 end cell row cell 0.65 end cell cell 0.85 end cell end table close parentheses 

(i)
Find the characteristic polynomial of A.

(ii)
Find the eigenvalues of A.

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1b
Sme Calculator4 marks

Let lambda subscript 1and lambda subscript 2 be the eigenvalues found in part (a)(ii), and let bold italic x subscript 1 and bold italic x subscript 2 be the eigenvectors of A corresponding to lambda subscript 1and lambda subscript 2respectively.

Find eigenvectors bold italic x subscript 1 and bold italic x subscript 2.

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1c
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Explain with justification whether the answers found in part (b) are unique.

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2a
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Consider the matrix A defined as

bold italic A equals open parentheses table row cell negative 1 end cell cell 0.75 end cell row 4 cell negative 1.5 end cell end table close parentheses 

Find the eigenvalues and corresponding eigenvectors of matrix A.

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2b
Sme Calculator5 marks

Now consider the matrix k bold italic Adefined as

 k bold italic A equals open parentheses table row cell negative k end cell cell 0.75 k end cell row cell 4 k end cell cell negative 1.5 k end cell end table close parentheses 

where k not equal to 0 is a real constant. 

Show that the eigenvectors found in part (a) are also eigenvectors of matrix k bold italic A and determine their corresponding eigenvalues.

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3
Sme Calculator6 marks

Consider the matrix bold italic B defined as

 bold italic B equals open parentheses table row 6 cell negative 2 end cell row 1 2 end table close parentheses 

Find the eigenvalues and corresponding eigenvectors of bold italic B.

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4a
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Find the eigenvalues for each of the following matrices:

bold italic C equals open parentheses table row 5 cell negative 4 end cell row cell 4.5 end cell cell negative 1 end cell end table close parentheses

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4b
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bold italic D equals open parentheses table row cell 5 k end cell cell negative 4 k end cell row cell 8.5 k end cell cell negative 5 k end cell end table close parentheses

where k not equal to 0 is a real constant.

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4c
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Show that open parentheses table row cell 2 plus 2 straight i end cell row 3 end table close parentheses    is an eigenvector of matrix bold italic C, and find the other eigenvector.

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5a
Sme Calculator6 marks

Consider the matrix bold italic Mdefined as

 bold italic M equals open parentheses table row cell negative 3 end cell k row 2 6 end table close parentheses 

where k element of straight real numbers is a constant. 

Given that -2 is an eigenvalue of bold italic M, 

find the remaining eigenvalue of bold italic M, as well as the eigenvectors that correspond to the two eigenvalues.

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5b
Sme Calculator2 marks

Hence diagonalise bold italic M by writing it in the form bold italic P bold italic D bold italic P to the power of bold minus bold 1 end exponent for appropriate matrices bold italic P and bold italic D.

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6a
Sme Calculator4 marks

It is given that, for n cross times n matrices bold italic A comma bold space bold italic B and bold italic C,

 bold italic A equals bold italic B bold italic C bold italic B to the power of negative 1 end exponent

Use the properties of matrices and matrix inverses to explain why  bold italic A to the power of n equals bold italic B bold italic C to the power of n bold italic B to the power of negative 1 end exponent.

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6b
Sme Calculator7 marks

Consider the matrix  bold italic M equals open parentheses table row p 1 row cell negative 2 end cell cell negative 3 end cell end table close parentheses,  where p element of straight real numbers is a constant and where it is given that open parentheses table row 1 row 1 end table close parentheses is an eigenvector of bold italic M.

By first finding the eigenvalues and the other eigenvector of bold italic M, write bold italic M in the form bold italic M bold equals bold italic P bold italic D bold italic P to the power of bold minus bold 1 end exponent   for appropriate matrices bold italic Pand bold italic D.

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6c
Sme Calculator5 marks
(i)
Use the result of part (b) to show that

 M to the power of n equals open parentheses negative 1 close parentheses to the power of n open parentheses table row cell 2 cross times 5 to the power of n end cell cell 4 to the power of n minus 5 to the power of n end cell row cell 2 open parentheses 5 to the power of n minus 4 to the power of n close parentheses end cell cell 2 cross times 4 to the power of n minus 5 to the power of n end cell end table close parentheses

(ii)
Show that the expression for bold italic M to the power of n in part (c)(i) gives the expected result when n equals 3. 

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7a
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Exobiologists are studying two species of animals in a region of the distant planet Dirion.  In the researchers’ models the population of Reddors (a prey species) is indicated by r, while the population of Sklyveths (a predator species that preys on Reddors) is indicated by s. 

If the respective populations at a particular point in time are r subscript n and s subscript n, then the researchers’ data suggest that the populations one year later may be modelled by the following system of coupled equations:

r subscript n plus 1 end subscript equals 1.3 r subscript n minus 0.25 s subscript n

s subscript n plus 1 end subscript equals 0.07 r subscript n plus 0.9 s subscript n

Represent the system of equations in the matrix form bold italic x subscript n plus 1 end subscript equals bold italic M bold italic x subscript n.

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7b
Sme Calculator8 marks

By first finding the eigenvalues and eigenvectors of bold italic M, write bold italic M in the form bold italic M equals bold italic P bold italic D bold italic P to the power of bold minus bold 1 end exponent for appropriate matrices bold italic P and bold italic D.

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7c
Sme Calculator3 marks

At the start of the study there were 2100 Reddors and 2850 Sklyveths in the region. 

Show that the respective populations after n years are predicted by the model to be r subscript n equals 75 open parentheses 1.25 to the power of n close parentheses plus 2025 open parentheses 0.95 to the power of n close parentheses and s subscript n equals 15 open parentheses 1.25 to the power of n close parentheses plus 2835 open parentheses 0.95 to the power of n close parentheses.

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7d
Sme Calculator4 marks
(i)
Determine the ratio of Reddors to Sklyveths that the model predicts will be in the region in the long term.  Be sure to justify your answer.

(ii)
Determine the number of years it will take after the start of the study for the population of Reddors to exceed the population of Sklyveths.

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1a
Sme Calculator6 marks

Find the eigenvalues and corresponding eigenvectors for each of the following matrices:

A equals open parentheses table row 2 cell negative 7 over 3 end cell row cell 13 over 6 end cell cell negative 5 over 2 end cell end table close parentheses

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1b
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B equals open parentheses table row cell 0.1 end cell cell 0.5 end cell row cell negative 0.02 end cell cell 0.3 end cell end table close parentheses

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2a
Sme Calculator7 marks

Find the eigenvalues and corresponding eigenvectors for each of the following matrices:

C equals open parentheses table row cell negative 3 end cell 17 row cell negative 2 end cell 3 end table close parentheses

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2b
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D equals open parentheses table row cell 1.5 end cell cell negative 2.5 end cell row cell 4.5 end cell cell 2.5 end cell end table close parentheses

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3a
Sme Calculator7 marks

Let M equals open parentheses table row a b row c d end table close parentheses be a 2 cross times 2 matrix with real-valued elements a comma space b comma space c and d.  Let lambda subscript 1 and lambda subscript 2 be the eigenvalues of matrix M.

In the case where lambda subscript 1 not equal to lambda subscript 2 ,  show that 

(i)
a plus d equals lambda subscript 1 plus lambda subscript 2

(ii)
det space M equals lambda subscript 1 lambda subscript 2

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3b
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In the case where lambda subscript 1 equals lambda subscript 2,  show that open parentheses a minus d close parentheses squared plus 4 b c equals 0.

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3c
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Hence show that the results of part (a) are also true when matrix M has a single repeated eigenvalue.

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4a
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Let  M equals open parentheses table row a b row c d end table close parentheses  be a  matrix with real-valued elements a comma space b comma space c and d which are such that  a plus c equals 1  and b plus d equals 1.

Show that the eigenvalues of M are 1 and open parentheses a plus d close parentheses minus 1. 

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4b
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In the case where M not equal to I ,  find the eigenvectors of M corresponding to the eigenvalues found in part (a).  Give your answers, where appropriate, in terms of a and d only. 

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4c
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In the case where M equals I ,  describe briefly the eigenvalues and eigenvectors of M.

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5
Sme Calculator8 marks

Consider the matrix M defined as

 M equals open parentheses table row cell 23 over 14 end cell cell 5 over 7 end cell row cell 15 over 14 end cell k end table close parentheses 

where k element of straight real numbers is a constant.  It is given that  negative 1 half is an eigenvalue of M. 

By first finding the value of k, diagonalise M by writing it in the form P D P to the power of negative 1 end exponent for appropriate matrices  P and D.

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6a
Sme Calculator9 marks

Consider the matrix M equals open parentheses table row p 1 row 2 q end table close parentheses,  where p comma space q element of straight real numbers are constants.

It is given that -6 is an eigenvalue of M, and also that open parentheses table row 1 row 2 end table close parentheses is an eigenvector of M which does not correspond to the eigenvalue -6.

By first finding the values of p and q, write M in the form M equals P D P to the power of negative 1 end exponent for appropriate matrices P and D. 

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6b
Sme Calculator4 marks

Hence show that

M to the power of n equals open parentheses negative 1 close parentheses to the power of n over 3 open parentheses table row cell 3 to the power of n plus 2 cross times 6 to the power of n end cell cell 3 to the power of n minus 6 to the power of n end cell row cell 2 open parentheses 3 to the power of n minus 6 to the power of n close parentheses end cell cell 2 cross times 3 to the power of n plus 6 to the power of n end cell end table close parentheses

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7a
Sme Calculator11 marks

Two towns, Avaricia and Covetton, are located on opposite sides of a national park. The two towns are heavily dependent on tourism, and they compete with one another both for the business of tourists coming to the park, and for residents to work in the tourism industry. 

Government officials studying the two towns indicate the population of Avaricia by a, and the population of Covetton by c.  If the respective populations at a particular point in time are a subscript nand c subscript n, then data suggest that the populations one year later may be modelled by the following system of coupled equations:

a subscript n plus 1 end subscript equals 1.025 a subscript n minus 0.075 c subscript n

c subscript n plus 1 end subscript equals negative 0.025 a subscript n plus 0.975 c subscript n

Let a subscript 0 and c subscript 0 indicate the respective populations of the two towns at the start of the study. 

Use a matrix method to show that the respective populations after n years are predicted by the model to be 

a subscript n equals 0.75 open parentheses a subscript 0 minus c subscript 0 close parentheses open parentheses 1.05 to the power of n close parentheses plus open parentheses 0.25 a subscript 0 plus 0.75 c subscript 0 close parentheses open parentheses 0.95 to the power of n close parentheses
c subscript n equals 0.25 open parentheses c subscript 0 minus a subscript 0 close parentheses open parentheses 1.05 to the power of n close parentheses plus open parentheses 0.25 a subscript 0 plus 0.75 c subscript 0 close parentheses open parentheses 0.95 to the power of n close parentheses

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7b
Sme Calculator7 marks

Describe what the model predicts in the long term for the populations of the two towns, for each of the following situations:

i)
a subscript 0 equals c subscript 0
ii)
a subscript 0 greater than c subscript 0
iii)
a subscript 0 less than c subscript 0

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