Find the values of $a$ and $b$ such that the following matrices are similar

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Find the values of $a$ and $b$ such that the following matrices are similar.

$$A=beginbmatrix -2& 0 & 0 \ 2 & a & 2 \ 3& 1 & 1 endbmatrix, qquad B=beginbmatrix -1& 0 & 0 \ 0 & 2 & 0 \ 0& 0 & b endbmatrix$$

I know that similar matrices must have the same ranks. Hence, I can take that value of $a$ as any real number and $b neq 0$. Is my answer is true or false? Any hints or solutions will be appreciated. Thank you.

linear-algebra matrices

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edited Jul 22 at 12:47

Rodrigo de Azevedo

12.6k41751

asked Jul 22 at 12:22

Messi fifa

1718

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $-2$ is an eigenvalue of $A$.
  – Lord Shark the Unknown
  Jul 22 at 12:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  ya,,,, -2 is the eigenvalue of A but what we can conclude From here ??
  – Messi fifa
  Jul 22 at 12:27
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $A$ and $B$ are meant to be similar....
  – Lord Shark the Unknown
  Jul 22 at 12:28
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

Find the values of $a$ and $b$ such that the following matrices are similar.

$$A=beginbmatrix -2& 0 & 0 \ 2 & a & 2 \ 3& 1 & 1 endbmatrix, qquad B=beginbmatrix -1& 0 & 0 \ 0 & 2 & 0 \ 0& 0 & b endbmatrix$$

I know that similar matrices must have the same ranks. Hence, I can take that value of $a$ as any real number and $b neq 0$. Is my answer is true or false? Any hints or solutions will be appreciated. Thank you.

linear-algebra matrices

share|cite|improve this question

edited Jul 22 at 12:47

Rodrigo de Azevedo

12.6k41751

asked Jul 22 at 12:22

Messi fifa

1718

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $-2$ is an eigenvalue of $A$.
  – Lord Shark the Unknown
  Jul 22 at 12:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  ya,,,, -2 is the eigenvalue of A but what we can conclude From here ??
  – Messi fifa
  Jul 22 at 12:27
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $A$ and $B$ are meant to be similar....
  – Lord Shark the Unknown
  Jul 22 at 12:28
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

Find the values of $a$ and $b$ such that the following matrices are similar.

$$A=beginbmatrix -2& 0 & 0 \ 2 & a & 2 \ 3& 1 & 1 endbmatrix, qquad B=beginbmatrix -1& 0 & 0 \ 0 & 2 & 0 \ 0& 0 & b endbmatrix$$

I know that similar matrices must have the same ranks. Hence, I can take that value of $a$ as any real number and $b neq 0$. Is my answer is true or false? Any hints or solutions will be appreciated. Thank you.

linear-algebra matrices

share|cite|improve this question

edited Jul 22 at 12:47

Rodrigo de Azevedo

12.6k41751

asked Jul 22 at 12:22

Messi fifa

1718

Find the values of $a$ and $b$ such that the following matrices are similar.

$$A=beginbmatrix -2& 0 & 0 \ 2 & a & 2 \ 3& 1 & 1 endbmatrix, qquad B=beginbmatrix -1& 0 & 0 \ 0 & 2 & 0 \ 0& 0 & b endbmatrix$$

I know that similar matrices must have the same ranks. Hence, I can take that value of $a$ as any real number and $b neq 0$. Is my answer is true or false? Any hints or solutions will be appreciated. Thank you.

linear-algebra matrices

share|cite|improve this question

edited Jul 22 at 12:47

Rodrigo de Azevedo

12.6k41751

asked Jul 22 at 12:22

Messi fifa

1718

share|cite|improve this question

share|cite|improve this question

edited Jul 22 at 12:47

Rodrigo de Azevedo

12.6k41751

edited Jul 22 at 12:47

Rodrigo de Azevedo

12.6k41751

edited Jul 22 at 12:47

Rodrigo de Azevedo

12.6k41751

12.6k41751

asked Jul 22 at 12:22

Messi fifa

1718

asked Jul 22 at 12:22

Messi fifa

1718

asked Jul 22 at 12:22

Messi fifa

1718

1718

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $-2$ is an eigenvalue of $A$.
  – Lord Shark the Unknown
  Jul 22 at 12:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  ya,,,, -2 is the eigenvalue of A but what we can conclude From here ??
  – Messi fifa
  Jul 22 at 12:27
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $A$ and $B$ are meant to be similar....
  – Lord Shark the Unknown
  Jul 22 at 12:28
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $-2$ is an eigenvalue of $A$.
  – Lord Shark the Unknown
  Jul 22 at 12:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  ya,,,, -2 is the eigenvalue of A but what we can conclude From here ??
  – Messi fifa
  Jul 22 at 12:27
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $A$ and $B$ are meant to be similar....
  – Lord Shark the Unknown
  Jul 22 at 12:28
  

  
  
  
  

2

2

$-2$ is an eigenvalue of $A$.
– Lord Shark the Unknown
Jul 22 at 12:25

$-2$ is an eigenvalue of $A$.
– Lord Shark the Unknown
Jul 22 at 12:25

ya,,,, -2 is the eigenvalue of A but what we can conclude From here ??
– Messi fifa
Jul 22 at 12:27

ya,,,, -2 is the eigenvalue of A but what we can conclude From here ??
– Messi fifa
Jul 22 at 12:27

1

1

$A$ and $B$ are meant to be similar....
– Lord Shark the Unknown
Jul 22 at 12:28

$A$ and $B$ are meant to be similar....
– Lord Shark the Unknown
Jul 22 at 12:28

add a comment | 

3 Answers
3

active

oldest

votes

up vote
4
down vote



accepted

If two matrices are similar, then they have the same characteristic polynomials. The characteristic polynomials of your matrices are$$-x^2+(a-1)x^2+(a+4)x+4-2atext and -x^3+(b+1)x^2+(-b+2)x-2b$$respectively. They are equal if and only if $a=0$ and $b=-2$. On the other hand, in this case the matrices have three distinct eigenvalues (the same for both). Therefore, they're similar.

share|cite|improve this answer

edited Jul 22 at 13:11

answered Jul 22 at 12:34

José Carlos Santos

113k1698177

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $b = a-2$ is not sufficient. They are similar if and only if $a = 0$ and $b = -2$.
  – mechanodroid
  Jul 22 at 12:45
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  How can you say (or imply) that for $b=-1$ or for $b=2$ the matrix $B$ has $3$ distinct eigenvalues?
  – Marc van Leeuwen
  Jul 22 at 12:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MarcvanLeeuwen I made a mistake while determining when the characteristic polynomials are equal. It is corrected now.
  – José Carlos Santos
  Jul 22 at 13:12
  

  
  
  
  

add a comment | 

up vote
2
down vote

Calculate the characteristic polynomials:

$$det (A - lambda I) = (-2-lambda) beginvmatrixa-2 & 2 \ 1 & 1-lambdaendvmatrix = (-2-lambda)(lambda^2 - lambda(a+1) + (a-2)) = (-2 - lambda)left(frac1+a-sqrta^2-2a+92 - lambdaright)left(frac1+a+sqrta^2-2a+92 - lambdaright)$$

$$det (B - lambda I) = (-1-lambda)(2 - lambda)(b - lambda)$$

If $A$ and $B$ are similar, then the characteristic polynomials are equal so uniqueness of factorization in $mathbbC[lambda]$ gives $b = -2$ and

$$leftfrac1+a-sqrta^2-2a+92, frac1+a+sqrta^2-2a+92right = -1, 2$$

for which the only solution is $a = 0$.

On the other hand, if $a = 0, b= 2$ then both $A$ and $B$ diagonalize with distinct eigenvalues $-1, 2, -2$ so they are similar.

share|cite|improve this answer

answered Jul 22 at 12:42

mechanodroid

22.2k52041

add a comment | 

up vote
2
down vote

Clearly (by block upper triangular form) $A$ has an eigenvalue $-2$ regardless of $a$, and $B$ only has such an eigenvalue if $b=-2$; that is therefore a necessary condition for similarity. But then the matrix $B$ has $3$ distinct eigenvalues, and will be similar to another matrix if and only if that matrix has the same characteristic polynomial as$~B$. For that to happen for$~A$, the bottom-right $2times2$ block of$~A$ must have characteristic polynomial equal to $(X+1)(X-2)=X^2-X-2$ which means it should have trace$~1$ and determinant$~-2$, and this happens if and only if $a=0$. So the matrices are similar if and only if $a=0$ and $b=-2$.

share|cite|improve this answer

answered Jul 22 at 12:54

Marc van Leeuwen

84.8k499212

add a comment | 

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
4
down vote



accepted

If two matrices are similar, then they have the same characteristic polynomials. The characteristic polynomials of your matrices are$$-x^2+(a-1)x^2+(a+4)x+4-2atext and -x^3+(b+1)x^2+(-b+2)x-2b$$respectively. They are equal if and only if $a=0$ and $b=-2$. On the other hand, in this case the matrices have three distinct eigenvalues (the same for both). Therefore, they're similar.

share|cite|improve this answer

edited Jul 22 at 13:11

answered Jul 22 at 12:34

José Carlos Santos

113k1698177

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $b = a-2$ is not sufficient. They are similar if and only if $a = 0$ and $b = -2$.
  – mechanodroid
  Jul 22 at 12:45
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  How can you say (or imply) that for $b=-1$ or for $b=2$ the matrix $B$ has $3$ distinct eigenvalues?
  – Marc van Leeuwen
  Jul 22 at 12:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MarcvanLeeuwen I made a mistake while determining when the characteristic polynomials are equal. It is corrected now.
  – José Carlos Santos
  Jul 22 at 13:12
  

  
  
  
  

add a comment | 

up vote
4
down vote



accepted

If two matrices are similar, then they have the same characteristic polynomials. The characteristic polynomials of your matrices are$$-x^2+(a-1)x^2+(a+4)x+4-2atext and -x^3+(b+1)x^2+(-b+2)x-2b$$respectively. They are equal if and only if $a=0$ and $b=-2$. On the other hand, in this case the matrices have three distinct eigenvalues (the same for both). Therefore, they're similar.

share|cite|improve this answer

edited Jul 22 at 13:11

answered Jul 22 at 12:34

José Carlos Santos

113k1698177

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $b = a-2$ is not sufficient. They are similar if and only if $a = 0$ and $b = -2$.
  – mechanodroid
  Jul 22 at 12:45
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  How can you say (or imply) that for $b=-1$ or for $b=2$ the matrix $B$ has $3$ distinct eigenvalues?
  – Marc van Leeuwen
  Jul 22 at 12:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MarcvanLeeuwen I made a mistake while determining when the characteristic polynomials are equal. It is corrected now.
  – José Carlos Santos
  Jul 22 at 13:12
  

  
  
  
  

add a comment | 

up vote
4
down vote



accepted

up vote
4
down vote



accepted

If two matrices are similar, then they have the same characteristic polynomials. The characteristic polynomials of your matrices are$$-x^2+(a-1)x^2+(a+4)x+4-2atext and -x^3+(b+1)x^2+(-b+2)x-2b$$respectively. They are equal if and only if $a=0$ and $b=-2$. On the other hand, in this case the matrices have three distinct eigenvalues (the same for both). Therefore, they're similar.

share|cite|improve this answer

edited Jul 22 at 13:11

answered Jul 22 at 12:34

José Carlos Santos

113k1698177

If two matrices are similar, then they have the same characteristic polynomials. The characteristic polynomials of your matrices are$$-x^2+(a-1)x^2+(a+4)x+4-2atext and -x^3+(b+1)x^2+(-b+2)x-2b$$respectively. They are equal if and only if $a=0$ and $b=-2$. On the other hand, in this case the matrices have three distinct eigenvalues (the same for both). Therefore, they're similar.

share|cite|improve this answer

edited Jul 22 at 13:11

answered Jul 22 at 12:34

José Carlos Santos

113k1698177

share|cite|improve this answer

share|cite|improve this answer

edited Jul 22 at 13:11

edited Jul 22 at 13:11

edited Jul 22 at 13:11

answered Jul 22 at 12:34

José Carlos Santos

113k1698177

answered Jul 22 at 12:34

José Carlos Santos

113k1698177

answered Jul 22 at 12:34

José Carlos Santos

113k1698177

113k1698177

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $b = a-2$ is not sufficient. They are similar if and only if $a = 0$ and $b = -2$.
  – mechanodroid
  Jul 22 at 12:45
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  How can you say (or imply) that for $b=-1$ or for $b=2$ the matrix $B$ has $3$ distinct eigenvalues?
  – Marc van Leeuwen
  Jul 22 at 12:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MarcvanLeeuwen I made a mistake while determining when the characteristic polynomials are equal. It is corrected now.
  – José Carlos Santos
  Jul 22 at 13:12
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $b = a-2$ is not sufficient. They are similar if and only if $a = 0$ and $b = -2$.
  – mechanodroid
  Jul 22 at 12:45
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  How can you say (or imply) that for $b=-1$ or for $b=2$ the matrix $B$ has $3$ distinct eigenvalues?
  – Marc van Leeuwen
  Jul 22 at 12:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MarcvanLeeuwen I made a mistake while determining when the characteristic polynomials are equal. It is corrected now.
  – José Carlos Santos
  Jul 22 at 13:12
  

  
  
  
  

$b = a-2$ is not sufficient. They are similar if and only if $a = 0$ and $b = -2$.
– mechanodroid
Jul 22 at 12:45

$b = a-2$ is not sufficient. They are similar if and only if $a = 0$ and $b = -2$.
– mechanodroid
Jul 22 at 12:45

1

1

How can you say (or imply) that for $b=-1$ or for $b=2$ the matrix $B$ has $3$ distinct eigenvalues?
– Marc van Leeuwen
Jul 22 at 12:45

How can you say (or imply) that for $b=-1$ or for $b=2$ the matrix $B$ has $3$ distinct eigenvalues?
– Marc van Leeuwen
Jul 22 at 12:45

@MarcvanLeeuwen I made a mistake while determining when the characteristic polynomials are equal. It is corrected now.
– José Carlos Santos
Jul 22 at 13:12

@MarcvanLeeuwen I made a mistake while determining when the characteristic polynomials are equal. It is corrected now.
– José Carlos Santos
Jul 22 at 13:12

add a comment | 

up vote
2
down vote

Calculate the characteristic polynomials:

$$det (A - lambda I) = (-2-lambda) beginvmatrixa-2 & 2 \ 1 & 1-lambdaendvmatrix = (-2-lambda)(lambda^2 - lambda(a+1) + (a-2)) = (-2 - lambda)left(frac1+a-sqrta^2-2a+92 - lambdaright)left(frac1+a+sqrta^2-2a+92 - lambdaright)$$

$$det (B - lambda I) = (-1-lambda)(2 - lambda)(b - lambda)$$

If $A$ and $B$ are similar, then the characteristic polynomials are equal so uniqueness of factorization in $mathbbC[lambda]$ gives $b = -2$ and

$$leftfrac1+a-sqrta^2-2a+92, frac1+a+sqrta^2-2a+92right = -1, 2$$

for which the only solution is $a = 0$.

On the other hand, if $a = 0, b= 2$ then both $A$ and $B$ diagonalize with distinct eigenvalues $-1, 2, -2$ so they are similar.

share|cite|improve this answer

answered Jul 22 at 12:42

mechanodroid

22.2k52041

add a comment | 

up vote
2
down vote

Calculate the characteristic polynomials:

$$det (A - lambda I) = (-2-lambda) beginvmatrixa-2 & 2 \ 1 & 1-lambdaendvmatrix = (-2-lambda)(lambda^2 - lambda(a+1) + (a-2)) = (-2 - lambda)left(frac1+a-sqrta^2-2a+92 - lambdaright)left(frac1+a+sqrta^2-2a+92 - lambdaright)$$

$$det (B - lambda I) = (-1-lambda)(2 - lambda)(b - lambda)$$

If $A$ and $B$ are similar, then the characteristic polynomials are equal so uniqueness of factorization in $mathbbC[lambda]$ gives $b = -2$ and

$$leftfrac1+a-sqrta^2-2a+92, frac1+a+sqrta^2-2a+92right = -1, 2$$

for which the only solution is $a = 0$.

On the other hand, if $a = 0, b= 2$ then both $A$ and $B$ diagonalize with distinct eigenvalues $-1, 2, -2$ so they are similar.

share|cite|improve this answer

answered Jul 22 at 12:42

mechanodroid

22.2k52041

add a comment | 

up vote
2
down vote

up vote
2
down vote

Calculate the characteristic polynomials:

$$det (A - lambda I) = (-2-lambda) beginvmatrixa-2 & 2 \ 1 & 1-lambdaendvmatrix = (-2-lambda)(lambda^2 - lambda(a+1) + (a-2)) = (-2 - lambda)left(frac1+a-sqrta^2-2a+92 - lambdaright)left(frac1+a+sqrta^2-2a+92 - lambdaright)$$

$$det (B - lambda I) = (-1-lambda)(2 - lambda)(b - lambda)$$

If $A$ and $B$ are similar, then the characteristic polynomials are equal so uniqueness of factorization in $mathbbC[lambda]$ gives $b = -2$ and

$$leftfrac1+a-sqrta^2-2a+92, frac1+a+sqrta^2-2a+92right = -1, 2$$

for which the only solution is $a = 0$.

On the other hand, if $a = 0, b= 2$ then both $A$ and $B$ diagonalize with distinct eigenvalues $-1, 2, -2$ so they are similar.

share|cite|improve this answer

answered Jul 22 at 12:42

mechanodroid

22.2k52041

Calculate the characteristic polynomials:

$$det (A - lambda I) = (-2-lambda) beginvmatrixa-2 & 2 \ 1 & 1-lambdaendvmatrix = (-2-lambda)(lambda^2 - lambda(a+1) + (a-2)) = (-2 - lambda)left(frac1+a-sqrta^2-2a+92 - lambdaright)left(frac1+a+sqrta^2-2a+92 - lambdaright)$$

$$det (B - lambda I) = (-1-lambda)(2 - lambda)(b - lambda)$$

If $A$ and $B$ are similar, then the characteristic polynomials are equal so uniqueness of factorization in $mathbbC[lambda]$ gives $b = -2$ and

$$leftfrac1+a-sqrta^2-2a+92, frac1+a+sqrta^2-2a+92right = -1, 2$$

for which the only solution is $a = 0$.

On the other hand, if $a = 0, b= 2$ then both $A$ and $B$ diagonalize with distinct eigenvalues $-1, 2, -2$ so they are similar.

share|cite|improve this answer

answered Jul 22 at 12:42

mechanodroid

22.2k52041

share|cite|improve this answer

share|cite|improve this answer

answered Jul 22 at 12:42

mechanodroid

22.2k52041

answered Jul 22 at 12:42

mechanodroid

22.2k52041

answered Jul 22 at 12:42

mechanodroid

22.2k52041

22.2k52041

add a comment | 

add a comment | 

up vote
2
down vote

Clearly (by block upper triangular form) $A$ has an eigenvalue $-2$ regardless of $a$, and $B$ only has such an eigenvalue if $b=-2$; that is therefore a necessary condition for similarity. But then the matrix $B$ has $3$ distinct eigenvalues, and will be similar to another matrix if and only if that matrix has the same characteristic polynomial as$~B$. For that to happen for$~A$, the bottom-right $2times2$ block of$~A$ must have characteristic polynomial equal to $(X+1)(X-2)=X^2-X-2$ which means it should have trace$~1$ and determinant$~-2$, and this happens if and only if $a=0$. So the matrices are similar if and only if $a=0$ and $b=-2$.

share|cite|improve this answer

answered Jul 22 at 12:54

Marc van Leeuwen

84.8k499212

add a comment | 

up vote
2
down vote

Clearly (by block upper triangular form) $A$ has an eigenvalue $-2$ regardless of $a$, and $B$ only has such an eigenvalue if $b=-2$; that is therefore a necessary condition for similarity. But then the matrix $B$ has $3$ distinct eigenvalues, and will be similar to another matrix if and only if that matrix has the same characteristic polynomial as$~B$. For that to happen for$~A$, the bottom-right $2times2$ block of$~A$ must have characteristic polynomial equal to $(X+1)(X-2)=X^2-X-2$ which means it should have trace$~1$ and determinant$~-2$, and this happens if and only if $a=0$. So the matrices are similar if and only if $a=0$ and $b=-2$.

share|cite|improve this answer

answered Jul 22 at 12:54

Marc van Leeuwen

84.8k499212

add a comment | 

up vote
2
down vote

up vote
2
down vote

Clearly (by block upper triangular form) $A$ has an eigenvalue $-2$ regardless of $a$, and $B$ only has such an eigenvalue if $b=-2$; that is therefore a necessary condition for similarity. But then the matrix $B$ has $3$ distinct eigenvalues, and will be similar to another matrix if and only if that matrix has the same characteristic polynomial as$~B$. For that to happen for$~A$, the bottom-right $2times2$ block of$~A$ must have characteristic polynomial equal to $(X+1)(X-2)=X^2-X-2$ which means it should have trace$~1$ and determinant$~-2$, and this happens if and only if $a=0$. So the matrices are similar if and only if $a=0$ and $b=-2$.

share|cite|improve this answer

answered Jul 22 at 12:54

Marc van Leeuwen

84.8k499212

Clearly (by block upper triangular form) $A$ has an eigenvalue $-2$ regardless of $a$, and $B$ only has such an eigenvalue if $b=-2$; that is therefore a necessary condition for similarity. But then the matrix $B$ has $3$ distinct eigenvalues, and will be similar to another matrix if and only if that matrix has the same characteristic polynomial as$~B$. For that to happen for$~A$, the bottom-right $2times2$ block of$~A$ must have characteristic polynomial equal to $(X+1)(X-2)=X^2-X-2$ which means it should have trace$~1$ and determinant$~-2$, and this happens if and only if $a=0$. So the matrices are similar if and only if $a=0$ and $b=-2$.

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answered Jul 22 at 12:54

Marc van Leeuwen

84.8k499212

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share|cite|improve this answer

answered Jul 22 at 12:54

Marc van Leeuwen

84.8k499212

answered Jul 22 at 12:54

Marc van Leeuwen

84.8k499212

answered Jul 22 at 12:54

Marc van Leeuwen

84.8k499212

84.8k499212

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