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A "Simple" Task

How many matrices satisfy this equality?

Find all matrices $A$ of order $2 times 2$ that satisfy the equation $A^2-5A+6I = O$Find matrices $S$ and $Q$ which satisfy $B = S^t AS$ and $A=Q^tQ$Matrices with eigenvalues 0 and 1Find all real matrices $A$ such that $A^2 = mathrmdiag(1,1,2,3,5,8,13)$Find all matrices that satisfy $mathrm B mathrm A = mathrm I_2$How to find basis for subspace of matricesFiguring out nilpotent $2 times 2$ matricesHow to find all the $2 times 2$ matrices that preserve the $|cdot|_infty$ norm?If $A$ and $B$ are $2times 2$ matrices and $AB=0$ then $A=0$ or $B=0$?Determine the transformation matrices in two different ways

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

3

1

$begingroup$

How many matrices $AinmathcalM_3times 3 (mathbbN)$ satisfy this equality?

$$beginpmatrix
1 2 4
endpmatrixcdot A=beginpmatrix
3 2 1
endpmatrix$$

I tried with examples and I found just one but I want to know how to approach this exercise.The right answer is $3$

linear-algebra matrices

share|cite|improve this question

edited 1 hour ago

hunter

16.7k33 gold badges2727 silver badges4444 bronze badges

asked 11 hours ago

DaniVajaDaniVaja

58611 silver badge88 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What are the entries of $A$? Integers, real numbers...?
  $endgroup$
  – B.Swan
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  natural numbers
  $endgroup$
  – DaniVaja
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I forgot about that.I edited my question
  $endgroup$
  – DaniVaja
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Yanko the matrix is multiplied from the left with a row vector, which makes perfect sense.
  $endgroup$
  – daw
  10 hours ago
  

  
  
  
  

add a comment | 

3

1

$begingroup$

How many matrices $AinmathcalM_3times 3 (mathbbN)$ satisfy this equality?

$$beginpmatrix
1 2 4
endpmatrixcdot A=beginpmatrix
3 2 1
endpmatrix$$

I tried with examples and I found just one but I want to know how to approach this exercise.The right answer is $3$

linear-algebra matrices

share|cite|improve this question

edited 1 hour ago

hunter

16.7k33 gold badges2727 silver badges4444 bronze badges

asked 11 hours ago

DaniVajaDaniVaja

58611 silver badge88 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What are the entries of $A$? Integers, real numbers...?
  $endgroup$
  – B.Swan
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  natural numbers
  $endgroup$
  – DaniVaja
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I forgot about that.I edited my question
  $endgroup$
  – DaniVaja
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Yanko the matrix is multiplied from the left with a row vector, which makes perfect sense.
  $endgroup$
  – daw
  10 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

How many matrices $AinmathcalM_3times 3 (mathbbN)$ satisfy this equality?

$$beginpmatrix
1 2 4
endpmatrixcdot A=beginpmatrix
3 2 1
endpmatrix$$

I tried with examples and I found just one but I want to know how to approach this exercise.The right answer is $3$

linear-algebra matrices

share|cite|improve this question

edited 1 hour ago

hunter

16.7k33 gold badges2727 silver badges4444 bronze badges

asked 11 hours ago

DaniVajaDaniVaja

58611 silver badge88 bronze badges

$endgroup$

share|cite|improve this question

edited 1 hour ago

hunter

16.7k33 gold badges2727 silver badges4444 bronze badges

asked 11 hours ago

DaniVajaDaniVaja

58611 silver badge88 bronze badges

share|cite|improve this question

edited 1 hour ago

hunter

16.7k33 gold badges2727 silver badges4444 bronze badges

asked 11 hours ago

DaniVajaDaniVaja

58611 silver badge88 bronze badges

edited 1 hour ago

hunter

16.7k33 gold badges2727 silver badges4444 bronze badges

edited 1 hour ago

hunter

16.7k33 gold badges2727 silver badges4444 bronze badges

asked 11 hours ago

DaniVajaDaniVaja

58611 silver badge88 bronze badges

asked 11 hours ago

DaniVajaDaniVaja

58611 silver badge88 bronze badges

1 Answer
1

active

oldest

votes

11

$begingroup$

Let $A=[x_i,j]$ Then we get

$$x_1,1 +2 x_2,1 +4 x_3,1 = 3$$

$$ x_1,2 + 2x_2,2 + 4 x_3,2 = 2 $$

$$x_1,3 + 2 x_2,3 + 4 x_3,3 = 1$$

We shall count the number of solutions for each equation.

First equation has 2 solutions:
Since $4>3$ we have that $x_3,1=0$. Similarly $x_2,1<2$. If $x_2,1=1$ then $x_1,1=1$ and if $x_2,1=0$ then $x_1,1=3$. Therefore there are two solutions for the first equation.

Second equation has 2 solutions:
As before, $x_3,2=0$. Also $x_2,2$ is at most $1$, hence if $x_2,2=1$ then $x_1,2=0$ and if not then $x_2,2=0$ and $x_1,2=1$.

Third equation has 1 solution:
$4,2>1$ so $x_2,3,x_3,3=0$ and so the only solution is $x_1,3=1$.

Therefore there is a total of $4$ solutions for all of the equations.

$$A_1 =beginbmatrix 3 & 0 &1 \ 0 & 1 & 0 \ 0 & 0 & 0endbmatrix, A_2=beginbmatrix 1 & 0 &1 \ 1 & 1 & 0 \ 0 & 0 & 0endbmatrix$$
$$A_3 =beginbmatrix 3 & 2 &1 \ 0 & 0 & 0 \ 0 & 0 & 0endbmatrix, A_4=beginbmatrix 1 & 2 &1 \ 1 & 0 & 0 \ 0 & 0 & 0endbmatrix$$

share|cite|improve this answer

edited 10 hours ago

answered 10 hours ago

YankoYanko

9,41122 gold badges1010 silver badges3131 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The OP forgot to mention that their definition of $mathbbN$ includes $0$ because otherwise there are zero solutions.
  $endgroup$
  – Peter Foreman
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Thank for your answer.If $x_2,1=1$ then $x_1,1=1$, right? Because otherwise the result will be 4
  $endgroup$
  – DaniVaja
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DaniVaja Correct. Thank you
  $endgroup$
  – Yanko
  10 hours ago
  

  
  
  
  

add a comment | 

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11

$begingroup$

Let $A=[x_i,j]$ Then we get

$$x_1,1 +2 x_2,1 +4 x_3,1 = 3$$

$$ x_1,2 + 2x_2,2 + 4 x_3,2 = 2 $$

$$x_1,3 + 2 x_2,3 + 4 x_3,3 = 1$$

We shall count the number of solutions for each equation.

First equation has 2 solutions:
Since $4>3$ we have that $x_3,1=0$. Similarly $x_2,1<2$. If $x_2,1=1$ then $x_1,1=1$ and if $x_2,1=0$ then $x_1,1=3$. Therefore there are two solutions for the first equation.

Second equation has 2 solutions:
As before, $x_3,2=0$. Also $x_2,2$ is at most $1$, hence if $x_2,2=1$ then $x_1,2=0$ and if not then $x_2,2=0$ and $x_1,2=1$.

Third equation has 1 solution:
$4,2>1$ so $x_2,3,x_3,3=0$ and so the only solution is $x_1,3=1$.

Therefore there is a total of $4$ solutions for all of the equations.

$$A_1 =beginbmatrix 3 & 0 &1 \ 0 & 1 & 0 \ 0 & 0 & 0endbmatrix, A_2=beginbmatrix 1 & 0 &1 \ 1 & 1 & 0 \ 0 & 0 & 0endbmatrix$$
$$A_3 =beginbmatrix 3 & 2 &1 \ 0 & 0 & 0 \ 0 & 0 & 0endbmatrix, A_4=beginbmatrix 1 & 2 &1 \ 1 & 0 & 0 \ 0 & 0 & 0endbmatrix$$

share|cite|improve this answer

edited 10 hours ago

answered 10 hours ago

YankoYanko

9,41122 gold badges1010 silver badges3131 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The OP forgot to mention that their definition of $mathbbN$ includes $0$ because otherwise there are zero solutions.
  $endgroup$
  – Peter Foreman
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Thank for your answer.If $x_2,1=1$ then $x_1,1=1$, right? Because otherwise the result will be 4
  $endgroup$
  – DaniVaja
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DaniVaja Correct. Thank you
  $endgroup$
  – Yanko
  10 hours ago
  

  
  
  
  

add a comment | 

11

$begingroup$

Let $A=[x_i,j]$ Then we get

$$x_1,1 +2 x_2,1 +4 x_3,1 = 3$$

$$ x_1,2 + 2x_2,2 + 4 x_3,2 = 2 $$

$$x_1,3 + 2 x_2,3 + 4 x_3,3 = 1$$

We shall count the number of solutions for each equation.

First equation has 2 solutions:
Since $4>3$ we have that $x_3,1=0$. Similarly $x_2,1<2$. If $x_2,1=1$ then $x_1,1=1$ and if $x_2,1=0$ then $x_1,1=3$. Therefore there are two solutions for the first equation.

Second equation has 2 solutions:
As before, $x_3,2=0$. Also $x_2,2$ is at most $1$, hence if $x_2,2=1$ then $x_1,2=0$ and if not then $x_2,2=0$ and $x_1,2=1$.

Third equation has 1 solution:
$4,2>1$ so $x_2,3,x_3,3=0$ and so the only solution is $x_1,3=1$.

Therefore there is a total of $4$ solutions for all of the equations.

$$A_1 =beginbmatrix 3 & 0 &1 \ 0 & 1 & 0 \ 0 & 0 & 0endbmatrix, A_2=beginbmatrix 1 & 0 &1 \ 1 & 1 & 0 \ 0 & 0 & 0endbmatrix$$
$$A_3 =beginbmatrix 3 & 2 &1 \ 0 & 0 & 0 \ 0 & 0 & 0endbmatrix, A_4=beginbmatrix 1 & 2 &1 \ 1 & 0 & 0 \ 0 & 0 & 0endbmatrix$$

share|cite|improve this answer

edited 10 hours ago

answered 10 hours ago

YankoYanko

9,41122 gold badges1010 silver badges3131 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The OP forgot to mention that their definition of $mathbbN$ includes $0$ because otherwise there are zero solutions.
  $endgroup$
  – Peter Foreman
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Thank for your answer.If $x_2,1=1$ then $x_1,1=1$, right? Because otherwise the result will be 4
  $endgroup$
  – DaniVaja
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DaniVaja Correct. Thank you
  $endgroup$
  – Yanko
  10 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Let $A=[x_i,j]$ Then we get

$$x_1,1 +2 x_2,1 +4 x_3,1 = 3$$

$$ x_1,2 + 2x_2,2 + 4 x_3,2 = 2 $$

$$x_1,3 + 2 x_2,3 + 4 x_3,3 = 1$$

We shall count the number of solutions for each equation.

First equation has 2 solutions:
Since $4>3$ we have that $x_3,1=0$. Similarly $x_2,1<2$. If $x_2,1=1$ then $x_1,1=1$ and if $x_2,1=0$ then $x_1,1=3$. Therefore there are two solutions for the first equation.

Second equation has 2 solutions:
As before, $x_3,2=0$. Also $x_2,2$ is at most $1$, hence if $x_2,2=1$ then $x_1,2=0$ and if not then $x_2,2=0$ and $x_1,2=1$.

Third equation has 1 solution:
$4,2>1$ so $x_2,3,x_3,3=0$ and so the only solution is $x_1,3=1$.

Therefore there is a total of $4$ solutions for all of the equations.

$$A_1 =beginbmatrix 3 & 0 &1 \ 0 & 1 & 0 \ 0 & 0 & 0endbmatrix, A_2=beginbmatrix 1 & 0 &1 \ 1 & 1 & 0 \ 0 & 0 & 0endbmatrix$$
$$A_3 =beginbmatrix 3 & 2 &1 \ 0 & 0 & 0 \ 0 & 0 & 0endbmatrix, A_4=beginbmatrix 1 & 2 &1 \ 1 & 0 & 0 \ 0 & 0 & 0endbmatrix$$

share|cite|improve this answer

edited 10 hours ago

answered 10 hours ago

YankoYanko

9,41122 gold badges1010 silver badges3131 bronze badges

$endgroup$

share|cite|improve this answer

edited 10 hours ago

answered 10 hours ago

YankoYanko

9,41122 gold badges1010 silver badges3131 bronze badges

answered 10 hours ago

YankoYanko

9,41122 gold badges1010 silver badges3131 bronze badges

answered 10 hours ago

YankoYanko

9,41122 gold badges1010 silver badges3131 bronze badges