Finding unknown values in matrix

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First of all i hope I'm at the right place to ask for help with this problem!

Let's assume a simple 2x2 grid in where all x and y values are not known. The total values are known though.

x1 | x2 | xTotal
y1 | y2 | yTotal
____________________
xy1Total | xy2Total

xTotal = x1 + x2

yTotal = y1 + y2

xy1Total = x1 + y1

xy2Total = x2 + y2


Example numbers for the known values:

xy1Total = 13

xy2Total = 17

xTotal = 15

yTotal = 15


Is there any way of "re-calculating" the unknown values? I stumbled upon the Gaussian elimination. I'm still not sure if this method can be applied here.
Thanks in advance.

matrices

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asked Jul 30 at 14:41

Lunatiic

31

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yeah, the elimination process totally works, 'cause this is a linear equation system involving 4 variables and 4 equations. But I don't mean to eliminate the 2 by 2 grid
  – xbh
  Jul 30 at 14:45
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

First of all i hope I'm at the right place to ask for help with this problem!

Let's assume a simple 2x2 grid in where all x and y values are not known. The total values are known though.

x1 | x2 | xTotal
y1 | y2 | yTotal
____________________
xy1Total | xy2Total

xTotal = x1 + x2

yTotal = y1 + y2

xy1Total = x1 + y1

xy2Total = x2 + y2


Example numbers for the known values:

xy1Total = 13

xy2Total = 17

xTotal = 15

yTotal = 15


Is there any way of "re-calculating" the unknown values? I stumbled upon the Gaussian elimination. I'm still not sure if this method can be applied here.
Thanks in advance.

matrices

share|cite|improve this question

asked Jul 30 at 14:41

Lunatiic

31

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yeah, the elimination process totally works, 'cause this is a linear equation system involving 4 variables and 4 equations. But I don't mean to eliminate the 2 by 2 grid
  – xbh
  Jul 30 at 14:45
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

First of all i hope I'm at the right place to ask for help with this problem!

Let's assume a simple 2x2 grid in where all x and y values are not known. The total values are known though.

x1 | x2 | xTotal
y1 | y2 | yTotal
____________________
xy1Total | xy2Total

xTotal = x1 + x2

yTotal = y1 + y2

xy1Total = x1 + y1

xy2Total = x2 + y2


Example numbers for the known values:

xy1Total = 13

xy2Total = 17

xTotal = 15

yTotal = 15


Is there any way of "re-calculating" the unknown values? I stumbled upon the Gaussian elimination. I'm still not sure if this method can be applied here.
Thanks in advance.

matrices

share|cite|improve this question

asked Jul 30 at 14:41

Lunatiic

31

First of all i hope I'm at the right place to ask for help with this problem!

Let's assume a simple 2x2 grid in where all x and y values are not known. The total values are known though.

x1 | x2 | xTotal
y1 | y2 | yTotal
____________________
xy1Total | xy2Total

xTotal = x1 + x2

yTotal = y1 + y2

xy1Total = x1 + y1

xy2Total = x2 + y2


Example numbers for the known values:

xy1Total = 13

xy2Total = 17

xTotal = 15

yTotal = 15


Is there any way of "re-calculating" the unknown values? I stumbled upon the Gaussian elimination. I'm still not sure if this method can be applied here.
Thanks in advance.

matrices

share|cite|improve this question

asked Jul 30 at 14:41

Lunatiic

31

share|cite|improve this question

share|cite|improve this question

asked Jul 30 at 14:41

Lunatiic

31

asked Jul 30 at 14:41

Lunatiic

31

asked Jul 30 at 14:41

Lunatiic

31

31

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yeah, the elimination process totally works, 'cause this is a linear equation system involving 4 variables and 4 equations. But I don't mean to eliminate the 2 by 2 grid
  – xbh
  Jul 30 at 14:45
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yeah, the elimination process totally works, 'cause this is a linear equation system involving 4 variables and 4 equations. But I don't mean to eliminate the 2 by 2 grid
  – xbh
  Jul 30 at 14:45
  
  

  
  
  
  

Yeah, the elimination process totally works, 'cause this is a linear equation system involving 4 variables and 4 equations. But I don't mean to eliminate the 2 by 2 grid
– xbh
Jul 30 at 14:45

Yeah, the elimination process totally works, 'cause this is a linear equation system involving 4 variables and 4 equations. But I don't mean to eliminate the 2 by 2 grid
– xbh
Jul 30 at 14:45

add a comment | 

3 Answers
3

active

oldest

votes

up vote
1
down vote



accepted

It suffices to consider the system

• $x_1 + x_2 =x_Total=13$


• $ y_1 + y_2 = y_Total=17$


• $x_1 + y_1 =xy1_Total=15$


• $ x_2 + y_2 = xy2_Total=15$

which is in the augmented matrix form

$$left[beginarrayc
1& 1& 0& 0& 13\
0& 0& 1& 1& 17\
1& 0& 1& 0& 15\
0& 1& 0& 1& 15\
endarrayright]$$

which can be solved by Gaussian elimination that is

$$left[beginarrayc
1& 1& 0& 0& 13\
1& 0& 1& 0& 15\
0& 1& 0& 1& 15\
0& 0& 1& 1& 17\
endarrayright]to left[beginarrayc
1& 1& 0& 0& 13\
0& -1& 1& 0& 2\
0& 1& 0& 1& 15\
0& 0& 1& 1& 17\
endarrayright]to left[beginarrayc
1& 1& 0& 0& 13\
0& -1& 1& 0& 2\
0& 0& 1& 1& 17\
0& 0& 1& 1& 17\
endarrayright]to left[beginarrayc
1& 1& 0& 0& 13\
0& -1& 1& 0& 2\
0& 0& 1& 1& 17\
0& 0& 0& 0& 0\
endarrayright]$$

therefore setting $y_2=k$ as free we obtain

• $y_1=17-k$


• $x_2=y_1-2=15-k$


• $x_1=13-x_1=k-2$

share|cite|improve this answer

edited Jul 30 at 14:53

answered Jul 30 at 14:45

gimusi

64.3k73480

add a comment | 

up vote
0
down vote

That is a simple linear system of equations, you could solve it in lot's of ways, one of them being Gaussian elimination. In your example you have $$begincasesx_1+x_2=15\ y_1+y_2=15\ x_1+y_1=13\ x_2+y_2=17endcases$$ four equations in four unknowns, to know from the start if the system has infinite, no or one solution you could use Rouchè-Capelli theorem. Gaussian elimination is a pretty involved process to find solutions to linear system of equations (it cannot be applied on non linear), in cases like this one you could simply use substitution, addition, subtraction, multiplication between rows etc.

If you really want to use gaussian elimination the system of equations can be put in matrix form like this $$left(beginmatrix1&1&0&0&15\0&0&1&1&15\1&0&1&0&13\0&1&0&1&17endmatrixright)$$ Your system seems like having infinite solutions, in this case you just choose one variable to explicit the others as a function of that

share|cite|improve this answer

answered Jul 30 at 14:50

Davide Morgante

1,693220

add a comment | 

up vote
0
down vote

The sums of the rows have a grand total, equal to the sum of all the $x_i$ and $y_i$. That must equal the grand total of all the column sums. Here, 13+17=15+15 so you are okay. If not, there will be no solutions.

Suppose the grand totals match, and you have a matrix with $M$ rows and $N$ columns. You can put in any numbers you like except for the last row and last column. Then work out numbers in the last column to make the row sums right, and numbers in the last row to make column sums right.

share|cite|improve this answer

answered Jul 30 at 15:09

Empy2

31.7k12059

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
1
down vote



accepted

It suffices to consider the system

• $x_1 + x_2 =x_Total=13$


• $ y_1 + y_2 = y_Total=17$


• $x_1 + y_1 =xy1_Total=15$


• $ x_2 + y_2 = xy2_Total=15$

which is in the augmented matrix form

$$left[beginarrayc
1& 1& 0& 0& 13\
0& 0& 1& 1& 17\
1& 0& 1& 0& 15\
0& 1& 0& 1& 15\
endarrayright]$$

which can be solved by Gaussian elimination that is

$$left[beginarrayc
1& 1& 0& 0& 13\
1& 0& 1& 0& 15\
0& 1& 0& 1& 15\
0& 0& 1& 1& 17\
endarrayright]to left[beginarrayc
1& 1& 0& 0& 13\
0& -1& 1& 0& 2\
0& 1& 0& 1& 15\
0& 0& 1& 1& 17\
endarrayright]to left[beginarrayc
1& 1& 0& 0& 13\
0& -1& 1& 0& 2\
0& 0& 1& 1& 17\
0& 0& 1& 1& 17\
endarrayright]to left[beginarrayc
1& 1& 0& 0& 13\
0& -1& 1& 0& 2\
0& 0& 1& 1& 17\
0& 0& 0& 0& 0\
endarrayright]$$

therefore setting $y_2=k$ as free we obtain

• $y_1=17-k$


• $x_2=y_1-2=15-k$


• $x_1=13-x_1=k-2$

share|cite|improve this answer

edited Jul 30 at 14:53

answered Jul 30 at 14:45

gimusi

64.3k73480

add a comment | 

up vote
1
down vote



accepted

It suffices to consider the system

• $x_1 + x_2 =x_Total=13$


• $ y_1 + y_2 = y_Total=17$


• $x_1 + y_1 =xy1_Total=15$


• $ x_2 + y_2 = xy2_Total=15$

which is in the augmented matrix form

$$left[beginarrayc
1& 1& 0& 0& 13\
0& 0& 1& 1& 17\
1& 0& 1& 0& 15\
0& 1& 0& 1& 15\
endarrayright]$$

which can be solved by Gaussian elimination that is

$$left[beginarrayc
1& 1& 0& 0& 13\
1& 0& 1& 0& 15\
0& 1& 0& 1& 15\
0& 0& 1& 1& 17\
endarrayright]to left[beginarrayc
1& 1& 0& 0& 13\
0& -1& 1& 0& 2\
0& 1& 0& 1& 15\
0& 0& 1& 1& 17\
endarrayright]to left[beginarrayc
1& 1& 0& 0& 13\
0& -1& 1& 0& 2\
0& 0& 1& 1& 17\
0& 0& 1& 1& 17\
endarrayright]to left[beginarrayc
1& 1& 0& 0& 13\
0& -1& 1& 0& 2\
0& 0& 1& 1& 17\
0& 0& 0& 0& 0\
endarrayright]$$

therefore setting $y_2=k$ as free we obtain

• $y_1=17-k$


• $x_2=y_1-2=15-k$


• $x_1=13-x_1=k-2$

share|cite|improve this answer

edited Jul 30 at 14:53

answered Jul 30 at 14:45

gimusi

64.3k73480

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

It suffices to consider the system

• $x_1 + x_2 =x_Total=13$


• $ y_1 + y_2 = y_Total=17$


• $x_1 + y_1 =xy1_Total=15$


• $ x_2 + y_2 = xy2_Total=15$

which is in the augmented matrix form

$$left[beginarrayc
1& 1& 0& 0& 13\
0& 0& 1& 1& 17\
1& 0& 1& 0& 15\
0& 1& 0& 1& 15\
endarrayright]$$

which can be solved by Gaussian elimination that is

$$left[beginarrayc
1& 1& 0& 0& 13\
1& 0& 1& 0& 15\
0& 1& 0& 1& 15\
0& 0& 1& 1& 17\
endarrayright]to left[beginarrayc
1& 1& 0& 0& 13\
0& -1& 1& 0& 2\
0& 1& 0& 1& 15\
0& 0& 1& 1& 17\
endarrayright]to left[beginarrayc
1& 1& 0& 0& 13\
0& -1& 1& 0& 2\
0& 0& 1& 1& 17\
0& 0& 1& 1& 17\
endarrayright]to left[beginarrayc
1& 1& 0& 0& 13\
0& -1& 1& 0& 2\
0& 0& 1& 1& 17\
0& 0& 0& 0& 0\
endarrayright]$$

therefore setting $y_2=k$ as free we obtain

• $y_1=17-k$


• $x_2=y_1-2=15-k$


• $x_1=13-x_1=k-2$

share|cite|improve this answer

edited Jul 30 at 14:53

answered Jul 30 at 14:45

gimusi

64.3k73480

It suffices to consider the system

• $x_1 + x_2 =x_Total=13$


• $ y_1 + y_2 = y_Total=17$


• $x_1 + y_1 =xy1_Total=15$


• $ x_2 + y_2 = xy2_Total=15$

which is in the augmented matrix form

$$left[beginarrayc
1& 1& 0& 0& 13\
0& 0& 1& 1& 17\
1& 0& 1& 0& 15\
0& 1& 0& 1& 15\
endarrayright]$$

which can be solved by Gaussian elimination that is

$$left[beginarrayc
1& 1& 0& 0& 13\
1& 0& 1& 0& 15\
0& 1& 0& 1& 15\
0& 0& 1& 1& 17\
endarrayright]to left[beginarrayc
1& 1& 0& 0& 13\
0& -1& 1& 0& 2\
0& 1& 0& 1& 15\
0& 0& 1& 1& 17\
endarrayright]to left[beginarrayc
1& 1& 0& 0& 13\
0& -1& 1& 0& 2\
0& 0& 1& 1& 17\
0& 0& 1& 1& 17\
endarrayright]to left[beginarrayc
1& 1& 0& 0& 13\
0& -1& 1& 0& 2\
0& 0& 1& 1& 17\
0& 0& 0& 0& 0\
endarrayright]$$

therefore setting $y_2=k$ as free we obtain

• $y_1=17-k$


• $x_2=y_1-2=15-k$


• $x_1=13-x_1=k-2$

share|cite|improve this answer

edited Jul 30 at 14:53

answered Jul 30 at 14:45

gimusi

64.3k73480

share|cite|improve this answer

share|cite|improve this answer

edited Jul 30 at 14:53

edited Jul 30 at 14:53

edited Jul 30 at 14:53

answered Jul 30 at 14:45

gimusi

64.3k73480

answered Jul 30 at 14:45

gimusi

64.3k73480

answered Jul 30 at 14:45

gimusi

64.3k73480

64.3k73480

add a comment | 

add a comment | 

up vote
0
down vote

That is a simple linear system of equations, you could solve it in lot's of ways, one of them being Gaussian elimination. In your example you have $$begincasesx_1+x_2=15\ y_1+y_2=15\ x_1+y_1=13\ x_2+y_2=17endcases$$ four equations in four unknowns, to know from the start if the system has infinite, no or one solution you could use Rouchè-Capelli theorem. Gaussian elimination is a pretty involved process to find solutions to linear system of equations (it cannot be applied on non linear), in cases like this one you could simply use substitution, addition, subtraction, multiplication between rows etc.

If you really want to use gaussian elimination the system of equations can be put in matrix form like this $$left(beginmatrix1&1&0&0&15\0&0&1&1&15\1&0&1&0&13\0&1&0&1&17endmatrixright)$$ Your system seems like having infinite solutions, in this case you just choose one variable to explicit the others as a function of that

share|cite|improve this answer

answered Jul 30 at 14:50

Davide Morgante

1,693220

add a comment | 

up vote
0
down vote

That is a simple linear system of equations, you could solve it in lot's of ways, one of them being Gaussian elimination. In your example you have $$begincasesx_1+x_2=15\ y_1+y_2=15\ x_1+y_1=13\ x_2+y_2=17endcases$$ four equations in four unknowns, to know from the start if the system has infinite, no or one solution you could use Rouchè-Capelli theorem. Gaussian elimination is a pretty involved process to find solutions to linear system of equations (it cannot be applied on non linear), in cases like this one you could simply use substitution, addition, subtraction, multiplication between rows etc.

If you really want to use gaussian elimination the system of equations can be put in matrix form like this $$left(beginmatrix1&1&0&0&15\0&0&1&1&15\1&0&1&0&13\0&1&0&1&17endmatrixright)$$ Your system seems like having infinite solutions, in this case you just choose one variable to explicit the others as a function of that

share|cite|improve this answer

answered Jul 30 at 14:50

Davide Morgante

1,693220

add a comment | 

up vote
0
down vote

up vote
0
down vote

That is a simple linear system of equations, you could solve it in lot's of ways, one of them being Gaussian elimination. In your example you have $$begincasesx_1+x_2=15\ y_1+y_2=15\ x_1+y_1=13\ x_2+y_2=17endcases$$ four equations in four unknowns, to know from the start if the system has infinite, no or one solution you could use Rouchè-Capelli theorem. Gaussian elimination is a pretty involved process to find solutions to linear system of equations (it cannot be applied on non linear), in cases like this one you could simply use substitution, addition, subtraction, multiplication between rows etc.

If you really want to use gaussian elimination the system of equations can be put in matrix form like this $$left(beginmatrix1&1&0&0&15\0&0&1&1&15\1&0&1&0&13\0&1&0&1&17endmatrixright)$$ Your system seems like having infinite solutions, in this case you just choose one variable to explicit the others as a function of that

share|cite|improve this answer

answered Jul 30 at 14:50

Davide Morgante

1,693220

That is a simple linear system of equations, you could solve it in lot's of ways, one of them being Gaussian elimination. In your example you have $$begincasesx_1+x_2=15\ y_1+y_2=15\ x_1+y_1=13\ x_2+y_2=17endcases$$ four equations in four unknowns, to know from the start if the system has infinite, no or one solution you could use Rouchè-Capelli theorem. Gaussian elimination is a pretty involved process to find solutions to linear system of equations (it cannot be applied on non linear), in cases like this one you could simply use substitution, addition, subtraction, multiplication between rows etc.

If you really want to use gaussian elimination the system of equations can be put in matrix form like this $$left(beginmatrix1&1&0&0&15\0&0&1&1&15\1&0&1&0&13\0&1&0&1&17endmatrixright)$$ Your system seems like having infinite solutions, in this case you just choose one variable to explicit the others as a function of that

share|cite|improve this answer

answered Jul 30 at 14:50

Davide Morgante

1,693220

share|cite|improve this answer

share|cite|improve this answer

answered Jul 30 at 14:50

Davide Morgante

1,693220

answered Jul 30 at 14:50

Davide Morgante

1,693220

answered Jul 30 at 14:50

Davide Morgante

1,693220

1,693220

add a comment | 

add a comment | 

up vote
0
down vote

The sums of the rows have a grand total, equal to the sum of all the $x_i$ and $y_i$. That must equal the grand total of all the column sums. Here, 13+17=15+15 so you are okay. If not, there will be no solutions.

Suppose the grand totals match, and you have a matrix with $M$ rows and $N$ columns. You can put in any numbers you like except for the last row and last column. Then work out numbers in the last column to make the row sums right, and numbers in the last row to make column sums right.

share|cite|improve this answer

answered Jul 30 at 15:09

Empy2

31.7k12059

add a comment | 

up vote
0
down vote

The sums of the rows have a grand total, equal to the sum of all the $x_i$ and $y_i$. That must equal the grand total of all the column sums. Here, 13+17=15+15 so you are okay. If not, there will be no solutions.

Suppose the grand totals match, and you have a matrix with $M$ rows and $N$ columns. You can put in any numbers you like except for the last row and last column. Then work out numbers in the last column to make the row sums right, and numbers in the last row to make column sums right.

share|cite|improve this answer

answered Jul 30 at 15:09

Empy2

31.7k12059

add a comment | 

up vote
0
down vote

up vote
0
down vote

The sums of the rows have a grand total, equal to the sum of all the $x_i$ and $y_i$. That must equal the grand total of all the column sums. Here, 13+17=15+15 so you are okay. If not, there will be no solutions.

Suppose the grand totals match, and you have a matrix with $M$ rows and $N$ columns. You can put in any numbers you like except for the last row and last column. Then work out numbers in the last column to make the row sums right, and numbers in the last row to make column sums right.

share|cite|improve this answer

answered Jul 30 at 15:09

Empy2

31.7k12059

The sums of the rows have a grand total, equal to the sum of all the $x_i$ and $y_i$. That must equal the grand total of all the column sums. Here, 13+17=15+15 so you are okay. If not, there will be no solutions.

Suppose the grand totals match, and you have a matrix with $M$ rows and $N$ columns. You can put in any numbers you like except for the last row and last column. Then work out numbers in the last column to make the row sums right, and numbers in the last row to make column sums right.

share|cite|improve this answer

answered Jul 30 at 15:09

Empy2

31.7k12059

share|cite|improve this answer

share|cite|improve this answer

answered Jul 30 at 15:09

Empy2

31.7k12059

answered Jul 30 at 15:09

Empy2

31.7k12059

answered Jul 30 at 15:09

Empy2

31.7k12059

31.7k12059

add a comment | 

add a comment | 

 

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