How to solve for unknown matrix?












1














How can I solve this?
$$ begin{bmatrix}
1 & 1 \
1 & 2 \
end{bmatrix}
X +
begin{bmatrix}
2 & -1\
-1 & 1\
end{bmatrix}
X
begin{bmatrix}
1 & 5 \
1 & 2 \
end{bmatrix}
=
begin{bmatrix}
1 & 1\
1 & 1\
end{bmatrix}
$$



I know there's similar question like:
Solve for unknown matrix.
But this one is much more complex as there are two separate terms with $X$.



Can I perform something similar like:
$Ax + Bx = C implies (A+B)x = C$? But at the second term, the $X$ is at the middle and that order is important in matrix.



Any help would be appreciated!






matrices matrix-equations







share|cite|improve this question
























  • Why don't you do exactly the same thing, done in that exercise? Like, do you know, how to multiply and add matrices? Is this the "much more complex" part? I mean, you can also left-multiply both sides by $ begin{bmatrix} 1 & 1 \ 1 & 2 \ end{bmatrix} $ and see what happens
    – Makina
    Dec 7 at 23:00












  • If you put $X=begin{bmatrix}x_1 & x_2 \ x_3 & x_4 end{bmatrix}$ and expand all the product, then it is still a set of linear equations.
    – Eclipse Sun
    Dec 7 at 23:31
















1














How can I solve this?
$$ begin{bmatrix}
1 & 1 \
1 & 2 \
end{bmatrix}
X +
begin{bmatrix}
2 & -1\
-1 & 1\
end{bmatrix}
X
begin{bmatrix}
1 & 5 \
1 & 2 \
end{bmatrix}
=
begin{bmatrix}
1 & 1\
1 & 1\
end{bmatrix}
$$



I know there's similar question like:
Solve for unknown matrix.
But this one is much more complex as there are two separate terms with $X$.



Can I perform something similar like:
$Ax + Bx = C implies (A+B)x = C$? But at the second term, the $X$ is at the middle and that order is important in matrix.



Any help would be appreciated!






matrices matrix-equations







share|cite|improve this question
























  • Why don't you do exactly the same thing, done in that exercise? Like, do you know, how to multiply and add matrices? Is this the "much more complex" part? I mean, you can also left-multiply both sides by $ begin{bmatrix} 1 & 1 \ 1 & 2 \ end{bmatrix} $ and see what happens
    – Makina
    Dec 7 at 23:00












  • If you put $X=begin{bmatrix}x_1 & x_2 \ x_3 & x_4 end{bmatrix}$ and expand all the product, then it is still a set of linear equations.
    – Eclipse Sun
    Dec 7 at 23:31














1












1








1







How can I solve this?
$$ begin{bmatrix}
1 & 1 \
1 & 2 \
end{bmatrix}
X +
begin{bmatrix}
2 & -1\
-1 & 1\
end{bmatrix}
X
begin{bmatrix}
1 & 5 \
1 & 2 \
end{bmatrix}
=
begin{bmatrix}
1 & 1\
1 & 1\
end{bmatrix}
$$



I know there's similar question like:
Solve for unknown matrix.
But this one is much more complex as there are two separate terms with $X$.



Can I perform something similar like:
$Ax + Bx = C implies (A+B)x = C$? But at the second term, the $X$ is at the middle and that order is important in matrix.



Any help would be appreciated!






matrices matrix-equations







share|cite|improve this question















How can I solve this?
$$ begin{bmatrix}
1 & 1 \
1 & 2 \
end{bmatrix}
X +
begin{bmatrix}
2 & -1\
-1 & 1\
end{bmatrix}
X
begin{bmatrix}
1 & 5 \
1 & 2 \
end{bmatrix}
=
begin{bmatrix}
1 & 1\
1 & 1\
end{bmatrix}
$$



I know there's similar question like:
Solve for unknown matrix.
But this one is much more complex as there are two separate terms with $X$.



Can I perform something similar like:
$Ax + Bx = C implies (A+B)x = C$? But at the second term, the $X$ is at the middle and that order is important in matrix.



Any help would be appreciated!






matrices matrix-equations




matrices matrix-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 8 at 0:06









Jean-Claude Arbaut

14.7k63464




14.7k63464










asked Dec 7 at 22:40









jeispyun

111




111












  • Why don't you do exactly the same thing, done in that exercise? Like, do you know, how to multiply and add matrices? Is this the "much more complex" part? I mean, you can also left-multiply both sides by $ begin{bmatrix} 1 & 1 \ 1 & 2 \ end{bmatrix} $ and see what happens
    – Makina
    Dec 7 at 23:00












  • If you put $X=begin{bmatrix}x_1 & x_2 \ x_3 & x_4 end{bmatrix}$ and expand all the product, then it is still a set of linear equations.
    – Eclipse Sun
    Dec 7 at 23:31


















  • Why don't you do exactly the same thing, done in that exercise? Like, do you know, how to multiply and add matrices? Is this the "much more complex" part? I mean, you can also left-multiply both sides by $ begin{bmatrix} 1 & 1 \ 1 & 2 \ end{bmatrix} $ and see what happens
    – Makina
    Dec 7 at 23:00












  • If you put $X=begin{bmatrix}x_1 & x_2 \ x_3 & x_4 end{bmatrix}$ and expand all the product, then it is still a set of linear equations.
    – Eclipse Sun
    Dec 7 at 23:31
















Why don't you do exactly the same thing, done in that exercise? Like, do you know, how to multiply and add matrices? Is this the "much more complex" part? I mean, you can also left-multiply both sides by $ begin{bmatrix} 1 & 1 \ 1 & 2 \ end{bmatrix} $ and see what happens
– Makina
Dec 7 at 23:00






Why don't you do exactly the same thing, done in that exercise? Like, do you know, how to multiply and add matrices? Is this the "much more complex" part? I mean, you can also left-multiply both sides by $ begin{bmatrix} 1 & 1 \ 1 & 2 \ end{bmatrix} $ and see what happens
– Makina
Dec 7 at 23:00














If you put $X=begin{bmatrix}x_1 & x_2 \ x_3 & x_4 end{bmatrix}$ and expand all the product, then it is still a set of linear equations.
– Eclipse Sun
Dec 7 at 23:31




If you put $X=begin{bmatrix}x_1 & x_2 \ x_3 & x_4 end{bmatrix}$ and expand all the product, then it is still a set of linear equations.
– Eclipse Sun
Dec 7 at 23:31










1 Answer
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1














If we can't see the trick suggested in the comments, by $X=begin{bmatrix}
a & b \
c & d \
end{bmatrix}$ we obtain



$$ begin{bmatrix}
1 & 1 \
1 & 2 \
end{bmatrix}X= begin{bmatrix}
a+b & a+2b \
c+d & c+2d \
end{bmatrix}$$



$$
begin{bmatrix}
2 & -1\
-1 & 1\
end{bmatrix}
X
begin{bmatrix}
1 & 5 \
1 & 2 \
end{bmatrix}
=$$
$$=begin{bmatrix}
2 & -1\
-1 & 1\
end{bmatrix}
begin{bmatrix}
a+b & 5a+2b\
c+d & 5c+2d\
end{bmatrix}
=begin{bmatrix}
2(a+b)-(c+d) & 2(5a+2b)-(5c+2d)\
-(a+b)+(c+d) & -(5a+2b)+(5c+2d)\
end{bmatrix}$$



then the system



$$begin{bmatrix}
3(a+b)-(c+d) & (11a+6b)-(5c+2d)\
-(a+b)+2(c+d) & -(5a+2b)+(6c+4d)\
end{bmatrix}= begin{bmatrix}
1 & 1 \
1 & 1 \
end{bmatrix}$$






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    If we can't see the trick suggested in the comments, by $X=begin{bmatrix}
    a & b \
    c & d \
    end{bmatrix}$     we obtain



    $$ begin{bmatrix}
    1 & 1 \
    1 & 2 \
    end{bmatrix}X= begin{bmatrix}
    a+b & a+2b \
    c+d & c+2d \
    end{bmatrix}$$



    $$
    begin{bmatrix}
    2 & -1\
    -1 & 1\
    end{bmatrix}
    X
    begin{bmatrix}
    1 & 5 \
    1 & 2 \
    end{bmatrix}
    =$$
    $$=begin{bmatrix}
    2 & -1\
    -1 & 1\
    end{bmatrix}
    begin{bmatrix}
    a+b & 5a+2b\
    c+d & 5c+2d\
    end{bmatrix}
    =begin{bmatrix}
    2(a+b)-(c+d) & 2(5a+2b)-(5c+2d)\
    -(a+b)+(c+d) & -(5a+2b)+(5c+2d)\
    end{bmatrix}$$



    then the system



    $$begin{bmatrix}
    3(a+b)-(c+d) & (11a+6b)-(5c+2d)\
    -(a+b)+2(c+d) & -(5a+2b)+(6c+4d)\
    end{bmatrix}= begin{bmatrix}
    1 & 1 \
    1 & 1 \
    end{bmatrix}$$






    share|cite|improve this answer




























      1














      If we can't see the trick suggested in the comments, by $X=begin{bmatrix}
      a & b \
      c & d \
      end{bmatrix}$       we obtain



      $$ begin{bmatrix}
      1 & 1 \
      1 & 2 \
      end{bmatrix}X= begin{bmatrix}
      a+b & a+2b \
      c+d & c+2d \
      end{bmatrix}$$



      $$
      begin{bmatrix}
      2 & -1\
      -1 & 1\
      end{bmatrix}
      X
      begin{bmatrix}
      1 & 5 \
      1 & 2 \
      end{bmatrix}
      =$$
      $$=begin{bmatrix}
      2 & -1\
      -1 & 1\
      end{bmatrix}
      begin{bmatrix}
      a+b & 5a+2b\
      c+d & 5c+2d\
      end{bmatrix}
      =begin{bmatrix}
      2(a+b)-(c+d) & 2(5a+2b)-(5c+2d)\
      -(a+b)+(c+d) & -(5a+2b)+(5c+2d)\
      end{bmatrix}$$



      then the system



      $$begin{bmatrix}
      3(a+b)-(c+d) & (11a+6b)-(5c+2d)\
      -(a+b)+2(c+d) & -(5a+2b)+(6c+4d)\
      end{bmatrix}= begin{bmatrix}
      1 & 1 \
      1 & 1 \
      end{bmatrix}$$






      share|cite|improve this answer


























        1












        1








        1






        If we can't see the trick suggested in the comments, by $X=begin{bmatrix}
        a & b \
        c & d \
        end{bmatrix}$         we obtain



        $$ begin{bmatrix}
        1 & 1 \
        1 & 2 \
        end{bmatrix}X= begin{bmatrix}
        a+b & a+2b \
        c+d & c+2d \
        end{bmatrix}$$



        $$
        begin{bmatrix}
        2 & -1\
        -1 & 1\
        end{bmatrix}
        X
        begin{bmatrix}
        1 & 5 \
        1 & 2 \
        end{bmatrix}
        =$$
        $$=begin{bmatrix}
        2 & -1\
        -1 & 1\
        end{bmatrix}
        begin{bmatrix}
        a+b & 5a+2b\
        c+d & 5c+2d\
        end{bmatrix}
        =begin{bmatrix}
        2(a+b)-(c+d) & 2(5a+2b)-(5c+2d)\
        -(a+b)+(c+d) & -(5a+2b)+(5c+2d)\
        end{bmatrix}$$



        then the system



        $$begin{bmatrix}
        3(a+b)-(c+d) & (11a+6b)-(5c+2d)\
        -(a+b)+2(c+d) & -(5a+2b)+(6c+4d)\
        end{bmatrix}= begin{bmatrix}
        1 & 1 \
        1 & 1 \
        end{bmatrix}$$






        share|cite|improve this answer














        If we can't see the trick suggested in the comments, by $X=begin{bmatrix}
        a & b \
        c & d \
        end{bmatrix}$         we obtain



        $$ begin{bmatrix}
        1 & 1 \
        1 & 2 \
        end{bmatrix}X= begin{bmatrix}
        a+b & a+2b \
        c+d & c+2d \
        end{bmatrix}$$



        $$
        begin{bmatrix}
        2 & -1\
        -1 & 1\
        end{bmatrix}
        X
        begin{bmatrix}
        1 & 5 \
        1 & 2 \
        end{bmatrix}
        =$$
        $$=begin{bmatrix}
        2 & -1\
        -1 & 1\
        end{bmatrix}
        begin{bmatrix}
        a+b & 5a+2b\
        c+d & 5c+2d\
        end{bmatrix}
        =begin{bmatrix}
        2(a+b)-(c+d) & 2(5a+2b)-(5c+2d)\
        -(a+b)+(c+d) & -(5a+2b)+(5c+2d)\
        end{bmatrix}$$



        then the system



        $$begin{bmatrix}
        3(a+b)-(c+d) & (11a+6b)-(5c+2d)\
        -(a+b)+2(c+d) & -(5a+2b)+(6c+4d)\
        end{bmatrix}= begin{bmatrix}
        1 & 1 \
        1 & 1 \
        end{bmatrix}$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 8 at 0:02

























        answered Dec 7 at 23:54









        gimusi

        1




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