Matrices of different dimensions in a matrix equation












1












$begingroup$


Given two matrices A, B, respectively, and an equation XA=B; how can we obtain X and discuss the system?



begin{pmatrix}
1 & a\
a & 1 \
end{pmatrix} begin{pmatrix}
a & 1 \
a^2 & 0 \
1 & 1 \
end{pmatrix}



I am not familiar with this structure and I can't even picture what the dimensions of X are. Any general matrix I write down looks uncomplete.






linear-algebra matrices matrix-equations matrix-calculus







share|cite|improve this question









$endgroup$












  • $begingroup$
    Do you intend that $A = begin{pmatrix} 1 & a \ a & 1 end{pmatrix}$ and $B$ is the other matrix?
    $endgroup$
    – Robert Lewis
    Dec 27 '18 at 20:14






  • 1




    $begingroup$
    @RobertLewis Yes, I do.
    $endgroup$
    – houda el fezzak
    Dec 28 '18 at 5:00










  • $begingroup$
    I thought so, and answered as such . . . thanks for the "acceptance". Cheers!
    $endgroup$
    – Robert Lewis
    Dec 28 '18 at 5:01










  • $begingroup$
    Oh, such a nice human being you are!
    $endgroup$
    – houda el fezzak
    Dec 29 '18 at 15:52
















1












$begingroup$


Given two matrices A, B, respectively, and an equation XA=B; how can we obtain X and discuss the system?



begin{pmatrix}
1 & a\
a & 1 \
end{pmatrix} begin{pmatrix}
a & 1 \
a^2 & 0 \
1 & 1 \
end{pmatrix}



I am not familiar with this structure and I can't even picture what the dimensions of X are. Any general matrix I write down looks uncomplete.






linear-algebra matrices matrix-equations matrix-calculus







share|cite|improve this question









$endgroup$












  • $begingroup$
    Do you intend that $A = begin{pmatrix} 1 & a \ a & 1 end{pmatrix}$ and $B$ is the other matrix?
    $endgroup$
    – Robert Lewis
    Dec 27 '18 at 20:14






  • 1




    $begingroup$
    @RobertLewis Yes, I do.
    $endgroup$
    – houda el fezzak
    Dec 28 '18 at 5:00










  • $begingroup$
    I thought so, and answered as such . . . thanks for the "acceptance". Cheers!
    $endgroup$
    – Robert Lewis
    Dec 28 '18 at 5:01










  • $begingroup$
    Oh, such a nice human being you are!
    $endgroup$
    – houda el fezzak
    Dec 29 '18 at 15:52














1












1








1


1



$begingroup$


Given two matrices A, B, respectively, and an equation XA=B; how can we obtain X and discuss the system?



begin{pmatrix}
1 & a\
a & 1 \
end{pmatrix} begin{pmatrix}
a & 1 \
a^2 & 0 \
1 & 1 \
end{pmatrix}



I am not familiar with this structure and I can't even picture what the dimensions of X are. Any general matrix I write down looks uncomplete.






linear-algebra matrices matrix-equations matrix-calculus







share|cite|improve this question









$endgroup$




Given two matrices A, B, respectively, and an equation XA=B; how can we obtain X and discuss the system?



begin{pmatrix}
1 & a\
a & 1 \
end{pmatrix} begin{pmatrix}
a & 1 \
a^2 & 0 \
1 & 1 \
end{pmatrix}



I am not familiar with this structure and I can't even picture what the dimensions of X are. Any general matrix I write down looks uncomplete.






linear-algebra matrices matrix-equations matrix-calculus




linear-algebra matrices matrix-equations matrix-calculus






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 27 '18 at 19:54









houda el fezzakhouda el fezzak

224




224












  • $begingroup$
    Do you intend that $A = begin{pmatrix} 1 & a \ a & 1 end{pmatrix}$ and $B$ is the other matrix?
    $endgroup$
    – Robert Lewis
    Dec 27 '18 at 20:14






  • 1




    $begingroup$
    @RobertLewis Yes, I do.
    $endgroup$
    – houda el fezzak
    Dec 28 '18 at 5:00










  • $begingroup$
    I thought so, and answered as such . . . thanks for the "acceptance". Cheers!
    $endgroup$
    – Robert Lewis
    Dec 28 '18 at 5:01










  • $begingroup$
    Oh, such a nice human being you are!
    $endgroup$
    – houda el fezzak
    Dec 29 '18 at 15:52


















  • $begingroup$
    Do you intend that $A = begin{pmatrix} 1 & a \ a & 1 end{pmatrix}$ and $B$ is the other matrix?
    $endgroup$
    – Robert Lewis
    Dec 27 '18 at 20:14






  • 1




    $begingroup$
    @RobertLewis Yes, I do.
    $endgroup$
    – houda el fezzak
    Dec 28 '18 at 5:00










  • $begingroup$
    I thought so, and answered as such . . . thanks for the "acceptance". Cheers!
    $endgroup$
    – Robert Lewis
    Dec 28 '18 at 5:01










  • $begingroup$
    Oh, such a nice human being you are!
    $endgroup$
    – houda el fezzak
    Dec 29 '18 at 15:52
















$begingroup$
Do you intend that $A = begin{pmatrix} 1 & a \ a & 1 end{pmatrix}$ and $B$ is the other matrix?
$endgroup$
– Robert Lewis
Dec 27 '18 at 20:14




$begingroup$
Do you intend that $A = begin{pmatrix} 1 & a \ a & 1 end{pmatrix}$ and $B$ is the other matrix?
$endgroup$
– Robert Lewis
Dec 27 '18 at 20:14




1




1




$begingroup$
@RobertLewis Yes, I do.
$endgroup$
– houda el fezzak
Dec 28 '18 at 5:00




$begingroup$
@RobertLewis Yes, I do.
$endgroup$
– houda el fezzak
Dec 28 '18 at 5:00












$begingroup$
I thought so, and answered as such . . . thanks for the "acceptance". Cheers!
$endgroup$
– Robert Lewis
Dec 28 '18 at 5:01




$begingroup$
I thought so, and answered as such . . . thanks for the "acceptance". Cheers!
$endgroup$
– Robert Lewis
Dec 28 '18 at 5:01












$begingroup$
Oh, such a nice human being you are!
$endgroup$
– houda el fezzak
Dec 29 '18 at 15:52




$begingroup$
Oh, such a nice human being you are!
$endgroup$
– houda el fezzak
Dec 29 '18 at 15:52










1 Answer
1






active

oldest

votes


















0












$begingroup$

$X$ must be a $3 times 2$ matrix; we may write



$X = begin{pmatrix} x_{11} & x_{12} \ x_{21} & x_{22} \ x_{31} & x_{32} end{pmatrix}; tag 1$



we may thus write out the system



$XA = B tag 2$



as



$begin{pmatrix} x_{11} & x_{12} \ x_{21} & x_{22} \ x_{31} & x_{32} end{pmatrix} begin{pmatrix} 1 & a \ a & 1 end{pmatrix} = begin{pmatrix} a & 1 \ a^2 & 0 \ 1 & 1 end{pmatrix}; tag 3$



here



$A = begin{pmatrix} 1 & a \ a & 1 end{pmatrix}, tag 4$



and



$B = begin{pmatrix} a & 1 \ a^2 & 0 \ 1 & 1 end{pmatrix}; tag 5$



we may write out the system (3) as a set of $6$ linear equations in two variables each:



$x_{11} + ax_{12} = a, tag 6$



$ax_{11} + x_{12} = 1; tag 7$



$x_{21} + ax_{22} = a^2, tag 8$



$ax_{21} + x_{22} = 0, tag 9$



$x_{31} + ax_{32} = 1, tag{10}$



$ax_{31} + x_{32} = 1; tag{11}$



the system of equations (6)-(11) is in fact easy to solve, one pair $(x_{i1}, x_{i2})$ at a time; for example, we multiply (6) by $a$:



$ax_{11} + a^2x_{12} = a^2, tag{12}$



and subtract (7) from (12):



$(a^2 - 1) x_{12} = a^2 - 1, tag{13}$



whence



$x_{12} = 1, tag{14}$



provided of course that



$a ne pm 1; tag{15}$



having $x_{12}$ we find via (6) that



$x_{11} = a(1 - x_{12}) = 0; tag{16}$



the remaining equations (8)-(11) may be similarly solved.



The system (2), (3) may also be addressed in a somewhat more matrix-oriented manner by setting



$Y = begin{pmatrix} x_{11} & x_{12} \ x_{21} & x_{22} end{pmatrix}, ; mathbf y = (x_{31}, x_{32}), ; C = begin{pmatrix} a & 1 \ a^2 & 0 end{pmatrix}; mathbf d = (1, 1); tag {17}$



then (3) may be written in the from



$begin{pmatrix} Y \ mathbf y end{pmatrix} A = begin{pmatrix} C \ mathbf d end{pmatrix}, tag{18}$



leading to



$YA = C, ; mathbf y A = mathbf d, tag{19}$



whence



$Y = CA^{-1}, ; mathbf y = mathbf d A^{-1}; tag{20}$



where it is easily seen that



$A^{-1} = dfrac{1}{1 - a^2} begin{pmatrix} 1 & -a \ -a & 1 end{pmatrix}. tag{21}$



We assume $a = pm 1$ in deriving (17)-(21); in the event that $a = pm 1$ we see that (8)-(9) become



$x_{21} pm x_{22} = 1, ; pm x_{21} + x_{22} = 0, tag{22}$



which lead to the contradicion



$pm 1 = 0; tag{23}$



thus we see there is no solution to (2) when $a = pm 1$.






share|cite|improve this answer









$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3054304%2fmatrices-of-different-dimensions-in-a-matrix-equation%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    $X$ must be a $3 times 2$ matrix; we may write



    $X = begin{pmatrix} x_{11} & x_{12} \ x_{21} & x_{22} \ x_{31} & x_{32} end{pmatrix}; tag 1$



    we may thus write out the system



    $XA = B tag 2$



    as



    $begin{pmatrix} x_{11} & x_{12} \ x_{21} & x_{22} \ x_{31} & x_{32} end{pmatrix} begin{pmatrix} 1 & a \ a & 1 end{pmatrix} = begin{pmatrix} a & 1 \ a^2 & 0 \ 1 & 1 end{pmatrix}; tag 3$



    here



    $A = begin{pmatrix} 1 & a \ a & 1 end{pmatrix}, tag 4$



    and



    $B = begin{pmatrix} a & 1 \ a^2 & 0 \ 1 & 1 end{pmatrix}; tag 5$



    we may write out the system (3) as a set of $6$ linear equations in two variables each:



    $x_{11} + ax_{12} = a, tag 6$



    $ax_{11} + x_{12} = 1; tag 7$



    $x_{21} + ax_{22} = a^2, tag 8$



    $ax_{21} + x_{22} = 0, tag 9$



    $x_{31} + ax_{32} = 1, tag{10}$



    $ax_{31} + x_{32} = 1; tag{11}$



    the system of equations (6)-(11) is in fact easy to solve, one pair $(x_{i1}, x_{i2})$ at a time; for example, we multiply (6) by $a$:



    $ax_{11} + a^2x_{12} = a^2, tag{12}$



    and subtract (7) from (12):



    $(a^2 - 1) x_{12} = a^2 - 1, tag{13}$



    whence



    $x_{12} = 1, tag{14}$



    provided of course that



    $a ne pm 1; tag{15}$



    having $x_{12}$ we find via (6) that



    $x_{11} = a(1 - x_{12}) = 0; tag{16}$



    the remaining equations (8)-(11) may be similarly solved.



    The system (2), (3) may also be addressed in a somewhat more matrix-oriented manner by setting



    $Y = begin{pmatrix} x_{11} & x_{12} \ x_{21} & x_{22} end{pmatrix}, ; mathbf y = (x_{31}, x_{32}), ; C = begin{pmatrix} a & 1 \ a^2 & 0 end{pmatrix}; mathbf d = (1, 1); tag {17}$



    then (3) may be written in the from



    $begin{pmatrix} Y \ mathbf y end{pmatrix} A = begin{pmatrix} C \ mathbf d end{pmatrix}, tag{18}$



    leading to



    $YA = C, ; mathbf y A = mathbf d, tag{19}$



    whence



    $Y = CA^{-1}, ; mathbf y = mathbf d A^{-1}; tag{20}$



    where it is easily seen that



    $A^{-1} = dfrac{1}{1 - a^2} begin{pmatrix} 1 & -a \ -a & 1 end{pmatrix}. tag{21}$



    We assume $a = pm 1$ in deriving (17)-(21); in the event that $a = pm 1$ we see that (8)-(9) become



    $x_{21} pm x_{22} = 1, ; pm x_{21} + x_{22} = 0, tag{22}$



    which lead to the contradicion



    $pm 1 = 0; tag{23}$



    thus we see there is no solution to (2) when $a = pm 1$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      $X$ must be a $3 times 2$ matrix; we may write



      $X = begin{pmatrix} x_{11} & x_{12} \ x_{21} & x_{22} \ x_{31} & x_{32} end{pmatrix}; tag 1$



      we may thus write out the system



      $XA = B tag 2$



      as



      $begin{pmatrix} x_{11} & x_{12} \ x_{21} & x_{22} \ x_{31} & x_{32} end{pmatrix} begin{pmatrix} 1 & a \ a & 1 end{pmatrix} = begin{pmatrix} a & 1 \ a^2 & 0 \ 1 & 1 end{pmatrix}; tag 3$



      here



      $A = begin{pmatrix} 1 & a \ a & 1 end{pmatrix}, tag 4$



      and



      $B = begin{pmatrix} a & 1 \ a^2 & 0 \ 1 & 1 end{pmatrix}; tag 5$



      we may write out the system (3) as a set of $6$ linear equations in two variables each:



      $x_{11} + ax_{12} = a, tag 6$



      $ax_{11} + x_{12} = 1; tag 7$



      $x_{21} + ax_{22} = a^2, tag 8$



      $ax_{21} + x_{22} = 0, tag 9$



      $x_{31} + ax_{32} = 1, tag{10}$



      $ax_{31} + x_{32} = 1; tag{11}$



      the system of equations (6)-(11) is in fact easy to solve, one pair $(x_{i1}, x_{i2})$ at a time; for example, we multiply (6) by $a$:



      $ax_{11} + a^2x_{12} = a^2, tag{12}$



      and subtract (7) from (12):



      $(a^2 - 1) x_{12} = a^2 - 1, tag{13}$



      whence



      $x_{12} = 1, tag{14}$



      provided of course that



      $a ne pm 1; tag{15}$



      having $x_{12}$ we find via (6) that



      $x_{11} = a(1 - x_{12}) = 0; tag{16}$



      the remaining equations (8)-(11) may be similarly solved.



      The system (2), (3) may also be addressed in a somewhat more matrix-oriented manner by setting



      $Y = begin{pmatrix} x_{11} & x_{12} \ x_{21} & x_{22} end{pmatrix}, ; mathbf y = (x_{31}, x_{32}), ; C = begin{pmatrix} a & 1 \ a^2 & 0 end{pmatrix}; mathbf d = (1, 1); tag {17}$



      then (3) may be written in the from



      $begin{pmatrix} Y \ mathbf y end{pmatrix} A = begin{pmatrix} C \ mathbf d end{pmatrix}, tag{18}$



      leading to



      $YA = C, ; mathbf y A = mathbf d, tag{19}$



      whence



      $Y = CA^{-1}, ; mathbf y = mathbf d A^{-1}; tag{20}$



      where it is easily seen that



      $A^{-1} = dfrac{1}{1 - a^2} begin{pmatrix} 1 & -a \ -a & 1 end{pmatrix}. tag{21}$



      We assume $a = pm 1$ in deriving (17)-(21); in the event that $a = pm 1$ we see that (8)-(9) become



      $x_{21} pm x_{22} = 1, ; pm x_{21} + x_{22} = 0, tag{22}$



      which lead to the contradicion



      $pm 1 = 0; tag{23}$



      thus we see there is no solution to (2) when $a = pm 1$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        $X$ must be a $3 times 2$ matrix; we may write



        $X = begin{pmatrix} x_{11} & x_{12} \ x_{21} & x_{22} \ x_{31} & x_{32} end{pmatrix}; tag 1$



        we may thus write out the system



        $XA = B tag 2$



        as



        $begin{pmatrix} x_{11} & x_{12} \ x_{21} & x_{22} \ x_{31} & x_{32} end{pmatrix} begin{pmatrix} 1 & a \ a & 1 end{pmatrix} = begin{pmatrix} a & 1 \ a^2 & 0 \ 1 & 1 end{pmatrix}; tag 3$



        here



        $A = begin{pmatrix} 1 & a \ a & 1 end{pmatrix}, tag 4$



        and



        $B = begin{pmatrix} a & 1 \ a^2 & 0 \ 1 & 1 end{pmatrix}; tag 5$



        we may write out the system (3) as a set of $6$ linear equations in two variables each:



        $x_{11} + ax_{12} = a, tag 6$



        $ax_{11} + x_{12} = 1; tag 7$



        $x_{21} + ax_{22} = a^2, tag 8$



        $ax_{21} + x_{22} = 0, tag 9$



        $x_{31} + ax_{32} = 1, tag{10}$



        $ax_{31} + x_{32} = 1; tag{11}$



        the system of equations (6)-(11) is in fact easy to solve, one pair $(x_{i1}, x_{i2})$ at a time; for example, we multiply (6) by $a$:



        $ax_{11} + a^2x_{12} = a^2, tag{12}$



        and subtract (7) from (12):



        $(a^2 - 1) x_{12} = a^2 - 1, tag{13}$



        whence



        $x_{12} = 1, tag{14}$



        provided of course that



        $a ne pm 1; tag{15}$



        having $x_{12}$ we find via (6) that



        $x_{11} = a(1 - x_{12}) = 0; tag{16}$



        the remaining equations (8)-(11) may be similarly solved.



        The system (2), (3) may also be addressed in a somewhat more matrix-oriented manner by setting



        $Y = begin{pmatrix} x_{11} & x_{12} \ x_{21} & x_{22} end{pmatrix}, ; mathbf y = (x_{31}, x_{32}), ; C = begin{pmatrix} a & 1 \ a^2 & 0 end{pmatrix}; mathbf d = (1, 1); tag {17}$



        then (3) may be written in the from



        $begin{pmatrix} Y \ mathbf y end{pmatrix} A = begin{pmatrix} C \ mathbf d end{pmatrix}, tag{18}$



        leading to



        $YA = C, ; mathbf y A = mathbf d, tag{19}$



        whence



        $Y = CA^{-1}, ; mathbf y = mathbf d A^{-1}; tag{20}$



        where it is easily seen that



        $A^{-1} = dfrac{1}{1 - a^2} begin{pmatrix} 1 & -a \ -a & 1 end{pmatrix}. tag{21}$



        We assume $a = pm 1$ in deriving (17)-(21); in the event that $a = pm 1$ we see that (8)-(9) become



        $x_{21} pm x_{22} = 1, ; pm x_{21} + x_{22} = 0, tag{22}$



        which lead to the contradicion



        $pm 1 = 0; tag{23}$



        thus we see there is no solution to (2) when $a = pm 1$.






        share|cite|improve this answer









        $endgroup$



        $X$ must be a $3 times 2$ matrix; we may write



        $X = begin{pmatrix} x_{11} & x_{12} \ x_{21} & x_{22} \ x_{31} & x_{32} end{pmatrix}; tag 1$



        we may thus write out the system



        $XA = B tag 2$



        as



        $begin{pmatrix} x_{11} & x_{12} \ x_{21} & x_{22} \ x_{31} & x_{32} end{pmatrix} begin{pmatrix} 1 & a \ a & 1 end{pmatrix} = begin{pmatrix} a & 1 \ a^2 & 0 \ 1 & 1 end{pmatrix}; tag 3$



        here



        $A = begin{pmatrix} 1 & a \ a & 1 end{pmatrix}, tag 4$



        and



        $B = begin{pmatrix} a & 1 \ a^2 & 0 \ 1 & 1 end{pmatrix}; tag 5$



        we may write out the system (3) as a set of $6$ linear equations in two variables each:



        $x_{11} + ax_{12} = a, tag 6$



        $ax_{11} + x_{12} = 1; tag 7$



        $x_{21} + ax_{22} = a^2, tag 8$



        $ax_{21} + x_{22} = 0, tag 9$



        $x_{31} + ax_{32} = 1, tag{10}$



        $ax_{31} + x_{32} = 1; tag{11}$



        the system of equations (6)-(11) is in fact easy to solve, one pair $(x_{i1}, x_{i2})$ at a time; for example, we multiply (6) by $a$:



        $ax_{11} + a^2x_{12} = a^2, tag{12}$



        and subtract (7) from (12):



        $(a^2 - 1) x_{12} = a^2 - 1, tag{13}$



        whence



        $x_{12} = 1, tag{14}$



        provided of course that



        $a ne pm 1; tag{15}$



        having $x_{12}$ we find via (6) that



        $x_{11} = a(1 - x_{12}) = 0; tag{16}$



        the remaining equations (8)-(11) may be similarly solved.



        The system (2), (3) may also be addressed in a somewhat more matrix-oriented manner by setting



        $Y = begin{pmatrix} x_{11} & x_{12} \ x_{21} & x_{22} end{pmatrix}, ; mathbf y = (x_{31}, x_{32}), ; C = begin{pmatrix} a & 1 \ a^2 & 0 end{pmatrix}; mathbf d = (1, 1); tag {17}$



        then (3) may be written in the from



        $begin{pmatrix} Y \ mathbf y end{pmatrix} A = begin{pmatrix} C \ mathbf d end{pmatrix}, tag{18}$



        leading to



        $YA = C, ; mathbf y A = mathbf d, tag{19}$



        whence



        $Y = CA^{-1}, ; mathbf y = mathbf d A^{-1}; tag{20}$



        where it is easily seen that



        $A^{-1} = dfrac{1}{1 - a^2} begin{pmatrix} 1 & -a \ -a & 1 end{pmatrix}. tag{21}$



        We assume $a = pm 1$ in deriving (17)-(21); in the event that $a = pm 1$ we see that (8)-(9) become



        $x_{21} pm x_{22} = 1, ; pm x_{21} + x_{22} = 0, tag{22}$



        which lead to the contradicion



        $pm 1 = 0; tag{23}$



        thus we see there is no solution to (2) when $a = pm 1$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 27 '18 at 22:21









        Robert LewisRobert Lewis

        46.7k23067




        46.7k23067






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3054304%2fmatrices-of-different-dimensions-in-a-matrix-equation%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Bressuire

            Image
            .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
            Read more

            Cabo Verde

            Image
              Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
            Read more

            Gyllenstierna

            Image
            .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
            Read more