Matrix equations, simplify them.












0












$begingroup$


I have some equations and I don't know am I doing the simplification right. Can someone check it?



For example, we have an equation $AX = 4X + B$, where $A$ and $B$ are matrices. So, what I have done with it is




  1. $AX - 4X = B $

  2. $X(A - 4) = B$

  3. $X = B(A-4)^{-1}$


And the second example. $(A+4X)^{-1} = B$




  1. $A^{-1} + frac{1}{4}X^{-1} = B$

  2. $frac{1}{4}X^{-1} = B - A^{-1}$

  3. $X^{-1} = 4(B - A^{-1})$

  4. $X = (4(B - A^{-1}))^{-1}$


I didn't find a lot of information about that, so maybe someone can share any links? Or just help me here pls.






linear-algebra matrices inverse matrix-equations







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$endgroup$








  • 1




    $begingroup$
    The expression $A-4$ is undefined: it’s the difference of a matrix and scalar. Get into the habit of writing $A-4I$ (or $4E$, if that’s what the identity matrix is called in your materials.)
    $endgroup$
    – amd
    Jan 16 at 17:53






  • 1




    $begingroup$
    $(A+4X)^{-1}$ is no more equal to $A^{-1}+frac14X^{-1}$ than $1/(a+4x)$ is equal to $1/a+1/(4x)$.
    $endgroup$
    – amd
    Jan 16 at 17:54


















0












$begingroup$


I have some equations and I don't know am I doing the simplification right. Can someone check it?



For example, we have an equation $AX = 4X + B$, where $A$ and $B$ are matrices. So, what I have done with it is




  1. $AX - 4X = B $

  2. $X(A - 4) = B$

  3. $X = B(A-4)^{-1}$


And the second example. $(A+4X)^{-1} = B$




  1. $A^{-1} + frac{1}{4}X^{-1} = B$

  2. $frac{1}{4}X^{-1} = B - A^{-1}$

  3. $X^{-1} = 4(B - A^{-1})$

  4. $X = (4(B - A^{-1}))^{-1}$


I didn't find a lot of information about that, so maybe someone can share any links? Or just help me here pls.






linear-algebra matrices inverse matrix-equations







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    The expression $A-4$ is undefined: it’s the difference of a matrix and scalar. Get into the habit of writing $A-4I$ (or $4E$, if that’s what the identity matrix is called in your materials.)
    $endgroup$
    – amd
    Jan 16 at 17:53






  • 1




    $begingroup$
    $(A+4X)^{-1}$ is no more equal to $A^{-1}+frac14X^{-1}$ than $1/(a+4x)$ is equal to $1/a+1/(4x)$.
    $endgroup$
    – amd
    Jan 16 at 17:54
















0












0








0





$begingroup$


I have some equations and I don't know am I doing the simplification right. Can someone check it?



For example, we have an equation $AX = 4X + B$, where $A$ and $B$ are matrices. So, what I have done with it is




  1. $AX - 4X = B $

  2. $X(A - 4) = B$

  3. $X = B(A-4)^{-1}$


And the second example. $(A+4X)^{-1} = B$




  1. $A^{-1} + frac{1}{4}X^{-1} = B$

  2. $frac{1}{4}X^{-1} = B - A^{-1}$

  3. $X^{-1} = 4(B - A^{-1})$

  4. $X = (4(B - A^{-1}))^{-1}$


I didn't find a lot of information about that, so maybe someone can share any links? Or just help me here pls.






linear-algebra matrices inverse matrix-equations







share|cite|improve this question











$endgroup$




I have some equations and I don't know am I doing the simplification right. Can someone check it?



For example, we have an equation $AX = 4X + B$, where $A$ and $B$ are matrices. So, what I have done with it is




  1. $AX - 4X = B $

  2. $X(A - 4) = B$

  3. $X = B(A-4)^{-1}$


And the second example. $(A+4X)^{-1} = B$




  1. $A^{-1} + frac{1}{4}X^{-1} = B$

  2. $frac{1}{4}X^{-1} = B - A^{-1}$

  3. $X^{-1} = 4(B - A^{-1})$

  4. $X = (4(B - A^{-1}))^{-1}$


I didn't find a lot of information about that, so maybe someone can share any links? Or just help me here pls.






linear-algebra matrices inverse matrix-equations




linear-algebra matrices inverse matrix-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 16 at 17:50









pwerth

3,320417




3,320417










asked Jan 16 at 17:44









Aliaksei KlimovichAliaksei Klimovich

536




536








  • 1




    $begingroup$
    The expression $A-4$ is undefined: it’s the difference of a matrix and scalar. Get into the habit of writing $A-4I$ (or $4E$, if that’s what the identity matrix is called in your materials.)
    $endgroup$
    – amd
    Jan 16 at 17:53






  • 1




    $begingroup$
    $(A+4X)^{-1}$ is no more equal to $A^{-1}+frac14X^{-1}$ than $1/(a+4x)$ is equal to $1/a+1/(4x)$.
    $endgroup$
    – amd
    Jan 16 at 17:54
















  • 1




    $begingroup$
    The expression $A-4$ is undefined: it’s the difference of a matrix and scalar. Get into the habit of writing $A-4I$ (or $4E$, if that’s what the identity matrix is called in your materials.)
    $endgroup$
    – amd
    Jan 16 at 17:53






  • 1




    $begingroup$
    $(A+4X)^{-1}$ is no more equal to $A^{-1}+frac14X^{-1}$ than $1/(a+4x)$ is equal to $1/a+1/(4x)$.
    $endgroup$
    – amd
    Jan 16 at 17:54










1




1




$begingroup$
The expression $A-4$ is undefined: it’s the difference of a matrix and scalar. Get into the habit of writing $A-4I$ (or $4E$, if that’s what the identity matrix is called in your materials.)
$endgroup$
– amd
Jan 16 at 17:53




$begingroup$
The expression $A-4$ is undefined: it’s the difference of a matrix and scalar. Get into the habit of writing $A-4I$ (or $4E$, if that’s what the identity matrix is called in your materials.)
$endgroup$
– amd
Jan 16 at 17:53




1




1




$begingroup$
$(A+4X)^{-1}$ is no more equal to $A^{-1}+frac14X^{-1}$ than $1/(a+4x)$ is equal to $1/a+1/(4x)$.
$endgroup$
– amd
Jan 16 at 17:54






$begingroup$
$(A+4X)^{-1}$ is no more equal to $A^{-1}+frac14X^{-1}$ than $1/(a+4x)$ is equal to $1/a+1/(4x)$.
$endgroup$
– amd
Jan 16 at 17:54












2 Answers
2






active

oldest

votes


















2












$begingroup$

Both examples contain some errors. For the first one, you have the right idea subtracting to get
$$AX-4X=B$$
however, you have to be careful about factoring matrices. Observe that
$$AX-4X=(A-4I)X$$
where $I$ is the identity matrix. Keep in mind that "$4$" does not denote a matrix (unless we are talking about $1times1$ matrices). To complete this, we have
$$X=(A-4I)^{-1}B$$



For the second example, your first step is wrong. In general, whether for real numbers or matrices $a$ and $b$ and integers $n$, it is not true that
$$(a+b)^{n}=a^{n}+b^{n}$$



What we can do is the following:
$$(A+4X)^{-1}=B$$
$$I=B(A+4X)=BA+4BX$$
$$I-BA=4BX$$
$$frac{1}{4}(I-BA)=BX$$
$$X=frac{1}{4}B^{-1}(I-BA)$$



I should mention that with all of the above manipulations, things only make sense if the corresponding matrices are invertible. For example, if $(A-4I)$ is not invertible then we cannot simplify the first example as shown.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    It should be $AX-4X=(A-4I)X$ else matrix multiplication need not be compatible.
    $endgroup$
    – Yadati Kiran
    Jan 16 at 18:05












  • $begingroup$
    Good point. Updated my answer
    $endgroup$
    – pwerth
    Jan 16 at 18:12



















0












$begingroup$

Your first example is almost correct. You just change your $4$ to $ 4I$



Your second example needs serious adjustments.



First take inverse of both sides , then try to isolate $X$ as you did in the first example.






share|cite|improve this answer









$endgroup$














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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

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    2












    $begingroup$

    Both examples contain some errors. For the first one, you have the right idea subtracting to get
    $$AX-4X=B$$
    however, you have to be careful about factoring matrices. Observe that
    $$AX-4X=(A-4I)X$$
    where $I$ is the identity matrix. Keep in mind that "$4$" does not denote a matrix (unless we are talking about $1times1$ matrices). To complete this, we have
    $$X=(A-4I)^{-1}B$$



    For the second example, your first step is wrong. In general, whether for real numbers or matrices $a$ and $b$ and integers $n$, it is not true that
    $$(a+b)^{n}=a^{n}+b^{n}$$



    What we can do is the following:
    $$(A+4X)^{-1}=B$$
    $$I=B(A+4X)=BA+4BX$$
    $$I-BA=4BX$$
    $$frac{1}{4}(I-BA)=BX$$
    $$X=frac{1}{4}B^{-1}(I-BA)$$



    I should mention that with all of the above manipulations, things only make sense if the corresponding matrices are invertible. For example, if $(A-4I)$ is not invertible then we cannot simplify the first example as shown.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      It should be $AX-4X=(A-4I)X$ else matrix multiplication need not be compatible.
      $endgroup$
      – Yadati Kiran
      Jan 16 at 18:05












    • $begingroup$
      Good point. Updated my answer
      $endgroup$
      – pwerth
      Jan 16 at 18:12
















    2












    $begingroup$

    Both examples contain some errors. For the first one, you have the right idea subtracting to get
    $$AX-4X=B$$
    however, you have to be careful about factoring matrices. Observe that
    $$AX-4X=(A-4I)X$$
    where $I$ is the identity matrix. Keep in mind that "$4$" does not denote a matrix (unless we are talking about $1times1$ matrices). To complete this, we have
    $$X=(A-4I)^{-1}B$$



    For the second example, your first step is wrong. In general, whether for real numbers or matrices $a$ and $b$ and integers $n$, it is not true that
    $$(a+b)^{n}=a^{n}+b^{n}$$



    What we can do is the following:
    $$(A+4X)^{-1}=B$$
    $$I=B(A+4X)=BA+4BX$$
    $$I-BA=4BX$$
    $$frac{1}{4}(I-BA)=BX$$
    $$X=frac{1}{4}B^{-1}(I-BA)$$



    I should mention that with all of the above manipulations, things only make sense if the corresponding matrices are invertible. For example, if $(A-4I)$ is not invertible then we cannot simplify the first example as shown.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      It should be $AX-4X=(A-4I)X$ else matrix multiplication need not be compatible.
      $endgroup$
      – Yadati Kiran
      Jan 16 at 18:05












    • $begingroup$
      Good point. Updated my answer
      $endgroup$
      – pwerth
      Jan 16 at 18:12














    2












    2








    2





    $begingroup$

    Both examples contain some errors. For the first one, you have the right idea subtracting to get
    $$AX-4X=B$$
    however, you have to be careful about factoring matrices. Observe that
    $$AX-4X=(A-4I)X$$
    where $I$ is the identity matrix. Keep in mind that "$4$" does not denote a matrix (unless we are talking about $1times1$ matrices). To complete this, we have
    $$X=(A-4I)^{-1}B$$



    For the second example, your first step is wrong. In general, whether for real numbers or matrices $a$ and $b$ and integers $n$, it is not true that
    $$(a+b)^{n}=a^{n}+b^{n}$$



    What we can do is the following:
    $$(A+4X)^{-1}=B$$
    $$I=B(A+4X)=BA+4BX$$
    $$I-BA=4BX$$
    $$frac{1}{4}(I-BA)=BX$$
    $$X=frac{1}{4}B^{-1}(I-BA)$$



    I should mention that with all of the above manipulations, things only make sense if the corresponding matrices are invertible. For example, if $(A-4I)$ is not invertible then we cannot simplify the first example as shown.






    share|cite|improve this answer











    $endgroup$



    Both examples contain some errors. For the first one, you have the right idea subtracting to get
    $$AX-4X=B$$
    however, you have to be careful about factoring matrices. Observe that
    $$AX-4X=(A-4I)X$$
    where $I$ is the identity matrix. Keep in mind that "$4$" does not denote a matrix (unless we are talking about $1times1$ matrices). To complete this, we have
    $$X=(A-4I)^{-1}B$$



    For the second example, your first step is wrong. In general, whether for real numbers or matrices $a$ and $b$ and integers $n$, it is not true that
    $$(a+b)^{n}=a^{n}+b^{n}$$



    What we can do is the following:
    $$(A+4X)^{-1}=B$$
    $$I=B(A+4X)=BA+4BX$$
    $$I-BA=4BX$$
    $$frac{1}{4}(I-BA)=BX$$
    $$X=frac{1}{4}B^{-1}(I-BA)$$



    I should mention that with all of the above manipulations, things only make sense if the corresponding matrices are invertible. For example, if $(A-4I)$ is not invertible then we cannot simplify the first example as shown.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 16 at 18:12

























    answered Jan 16 at 17:56









    pwerthpwerth

    3,320417




    3,320417












    • $begingroup$
      It should be $AX-4X=(A-4I)X$ else matrix multiplication need not be compatible.
      $endgroup$
      – Yadati Kiran
      Jan 16 at 18:05












    • $begingroup$
      Good point. Updated my answer
      $endgroup$
      – pwerth
      Jan 16 at 18:12


















    • $begingroup$
      It should be $AX-4X=(A-4I)X$ else matrix multiplication need not be compatible.
      $endgroup$
      – Yadati Kiran
      Jan 16 at 18:05












    • $begingroup$
      Good point. Updated my answer
      $endgroup$
      – pwerth
      Jan 16 at 18:12
















    $begingroup$
    It should be $AX-4X=(A-4I)X$ else matrix multiplication need not be compatible.
    $endgroup$
    – Yadati Kiran
    Jan 16 at 18:05






    $begingroup$
    It should be $AX-4X=(A-4I)X$ else matrix multiplication need not be compatible.
    $endgroup$
    – Yadati Kiran
    Jan 16 at 18:05














    $begingroup$
    Good point. Updated my answer
    $endgroup$
    – pwerth
    Jan 16 at 18:12




    $begingroup$
    Good point. Updated my answer
    $endgroup$
    – pwerth
    Jan 16 at 18:12











    0












    $begingroup$

    Your first example is almost correct. You just change your $4$ to $ 4I$



    Your second example needs serious adjustments.



    First take inverse of both sides , then try to isolate $X$ as you did in the first example.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Your first example is almost correct. You just change your $4$ to $ 4I$



      Your second example needs serious adjustments.



      First take inverse of both sides , then try to isolate $X$ as you did in the first example.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Your first example is almost correct. You just change your $4$ to $ 4I$



        Your second example needs serious adjustments.



        First take inverse of both sides , then try to isolate $X$ as you did in the first example.






        share|cite|improve this answer









        $endgroup$



        Your first example is almost correct. You just change your $4$ to $ 4I$



        Your second example needs serious adjustments.



        First take inverse of both sides , then try to isolate $X$ as you did in the first example.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 16 at 17:59









        Mohammad Riazi-KermaniMohammad Riazi-Kermani

        41.6k42061




        41.6k42061






























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