Matrices Equations: is it Okay to Transpose Both Sides?












1












$begingroup$


I have this matrices equation:
$$(AX+I)^T=2I$$



Is it possible to transpose both sides to get this?
$$
((AX+I)^T)^T=2(I)^T
\
AX+I=2I$$
And then
$$AX=I
\X=A^{-1}
$$



Thank you.






linear-algebra matrices matrix-equations







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Its possible. In the last step, $A$ must be invertible.
    $endgroup$
    – Wuestenfux
    Dec 20 '18 at 11:46










  • $begingroup$
    It is the first property of matrix transposition.
    $endgroup$
    – farruhota
    Dec 20 '18 at 13:39
















1












$begingroup$


I have this matrices equation:
$$(AX+I)^T=2I$$



Is it possible to transpose both sides to get this?
$$
((AX+I)^T)^T=2(I)^T
\
AX+I=2I$$
And then
$$AX=I
\X=A^{-1}
$$



Thank you.






linear-algebra matrices matrix-equations







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Its possible. In the last step, $A$ must be invertible.
    $endgroup$
    – Wuestenfux
    Dec 20 '18 at 11:46










  • $begingroup$
    It is the first property of matrix transposition.
    $endgroup$
    – farruhota
    Dec 20 '18 at 13:39














1












1








1





$begingroup$


I have this matrices equation:
$$(AX+I)^T=2I$$



Is it possible to transpose both sides to get this?
$$
((AX+I)^T)^T=2(I)^T
\
AX+I=2I$$
And then
$$AX=I
\X=A^{-1}
$$



Thank you.






linear-algebra matrices matrix-equations







share|cite|improve this question









$endgroup$




I have this matrices equation:
$$(AX+I)^T=2I$$



Is it possible to transpose both sides to get this?
$$
((AX+I)^T)^T=2(I)^T
\
AX+I=2I$$
And then
$$AX=I
\X=A^{-1}
$$



Thank you.






linear-algebra matrices matrix-equations




linear-algebra matrices matrix-equations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 20 '18 at 11:44









NetanelNetanel

974




974








  • 2




    $begingroup$
    Its possible. In the last step, $A$ must be invertible.
    $endgroup$
    – Wuestenfux
    Dec 20 '18 at 11:46










  • $begingroup$
    It is the first property of matrix transposition.
    $endgroup$
    – farruhota
    Dec 20 '18 at 13:39














  • 2




    $begingroup$
    Its possible. In the last step, $A$ must be invertible.
    $endgroup$
    – Wuestenfux
    Dec 20 '18 at 11:46










  • $begingroup$
    It is the first property of matrix transposition.
    $endgroup$
    – farruhota
    Dec 20 '18 at 13:39








2




2




$begingroup$
Its possible. In the last step, $A$ must be invertible.
$endgroup$
– Wuestenfux
Dec 20 '18 at 11:46




$begingroup$
Its possible. In the last step, $A$ must be invertible.
$endgroup$
– Wuestenfux
Dec 20 '18 at 11:46












$begingroup$
It is the first property of matrix transposition.
$endgroup$
– farruhota
Dec 20 '18 at 13:39




$begingroup$
It is the first property of matrix transposition.
$endgroup$
– farruhota
Dec 20 '18 at 13:39










1 Answer
1






active

oldest

votes


















1












$begingroup$

All your steps are correct ! Your result $X=A^{-1}$ is also correct.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    $A$ must be invertible.
    $endgroup$
    – Thinking
    Dec 20 '18 at 11:46












  • $begingroup$
    If all matrices are square, then from $AX=I$ we get that $A$ is invertble !
    $endgroup$
    – Fred
    Dec 20 '18 at 11:50










  • $begingroup$
    It is necessary for $A$ to be invertible for the equation to have a solution indeed. But it is important to note that the equation has no solution if $A$ is not invertible.
    $endgroup$
    – nicomezi
    Dec 20 '18 at 12:05










  • $begingroup$
    What is your problem ? We have $(AX+I)^T=2I iff AX=I$. Hence, the equation $(AX+I)^T=2I $ implies that $A$ is invertible.
    $endgroup$
    – Fred
    Dec 20 '18 at 12:08










  • $begingroup$
    You are making the assumption that the equation has a solution in order to conclude that $A$ is invertible. That is all is being said here.
    $endgroup$
    – nicomezi
    Dec 20 '18 at 12:13











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

All your steps are correct ! Your result $X=A^{-1}$ is also correct.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    $A$ must be invertible.
    $endgroup$
    – Thinking
    Dec 20 '18 at 11:46












  • $begingroup$
    If all matrices are square, then from $AX=I$ we get that $A$ is invertble !
    $endgroup$
    – Fred
    Dec 20 '18 at 11:50










  • $begingroup$
    It is necessary for $A$ to be invertible for the equation to have a solution indeed. But it is important to note that the equation has no solution if $A$ is not invertible.
    $endgroup$
    – nicomezi
    Dec 20 '18 at 12:05










  • $begingroup$
    What is your problem ? We have $(AX+I)^T=2I iff AX=I$. Hence, the equation $(AX+I)^T=2I $ implies that $A$ is invertible.
    $endgroup$
    – Fred
    Dec 20 '18 at 12:08










  • $begingroup$
    You are making the assumption that the equation has a solution in order to conclude that $A$ is invertible. That is all is being said here.
    $endgroup$
    – nicomezi
    Dec 20 '18 at 12:13
















1












$begingroup$

All your steps are correct ! Your result $X=A^{-1}$ is also correct.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    $A$ must be invertible.
    $endgroup$
    – Thinking
    Dec 20 '18 at 11:46












  • $begingroup$
    If all matrices are square, then from $AX=I$ we get that $A$ is invertble !
    $endgroup$
    – Fred
    Dec 20 '18 at 11:50










  • $begingroup$
    It is necessary for $A$ to be invertible for the equation to have a solution indeed. But it is important to note that the equation has no solution if $A$ is not invertible.
    $endgroup$
    – nicomezi
    Dec 20 '18 at 12:05










  • $begingroup$
    What is your problem ? We have $(AX+I)^T=2I iff AX=I$. Hence, the equation $(AX+I)^T=2I $ implies that $A$ is invertible.
    $endgroup$
    – Fred
    Dec 20 '18 at 12:08










  • $begingroup$
    You are making the assumption that the equation has a solution in order to conclude that $A$ is invertible. That is all is being said here.
    $endgroup$
    – nicomezi
    Dec 20 '18 at 12:13














1












1








1





$begingroup$

All your steps are correct ! Your result $X=A^{-1}$ is also correct.






share|cite|improve this answer









$endgroup$



All your steps are correct ! Your result $X=A^{-1}$ is also correct.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 20 '18 at 11:46









FredFred

46.8k1848




46.8k1848








  • 1




    $begingroup$
    $A$ must be invertible.
    $endgroup$
    – Thinking
    Dec 20 '18 at 11:46












  • $begingroup$
    If all matrices are square, then from $AX=I$ we get that $A$ is invertble !
    $endgroup$
    – Fred
    Dec 20 '18 at 11:50










  • $begingroup$
    It is necessary for $A$ to be invertible for the equation to have a solution indeed. But it is important to note that the equation has no solution if $A$ is not invertible.
    $endgroup$
    – nicomezi
    Dec 20 '18 at 12:05










  • $begingroup$
    What is your problem ? We have $(AX+I)^T=2I iff AX=I$. Hence, the equation $(AX+I)^T=2I $ implies that $A$ is invertible.
    $endgroup$
    – Fred
    Dec 20 '18 at 12:08










  • $begingroup$
    You are making the assumption that the equation has a solution in order to conclude that $A$ is invertible. That is all is being said here.
    $endgroup$
    – nicomezi
    Dec 20 '18 at 12:13














  • 1




    $begingroup$
    $A$ must be invertible.
    $endgroup$
    – Thinking
    Dec 20 '18 at 11:46












  • $begingroup$
    If all matrices are square, then from $AX=I$ we get that $A$ is invertble !
    $endgroup$
    – Fred
    Dec 20 '18 at 11:50










  • $begingroup$
    It is necessary for $A$ to be invertible for the equation to have a solution indeed. But it is important to note that the equation has no solution if $A$ is not invertible.
    $endgroup$
    – nicomezi
    Dec 20 '18 at 12:05










  • $begingroup$
    What is your problem ? We have $(AX+I)^T=2I iff AX=I$. Hence, the equation $(AX+I)^T=2I $ implies that $A$ is invertible.
    $endgroup$
    – Fred
    Dec 20 '18 at 12:08










  • $begingroup$
    You are making the assumption that the equation has a solution in order to conclude that $A$ is invertible. That is all is being said here.
    $endgroup$
    – nicomezi
    Dec 20 '18 at 12:13








1




1




$begingroup$
$A$ must be invertible.
$endgroup$
– Thinking
Dec 20 '18 at 11:46






$begingroup$
$A$ must be invertible.
$endgroup$
– Thinking
Dec 20 '18 at 11:46














$begingroup$
If all matrices are square, then from $AX=I$ we get that $A$ is invertble !
$endgroup$
– Fred
Dec 20 '18 at 11:50




$begingroup$
If all matrices are square, then from $AX=I$ we get that $A$ is invertble !
$endgroup$
– Fred
Dec 20 '18 at 11:50












$begingroup$
It is necessary for $A$ to be invertible for the equation to have a solution indeed. But it is important to note that the equation has no solution if $A$ is not invertible.
$endgroup$
– nicomezi
Dec 20 '18 at 12:05




$begingroup$
It is necessary for $A$ to be invertible for the equation to have a solution indeed. But it is important to note that the equation has no solution if $A$ is not invertible.
$endgroup$
– nicomezi
Dec 20 '18 at 12:05












$begingroup$
What is your problem ? We have $(AX+I)^T=2I iff AX=I$. Hence, the equation $(AX+I)^T=2I $ implies that $A$ is invertible.
$endgroup$
– Fred
Dec 20 '18 at 12:08




$begingroup$
What is your problem ? We have $(AX+I)^T=2I iff AX=I$. Hence, the equation $(AX+I)^T=2I $ implies that $A$ is invertible.
$endgroup$
– Fred
Dec 20 '18 at 12:08












$begingroup$
You are making the assumption that the equation has a solution in order to conclude that $A$ is invertible. That is all is being said here.
$endgroup$
– nicomezi
Dec 20 '18 at 12:13




$begingroup$
You are making the assumption that the equation has a solution in order to conclude that $A$ is invertible. That is all is being said here.
$endgroup$
– nicomezi
Dec 20 '18 at 12:13


















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