Invertible matrix properties of a matrix












1












$begingroup$


I have here the following question:




Let $X$ be the $5 times 5$ matrix "full of ones":



$X = begin{pmatrix}1 & 1 & 1 & 1 & 1 \ 1 & 1 & 1 & 1 & 1 \ 1 & 1 & 1 & 1 & 1 \ 1 & 1 & 1 & 1 & 1\ 1 & 1 & 1 & 1 & 1end{pmatrix}$



$(a)$ Is $X$ invertible? Explain.



$(b)$ Find a number $c$ such that $X^2=cX$.



$(c)$ Compute $(X-aI_5)(X+(a-c)I_5)$, where $a$ is a real number, $c$ is the constant from part $(b)$, $aneq 0$, and $aneq c$. If $M=(X-aI_5)$, what is $M^{-1}$? (You may express your answer in terms of $X,I_5,a$, and $c.$)




I already did part $(a)$ and $(b)$.



$(a)$ No. It's not invertible since there are two or more identical rows or columns so the determinant would be $0$ and hence it is not invertible.



$(b)$ $X^2 = begin{pmatrix}5 & 5 & 5 & 5 & 5 \ 5 & 5 & 5 & 5 & 5 \ 5 & 5 & 5 & 5 & 5 \ 5 & 5 & 5 & 5 & 5\ 5 & 5 & 5 & 5 & 5 end{pmatrix}$



Thus, $c=5$.



$(c)$ I almost got part $(c)$ too. It's the last part I can't figure out.



$=(X-aI_5)(X+(a-c)I_5)$



$=-a(a-c)I_5$



If $M=(X-aI_5)$, that would mean if I multiply both sides by $(X+(a-c)I_5)$, I get that:



$$M(X+(a-c)I_5)=(X-aI_5)(X+(a-c)I_5)$$
$$M(X+(a-c)I_5)=-a(a-c)I_5$$



I can literally see the answer in front of me, but it's not quite there. How do I proceed from here?



For reference, the answer should be:



$M^{-1}=frac{X+(a-c)I_5}{-a(a-c)}$










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    HINT:$MM^{-1}=M^{-1}M=I $.
    $endgroup$
    – Thomas Shelby
    Dec 15 '18 at 8:18










  • $begingroup$
    Yes I know that but how does that help? My problem is that I need to divide what are essentially matrices, and I don't think that's allowed.
    $endgroup$
    – Future Math person
    Dec 15 '18 at 8:21






  • 1




    $begingroup$
    Note that $-a(a-c) in Bbb R $.
    $endgroup$
    – Thomas Shelby
    Dec 15 '18 at 8:26






  • 1




    $begingroup$
    Never mind I get it! Thanks a lot!!
    $endgroup$
    – Future Math person
    Dec 15 '18 at 8:29






  • 1




    $begingroup$
    Just divide the whole equation by the scalar and compare.
    $endgroup$
    – Thomas Shelby
    Dec 15 '18 at 8:29
















1












$begingroup$


I have here the following question:




Let $X$ be the $5 times 5$ matrix "full of ones":



$X = begin{pmatrix}1 & 1 & 1 & 1 & 1 \ 1 & 1 & 1 & 1 & 1 \ 1 & 1 & 1 & 1 & 1 \ 1 & 1 & 1 & 1 & 1\ 1 & 1 & 1 & 1 & 1end{pmatrix}$



$(a)$ Is $X$ invertible? Explain.



$(b)$ Find a number $c$ such that $X^2=cX$.



$(c)$ Compute $(X-aI_5)(X+(a-c)I_5)$, where $a$ is a real number, $c$ is the constant from part $(b)$, $aneq 0$, and $aneq c$. If $M=(X-aI_5)$, what is $M^{-1}$? (You may express your answer in terms of $X,I_5,a$, and $c.$)




I already did part $(a)$ and $(b)$.



$(a)$ No. It's not invertible since there are two or more identical rows or columns so the determinant would be $0$ and hence it is not invertible.



$(b)$ $X^2 = begin{pmatrix}5 & 5 & 5 & 5 & 5 \ 5 & 5 & 5 & 5 & 5 \ 5 & 5 & 5 & 5 & 5 \ 5 & 5 & 5 & 5 & 5\ 5 & 5 & 5 & 5 & 5 end{pmatrix}$



Thus, $c=5$.



$(c)$ I almost got part $(c)$ too. It's the last part I can't figure out.



$=(X-aI_5)(X+(a-c)I_5)$



$=-a(a-c)I_5$



If $M=(X-aI_5)$, that would mean if I multiply both sides by $(X+(a-c)I_5)$, I get that:



$$M(X+(a-c)I_5)=(X-aI_5)(X+(a-c)I_5)$$
$$M(X+(a-c)I_5)=-a(a-c)I_5$$



I can literally see the answer in front of me, but it's not quite there. How do I proceed from here?



For reference, the answer should be:



$M^{-1}=frac{X+(a-c)I_5}{-a(a-c)}$










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    HINT:$MM^{-1}=M^{-1}M=I $.
    $endgroup$
    – Thomas Shelby
    Dec 15 '18 at 8:18










  • $begingroup$
    Yes I know that but how does that help? My problem is that I need to divide what are essentially matrices, and I don't think that's allowed.
    $endgroup$
    – Future Math person
    Dec 15 '18 at 8:21






  • 1




    $begingroup$
    Note that $-a(a-c) in Bbb R $.
    $endgroup$
    – Thomas Shelby
    Dec 15 '18 at 8:26






  • 1




    $begingroup$
    Never mind I get it! Thanks a lot!!
    $endgroup$
    – Future Math person
    Dec 15 '18 at 8:29






  • 1




    $begingroup$
    Just divide the whole equation by the scalar and compare.
    $endgroup$
    – Thomas Shelby
    Dec 15 '18 at 8:29














1












1








1


1



$begingroup$


I have here the following question:




Let $X$ be the $5 times 5$ matrix "full of ones":



$X = begin{pmatrix}1 & 1 & 1 & 1 & 1 \ 1 & 1 & 1 & 1 & 1 \ 1 & 1 & 1 & 1 & 1 \ 1 & 1 & 1 & 1 & 1\ 1 & 1 & 1 & 1 & 1end{pmatrix}$



$(a)$ Is $X$ invertible? Explain.



$(b)$ Find a number $c$ such that $X^2=cX$.



$(c)$ Compute $(X-aI_5)(X+(a-c)I_5)$, where $a$ is a real number, $c$ is the constant from part $(b)$, $aneq 0$, and $aneq c$. If $M=(X-aI_5)$, what is $M^{-1}$? (You may express your answer in terms of $X,I_5,a$, and $c.$)




I already did part $(a)$ and $(b)$.



$(a)$ No. It's not invertible since there are two or more identical rows or columns so the determinant would be $0$ and hence it is not invertible.



$(b)$ $X^2 = begin{pmatrix}5 & 5 & 5 & 5 & 5 \ 5 & 5 & 5 & 5 & 5 \ 5 & 5 & 5 & 5 & 5 \ 5 & 5 & 5 & 5 & 5\ 5 & 5 & 5 & 5 & 5 end{pmatrix}$



Thus, $c=5$.



$(c)$ I almost got part $(c)$ too. It's the last part I can't figure out.



$=(X-aI_5)(X+(a-c)I_5)$



$=-a(a-c)I_5$



If $M=(X-aI_5)$, that would mean if I multiply both sides by $(X+(a-c)I_5)$, I get that:



$$M(X+(a-c)I_5)=(X-aI_5)(X+(a-c)I_5)$$
$$M(X+(a-c)I_5)=-a(a-c)I_5$$



I can literally see the answer in front of me, but it's not quite there. How do I proceed from here?



For reference, the answer should be:



$M^{-1}=frac{X+(a-c)I_5}{-a(a-c)}$










share|cite|improve this question











$endgroup$




I have here the following question:




Let $X$ be the $5 times 5$ matrix "full of ones":



$X = begin{pmatrix}1 & 1 & 1 & 1 & 1 \ 1 & 1 & 1 & 1 & 1 \ 1 & 1 & 1 & 1 & 1 \ 1 & 1 & 1 & 1 & 1\ 1 & 1 & 1 & 1 & 1end{pmatrix}$



$(a)$ Is $X$ invertible? Explain.



$(b)$ Find a number $c$ such that $X^2=cX$.



$(c)$ Compute $(X-aI_5)(X+(a-c)I_5)$, where $a$ is a real number, $c$ is the constant from part $(b)$, $aneq 0$, and $aneq c$. If $M=(X-aI_5)$, what is $M^{-1}$? (You may express your answer in terms of $X,I_5,a$, and $c.$)




I already did part $(a)$ and $(b)$.



$(a)$ No. It's not invertible since there are two or more identical rows or columns so the determinant would be $0$ and hence it is not invertible.



$(b)$ $X^2 = begin{pmatrix}5 & 5 & 5 & 5 & 5 \ 5 & 5 & 5 & 5 & 5 \ 5 & 5 & 5 & 5 & 5 \ 5 & 5 & 5 & 5 & 5\ 5 & 5 & 5 & 5 & 5 end{pmatrix}$



Thus, $c=5$.



$(c)$ I almost got part $(c)$ too. It's the last part I can't figure out.



$=(X-aI_5)(X+(a-c)I_5)$



$=-a(a-c)I_5$



If $M=(X-aI_5)$, that would mean if I multiply both sides by $(X+(a-c)I_5)$, I get that:



$$M(X+(a-c)I_5)=(X-aI_5)(X+(a-c)I_5)$$
$$M(X+(a-c)I_5)=-a(a-c)I_5$$



I can literally see the answer in front of me, but it's not quite there. How do I proceed from here?



For reference, the answer should be:



$M^{-1}=frac{X+(a-c)I_5}{-a(a-c)}$







linear-algebra matrices inverse






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 15 '18 at 8:26







Future Math person

















asked Dec 15 '18 at 8:14









Future Math personFuture Math person

977817




977817








  • 1




    $begingroup$
    HINT:$MM^{-1}=M^{-1}M=I $.
    $endgroup$
    – Thomas Shelby
    Dec 15 '18 at 8:18










  • $begingroup$
    Yes I know that but how does that help? My problem is that I need to divide what are essentially matrices, and I don't think that's allowed.
    $endgroup$
    – Future Math person
    Dec 15 '18 at 8:21






  • 1




    $begingroup$
    Note that $-a(a-c) in Bbb R $.
    $endgroup$
    – Thomas Shelby
    Dec 15 '18 at 8:26






  • 1




    $begingroup$
    Never mind I get it! Thanks a lot!!
    $endgroup$
    – Future Math person
    Dec 15 '18 at 8:29






  • 1




    $begingroup$
    Just divide the whole equation by the scalar and compare.
    $endgroup$
    – Thomas Shelby
    Dec 15 '18 at 8:29














  • 1




    $begingroup$
    HINT:$MM^{-1}=M^{-1}M=I $.
    $endgroup$
    – Thomas Shelby
    Dec 15 '18 at 8:18










  • $begingroup$
    Yes I know that but how does that help? My problem is that I need to divide what are essentially matrices, and I don't think that's allowed.
    $endgroup$
    – Future Math person
    Dec 15 '18 at 8:21






  • 1




    $begingroup$
    Note that $-a(a-c) in Bbb R $.
    $endgroup$
    – Thomas Shelby
    Dec 15 '18 at 8:26






  • 1




    $begingroup$
    Never mind I get it! Thanks a lot!!
    $endgroup$
    – Future Math person
    Dec 15 '18 at 8:29






  • 1




    $begingroup$
    Just divide the whole equation by the scalar and compare.
    $endgroup$
    – Thomas Shelby
    Dec 15 '18 at 8:29








1




1




$begingroup$
HINT:$MM^{-1}=M^{-1}M=I $.
$endgroup$
– Thomas Shelby
Dec 15 '18 at 8:18




$begingroup$
HINT:$MM^{-1}=M^{-1}M=I $.
$endgroup$
– Thomas Shelby
Dec 15 '18 at 8:18












$begingroup$
Yes I know that but how does that help? My problem is that I need to divide what are essentially matrices, and I don't think that's allowed.
$endgroup$
– Future Math person
Dec 15 '18 at 8:21




$begingroup$
Yes I know that but how does that help? My problem is that I need to divide what are essentially matrices, and I don't think that's allowed.
$endgroup$
– Future Math person
Dec 15 '18 at 8:21




1




1




$begingroup$
Note that $-a(a-c) in Bbb R $.
$endgroup$
– Thomas Shelby
Dec 15 '18 at 8:26




$begingroup$
Note that $-a(a-c) in Bbb R $.
$endgroup$
– Thomas Shelby
Dec 15 '18 at 8:26




1




1




$begingroup$
Never mind I get it! Thanks a lot!!
$endgroup$
– Future Math person
Dec 15 '18 at 8:29




$begingroup$
Never mind I get it! Thanks a lot!!
$endgroup$
– Future Math person
Dec 15 '18 at 8:29




1




1




$begingroup$
Just divide the whole equation by the scalar and compare.
$endgroup$
– Thomas Shelby
Dec 15 '18 at 8:29




$begingroup$
Just divide the whole equation by the scalar and compare.
$endgroup$
– Thomas Shelby
Dec 15 '18 at 8:29










1 Answer
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$begingroup$

Hint. You have $M(X+(a-c)I_5)=-a(a-c)I_5$. Pre-multiply by $M^{-1}$.






share|cite|improve this answer









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    $begingroup$

    Hint. You have $M(X+(a-c)I_5)=-a(a-c)I_5$. Pre-multiply by $M^{-1}$.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Hint. You have $M(X+(a-c)I_5)=-a(a-c)I_5$. Pre-multiply by $M^{-1}$.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Hint. You have $M(X+(a-c)I_5)=-a(a-c)I_5$. Pre-multiply by $M^{-1}$.






        share|cite|improve this answer









        $endgroup$



        Hint. You have $M(X+(a-c)I_5)=-a(a-c)I_5$. Pre-multiply by $M^{-1}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 15 '18 at 8:28









        Shubham JohriShubham Johri

        5,204718




        5,204718






























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