Trick for Inverse Hollow Matrix Calculation (Self-Answered)












1












$begingroup$


Let $A$ be the hollow matrix :
$$
A=begin{pmatrix}
0&1&1&1\
1&0&1&1\
1&1&0&1\
1&1&1&0
end{pmatrix}
$$



Find the inverse matrix $A^{-1}$ without using any elementary row/ column operations.






linear-algebra matrices inverse







share|cite|improve this question









$endgroup$












  • $begingroup$
    What do you mean by "a hollow matrix" ?
    $endgroup$
    – Jean Marie
    Jan 7 at 20:49
















1












$begingroup$


Let $A$ be the hollow matrix :
$$
A=begin{pmatrix}
0&1&1&1\
1&0&1&1\
1&1&0&1\
1&1&1&0
end{pmatrix}
$$



Find the inverse matrix $A^{-1}$ without using any elementary row/ column operations.






linear-algebra matrices inverse







share|cite|improve this question









$endgroup$












  • $begingroup$
    What do you mean by "a hollow matrix" ?
    $endgroup$
    – Jean Marie
    Jan 7 at 20:49














1












1








1





$begingroup$


Let $A$ be the hollow matrix :
$$
A=begin{pmatrix}
0&1&1&1\
1&0&1&1\
1&1&0&1\
1&1&1&0
end{pmatrix}
$$



Find the inverse matrix $A^{-1}$ without using any elementary row/ column operations.






linear-algebra matrices inverse







share|cite|improve this question









$endgroup$




Let $A$ be the hollow matrix :
$$
A=begin{pmatrix}
0&1&1&1\
1&0&1&1\
1&1&0&1\
1&1&1&0
end{pmatrix}
$$



Find the inverse matrix $A^{-1}$ without using any elementary row/ column operations.






linear-algebra matrices inverse




linear-algebra matrices inverse






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 5 at 15:20









NetUser5y62NetUser5y62

525215




525215












  • $begingroup$
    What do you mean by "a hollow matrix" ?
    $endgroup$
    – Jean Marie
    Jan 7 at 20:49


















  • $begingroup$
    What do you mean by "a hollow matrix" ?
    $endgroup$
    – Jean Marie
    Jan 7 at 20:49
















$begingroup$
What do you mean by "a hollow matrix" ?
$endgroup$
– Jean Marie
Jan 7 at 20:49




$begingroup$
What do you mean by "a hollow matrix" ?
$endgroup$
– Jean Marie
Jan 7 at 20:49










1 Answer
1






active

oldest

votes


















2












$begingroup$

Notice that
$$
A=
begin{pmatrix}
1&1&1&1\
1&1&1&1\
1&1&1&1\
1&1&1&1
end{pmatrix} - I_{4times4}
$$



Let
$$
B=
begin{pmatrix}
1&1&1&1\
1&1&1&1\
1&1&1&1\
1&1&1&1
end{pmatrix}
$$



then $B=A+I$. Observe that $B^2 = (A+I)^2=4B$. Expanding,
$$
A^2+2A+I = 4(A+I) implies A^2-2A=3I
$$
Therefore $A^{-1}=frac{1}{3}(A-2I)$.



More generally, this method gives us an easy way to express the inverse of a hollow matrix $A$ (of the particular stated form) of any order $n$, which is the main reason for posting this question.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    You can generalize this to $alpha I+beta B$.
    $endgroup$
    – amd
    Jan 5 at 20:00












  • $begingroup$
    This is in fact a classical method, with the use of Cayley-Hamilton theorem (see yutsumura.com/…) in general, or the minimal polynomial.
    $endgroup$
    – Jean Marie
    Jan 7 at 20:48












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1 Answer
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active

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Notice that
$$
A=
begin{pmatrix}
1&1&1&1\
1&1&1&1\
1&1&1&1\
1&1&1&1
end{pmatrix} - I_{4times4}
$$



Let
$$
B=
begin{pmatrix}
1&1&1&1\
1&1&1&1\
1&1&1&1\
1&1&1&1
end{pmatrix}
$$



then $B=A+I$. Observe that $B^2 = (A+I)^2=4B$. Expanding,
$$
A^2+2A+I = 4(A+I) implies A^2-2A=3I
$$
Therefore $A^{-1}=frac{1}{3}(A-2I)$.



More generally, this method gives us an easy way to express the inverse of a hollow matrix $A$ (of the particular stated form) of any order $n$, which is the main reason for posting this question.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    You can generalize this to $alpha I+beta B$.
    $endgroup$
    – amd
    Jan 5 at 20:00












  • $begingroup$
    This is in fact a classical method, with the use of Cayley-Hamilton theorem (see yutsumura.com/…) in general, or the minimal polynomial.
    $endgroup$
    – Jean Marie
    Jan 7 at 20:48
















2












$begingroup$

Notice that
$$
A=
begin{pmatrix}
1&1&1&1\
1&1&1&1\
1&1&1&1\
1&1&1&1
end{pmatrix} - I_{4times4}
$$



Let
$$
B=
begin{pmatrix}
1&1&1&1\
1&1&1&1\
1&1&1&1\
1&1&1&1
end{pmatrix}
$$



then $B=A+I$. Observe that $B^2 = (A+I)^2=4B$. Expanding,
$$
A^2+2A+I = 4(A+I) implies A^2-2A=3I
$$
Therefore $A^{-1}=frac{1}{3}(A-2I)$.



More generally, this method gives us an easy way to express the inverse of a hollow matrix $A$ (of the particular stated form) of any order $n$, which is the main reason for posting this question.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    You can generalize this to $alpha I+beta B$.
    $endgroup$
    – amd
    Jan 5 at 20:00












  • $begingroup$
    This is in fact a classical method, with the use of Cayley-Hamilton theorem (see yutsumura.com/…) in general, or the minimal polynomial.
    $endgroup$
    – Jean Marie
    Jan 7 at 20:48














2












2








2





$begingroup$

Notice that
$$
A=
begin{pmatrix}
1&1&1&1\
1&1&1&1\
1&1&1&1\
1&1&1&1
end{pmatrix} - I_{4times4}
$$



Let
$$
B=
begin{pmatrix}
1&1&1&1\
1&1&1&1\
1&1&1&1\
1&1&1&1
end{pmatrix}
$$



then $B=A+I$. Observe that $B^2 = (A+I)^2=4B$. Expanding,
$$
A^2+2A+I = 4(A+I) implies A^2-2A=3I
$$
Therefore $A^{-1}=frac{1}{3}(A-2I)$.



More generally, this method gives us an easy way to express the inverse of a hollow matrix $A$ (of the particular stated form) of any order $n$, which is the main reason for posting this question.






share|cite|improve this answer











$endgroup$



Notice that
$$
A=
begin{pmatrix}
1&1&1&1\
1&1&1&1\
1&1&1&1\
1&1&1&1
end{pmatrix} - I_{4times4}
$$



Let
$$
B=
begin{pmatrix}
1&1&1&1\
1&1&1&1\
1&1&1&1\
1&1&1&1
end{pmatrix}
$$



then $B=A+I$. Observe that $B^2 = (A+I)^2=4B$. Expanding,
$$
A^2+2A+I = 4(A+I) implies A^2-2A=3I
$$
Therefore $A^{-1}=frac{1}{3}(A-2I)$.



More generally, this method gives us an easy way to express the inverse of a hollow matrix $A$ (of the particular stated form) of any order $n$, which is the main reason for posting this question.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 5 at 15:38

























answered Jan 5 at 15:20









NetUser5y62NetUser5y62

525215




525215












  • $begingroup$
    You can generalize this to $alpha I+beta B$.
    $endgroup$
    – amd
    Jan 5 at 20:00












  • $begingroup$
    This is in fact a classical method, with the use of Cayley-Hamilton theorem (see yutsumura.com/…) in general, or the minimal polynomial.
    $endgroup$
    – Jean Marie
    Jan 7 at 20:48


















  • $begingroup$
    You can generalize this to $alpha I+beta B$.
    $endgroup$
    – amd
    Jan 5 at 20:00












  • $begingroup$
    This is in fact a classical method, with the use of Cayley-Hamilton theorem (see yutsumura.com/…) in general, or the minimal polynomial.
    $endgroup$
    – Jean Marie
    Jan 7 at 20:48
















$begingroup$
You can generalize this to $alpha I+beta B$.
$endgroup$
– amd
Jan 5 at 20:00






$begingroup$
You can generalize this to $alpha I+beta B$.
$endgroup$
– amd
Jan 5 at 20:00














$begingroup$
This is in fact a classical method, with the use of Cayley-Hamilton theorem (see yutsumura.com/…) in general, or the minimal polynomial.
$endgroup$
– Jean Marie
Jan 7 at 20:48




$begingroup$
This is in fact a classical method, with the use of Cayley-Hamilton theorem (see yutsumura.com/…) in general, or the minimal polynomial.
$endgroup$
– Jean Marie
Jan 7 at 20:48


















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