Trick for Inverse Hollow Matrix Calculation (Self-Answered)












1














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






















  • What do you mean by "a hollow matrix" ?
    – Jean Marie
    5 hours ago
















1














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






















  • What do you mean by "a hollow matrix" ?
    – Jean Marie
    5 hours ago














1












1








1







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













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 2 days ago









NetUser5y62NetUser5y62

418214




418214












  • What do you mean by "a hollow matrix" ?
    – Jean Marie
    5 hours ago


















  • What do you mean by "a hollow matrix" ?
    – Jean Marie
    5 hours ago
















What do you mean by "a hollow matrix" ?
– Jean Marie
5 hours ago




What do you mean by "a hollow matrix" ?
– Jean Marie
5 hours ago










1 Answer
1






active

oldest

votes


















2














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























  • You can generalize this to $alpha I+beta B$.
    – amd
    2 days ago












  • This is in fact a classical method, with the use of Cayley-Hamilton theorem (see yutsumura.com/…) in general, or the minimal polynomial.
    – Jean Marie
    5 hours ago











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1 Answer
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1






active

oldest

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oldest

votes






active

oldest

votes









2














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























  • You can generalize this to $alpha I+beta B$.
    – amd
    2 days ago












  • This is in fact a classical method, with the use of Cayley-Hamilton theorem (see yutsumura.com/…) in general, or the minimal polynomial.
    – Jean Marie
    5 hours ago
















2














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























  • You can generalize this to $alpha I+beta B$.
    – amd
    2 days ago












  • This is in fact a classical method, with the use of Cayley-Hamilton theorem (see yutsumura.com/…) in general, or the minimal polynomial.
    – Jean Marie
    5 hours ago














2












2








2






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














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 2 days ago

























answered 2 days ago









NetUser5y62NetUser5y62

418214




418214












  • You can generalize this to $alpha I+beta B$.
    – amd
    2 days ago












  • This is in fact a classical method, with the use of Cayley-Hamilton theorem (see yutsumura.com/…) in general, or the minimal polynomial.
    – Jean Marie
    5 hours ago


















  • You can generalize this to $alpha I+beta B$.
    – amd
    2 days ago












  • This is in fact a classical method, with the use of Cayley-Hamilton theorem (see yutsumura.com/…) in general, or the minimal polynomial.
    – Jean Marie
    5 hours ago
















You can generalize this to $alpha I+beta B$.
– amd
2 days ago






You can generalize this to $alpha I+beta B$.
– amd
2 days ago














This is in fact a classical method, with the use of Cayley-Hamilton theorem (see yutsumura.com/…) in general, or the minimal polynomial.
– Jean Marie
5 hours ago




This is in fact a classical method, with the use of Cayley-Hamilton theorem (see yutsumura.com/…) in general, or the minimal polynomial.
– Jean Marie
5 hours ago


















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