Mixing
Finding inverse of Elementary matrices
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I wish to prove that for an identity matrix I of order n , n is a + ve integer, inverse of a matrix A = [a_ij] obtained by interchanging any two rows / columns of identity matrix I is again the same matrix A.
i.e. Inv( E_ij)= E_ij , 1<= i,j<=n
where E_ij is elementary matrix obtained by interchanging R_i with R_j or by interchanging C_i withC_j.
What I have done so far?
I got that det(E_ij) = -1 as interchanging any two rows / columns only changes the sign of a determinant.
I tried finding out adj( E_ij)
,but unable to calculate cofactors of two elements a_ij and a_ji ,both of which are unity. Any help to find adj(E_ij).
matrices
asked Jan 20 at 2:32
Shivam KumarShivam Kumar
105
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add a comment |
$begingroup$
I wish to prove that for an identity matrix I of order n , n is a + ve integer, inverse of a matrix A = [a_ij] obtained by interchanging any two rows / columns of identity matrix I is again the same matrix A.
i.e. Inv( E_ij)= E_ij , 1<= i,j<=n
where E_ij is elementary matrix obtained by interchanging R_i with R_j or by interchanging C_i withC_j.
What I have done so far?
I got that det(E_ij) = -1 as interchanging any two rows / columns only changes the sign of a determinant.
I tried finding out adj( E_ij)
,but unable to calculate cofactors of two elements a_ij and a_ji ,both of which are unity. Any help to find adj(E_ij).
matrices
asked Jan 20 at 2:32
Shivam KumarShivam Kumar
105
$endgroup$
$begingroup$
What does the identity matrix $I$ have to do with the inverse of $A$, or $E$. Can you please give a simple example with $n=3$ in order to understand what the thought process is here.
$endgroup$
– ja72
Jan 20 at 2:40
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Sir, A or E_ij is the matrix obtained from Identity matrix I by interchanging i'th and j 'th rows of I
$endgroup$
– Shivam Kumar
Jan 20 at 2:57
$begingroup$
So like $$ begin{bmatrix} & & 1\ 1\ & 1 end{bmatrix}begin{bmatrix} & 1\ & & 1\ 1 end{bmatrix}=begin{bmatrix}1\ & 1\ & & 1 end{bmatrix} $$ ?
$endgroup$
– ja72
Jan 20 at 19:37
add a comment |
$begingroup$
I wish to prove that for an identity matrix I of order n , n is a + ve integer, inverse of a matrix A = [a_ij] obtained by interchanging any two rows / columns of identity matrix I is again the same matrix A.
i.e. Inv( E_ij)= E_ij , 1<= i,j<=n
where E_ij is elementary matrix obtained by interchanging R_i with R_j or by interchanging C_i withC_j.
What I have done so far?
I got that det(E_ij) = -1 as interchanging any two rows / columns only changes the sign of a determinant.
I tried finding out adj( E_ij)
,but unable to calculate cofactors of two elements a_ij and a_ji ,both of which are unity. Any help to find adj(E_ij).
matrices
asked Jan 20 at 2:32
Shivam KumarShivam Kumar
105
$endgroup$
I wish to prove that for an identity matrix I of order n , n is a + ve integer, inverse of a matrix A = [a_ij] obtained by interchanging any two rows / columns of identity matrix I is again the same matrix A.
i.e. Inv( E_ij)= E_ij , 1<= i,j<=n
where E_ij is elementary matrix obtained by interchanging R_i with R_j or by interchanging C_i withC_j.
What I have done so far?
I got that det(E_ij) = -1 as interchanging any two rows / columns only changes the sign of a determinant.
I tried finding out adj( E_ij)
,but unable to calculate cofactors of two elements a_ij and a_ji ,both of which are unity. Any help to find adj(E_ij).
matrices
matrices
asked Jan 20 at 2:32
Shivam KumarShivam Kumar
105
asked Jan 20 at 2:32
Shivam KumarShivam Kumar
105
edited Jan 20 at 2:54
Shivam Kumar
asked Jan 20 at 2:32
Shivam KumarShivam Kumar
105
asked Jan 20 at 2:32
Shivam KumarShivam Kumar
105
asked Jan 20 at 2:32
Shivam KumarShivam Kumar
105
105
$begingroup$
What does the identity matrix $I$ have to do with the inverse of $A$, or $E$. Can you please give a simple example with $n=3$ in order to understand what the thought process is here.
$endgroup$
– ja72
Jan 20 at 2:40
$begingroup$
Sir, A or E_ij is the matrix obtained from Identity matrix I by interchanging i'th and j 'th rows of I
$endgroup$
– Shivam Kumar
Jan 20 at 2:57
$begingroup$
So like $$ begin{bmatrix} & & 1\ 1\ & 1 end{bmatrix}begin{bmatrix} & 1\ & & 1\ 1 end{bmatrix}=begin{bmatrix}1\ & 1\ & & 1 end{bmatrix} $$ ?
$endgroup$
– ja72
Jan 20 at 19:37
add a comment |
$begingroup$
What does the identity matrix $I$ have to do with the inverse of $A$, or $E$. Can you please give a simple example with $n=3$ in order to understand what the thought process is here.
$endgroup$
– ja72
Jan 20 at 2:40
$begingroup$
Sir, A or E_ij is the matrix obtained from Identity matrix I by interchanging i'th and j 'th rows of I
$endgroup$
– Shivam Kumar
Jan 20 at 2:57
$begingroup$
So like $$ begin{bmatrix} & & 1\ 1\ & 1 end{bmatrix}begin{bmatrix} & 1\ & & 1\ 1 end{bmatrix}=begin{bmatrix}1\ & 1\ & & 1 end{bmatrix} $$ ?
$endgroup$
– ja72
Jan 20 at 19:37
$begingroup$
What does the identity matrix $I$ have to do with the inverse of $A$, or $E$. Can you please give a simple example with $n=3$ in order to understand what the thought process is here.
$endgroup$
– ja72
Jan 20 at 2:40
$begingroup$
What does the identity matrix $I$ have to do with the inverse of $A$, or $E$. Can you please give a simple example with $n=3$ in order to understand what the thought process is here.
$endgroup$
– ja72
Jan 20 at 2:40
$begingroup$
Sir, A or E_ij is the matrix obtained from Identity matrix I by interchanging i'th and j 'th rows of I
$endgroup$
– Shivam Kumar
Jan 20 at 2:57
$begingroup$
Sir, A or E_ij is the matrix obtained from Identity matrix I by interchanging i'th and j 'th rows of I
$endgroup$
– Shivam Kumar
Jan 20 at 2:57
$begingroup$
So like $$ begin{bmatrix} & & 1\ 1\ & 1 end{bmatrix}begin{bmatrix} & 1\ & & 1\ 1 end{bmatrix}=begin{bmatrix}1\ & 1\ & & 1 end{bmatrix} $$ ?
$endgroup$
– ja72
Jan 20 at 19:37
$begingroup$
So like $$ begin{bmatrix} & & 1\ 1\ & 1 end{bmatrix}begin{bmatrix} & 1\ & & 1\ 1 end{bmatrix}=begin{bmatrix}1\ & 1\ & & 1 end{bmatrix} $$ ?
$endgroup$
– ja72
Jan 20 at 19:37
add a comment |
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$begingroup$
What does the identity matrix $I$ have to do with the inverse of $A$, or $E$. Can you please give a simple example with $n=3$ in order to understand what the thought process is here.
$endgroup$
– ja72
Jan 20 at 2:40
$begingroup$
Sir, A or E_ij is the matrix obtained from Identity matrix I by interchanging i'th and j 'th rows of I
$endgroup$
– Shivam Kumar
Jan 20 at 2:57
$begingroup$
So like $$ begin{bmatrix} & & 1\ 1\ & 1 end{bmatrix}begin{bmatrix} & 1\ & & 1\ 1 end{bmatrix}=begin{bmatrix}1\ & 1\ & & 1 end{bmatrix} $$ ?
$endgroup$
– ja72
Jan 20 at 19:37