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How to prove that a cofactor of a matrix is $A$ is $(-1)^i+j times $ a minor

0

$begingroup$

Let $A in M_n(mathbbR)$ :

$A'_ij$ have the same columns as $A$ except the $j$th one which is a column full of zero except on the $i$th entry where it is a $1$.

$A''_ij$ is the matrix we get from $A$ by deleting it's $i$th row and its $j$th column.

Then I don't understand why we have the following equality :

$$det(A'_ij) = (-1)^i+1 det(A''_ij)$$

The thing is that when I am trying to manipulate somehting of the form : $det(C_1, ..., C_n)$ where $C_i$ is a vector of size $n$ I don't know how I can't get something equivalent of the form : $det(C'_1, ..., C'_n-1)$ where $C'_i$ is a vector of size $n-1$.

Thank you !

linear-algebra determinant

share|cite|improve this question

asked Mar 24 at 17:04

15978462548991597846254899

295

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Simply using the formula for the expansion of $det A'_ij$ along the $j$th column.
  $endgroup$
  – Bernard
  Mar 24 at 17:08
  
  

  
  
  
  

add a comment | 

0

$begingroup$

Let $A in M_n(mathbbR)$ :

$A'_ij$ have the same columns as $A$ except the $j$th one which is a column full of zero except on the $i$th entry where it is a $1$.

$A''_ij$ is the matrix we get from $A$ by deleting it's $i$th row and its $j$th column.

Then I don't understand why we have the following equality :

$$det(A'_ij) = (-1)^i+1 det(A''_ij)$$

The thing is that when I am trying to manipulate somehting of the form : $det(C_1, ..., C_n)$ where $C_i$ is a vector of size $n$ I don't know how I can't get something equivalent of the form : $det(C'_1, ..., C'_n-1)$ where $C'_i$ is a vector of size $n-1$.

Thank you !

linear-algebra determinant

share|cite|improve this question

asked Mar 24 at 17:04

15978462548991597846254899

295

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Simply using the formula for the expansion of $det A'_ij$ along the $j$th column.
  $endgroup$
  – Bernard
  Mar 24 at 17:08
  
  

  
  
  
  

add a comment | 

$begingroup$

Let $A in M_n(mathbbR)$ :

$A'_ij$ have the same columns as $A$ except the $j$th one which is a column full of zero except on the $i$th entry where it is a $1$.

$A''_ij$ is the matrix we get from $A$ by deleting it's $i$th row and its $j$th column.

Then I don't understand why we have the following equality :

$$det(A'_ij) = (-1)^i+1 det(A''_ij)$$

The thing is that when I am trying to manipulate somehting of the form : $det(C_1, ..., C_n)$ where $C_i$ is a vector of size $n$ I don't know how I can't get something equivalent of the form : $det(C'_1, ..., C'_n-1)$ where $C'_i$ is a vector of size $n-1$.

Thank you !

linear-algebra determinant

share|cite|improve this question

asked Mar 24 at 17:04

15978462548991597846254899

295

$endgroup$

share|cite|improve this question

asked Mar 24 at 17:04

15978462548991597846254899

295

share|cite|improve this question

asked Mar 24 at 17:04

15978462548991597846254899

295

asked Mar 24 at 17:04

15978462548991597846254899

295

asked Mar 24 at 17:04

15978462548991597846254899

295