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inverse of an adjoint matrix

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

11 12 1
21 22 2
1 11 2 21 1 11 1 1 2 2 1 1
... ...
n
n
n n n n n n n
a a a
a a a
A
ra r a r a ra r a r a
=
+ + + + + +


( )
Adj A



1
r



2
r
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1
th
n

1n
r

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!%$
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+(,(
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3 3×

11 22 21 12
a a a a
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+n.
( )
( )
1
det
n
Adj A A
=

( )
( )
1
det
adj A
A
A
=

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/
( )
1
Adj A A A
=
 
('
( )
( )
( )
( )
1
det
Adj A djA
Adj A
AdjA
=
#
( ) ( )
1
det det
n
AdjA A
=
+
( ) ( ) ( )
( )
1 1
det
n
AdjA A Adj Adj A
=
 
0"
( )
( )
( )
( )
1
1
1
1
det det
A
AdjA A
A A
= =
/1
( )
( ) ( )
1
det
det
n
A
A Adj A djA
A
=
&
( ) ( )
2
det
n
Adj AdjA A A
= ×

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4a) in view of solution (3) above, without loss of generality, suppose the last row of the given matrix is a linear combination of all its previous rows as , then we can see as in result (3a) that has 1st column is the multiple of the last column, 2nd column is the multiple of the last column and so on upto column is the multiple of last column.(b) This forces that while picking out the scalar multiples of the columns from the rows or columns in view of properties of determinants, we get each column is identical to last column when the respective scalar multiple is taken out from the determi ...
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Anonymous
Excellent resource! Really helped me get the gist of things.

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