1.Algebra Booster

Matrices and Determinants 7.7
11. INTEGRATION OF DETERMINANT
If
then
f() x g() x h()
x
Fx () = a b c ,
l m n
b
Ú
a
b b b
Ú Ú Ú
f () x dx g() x dx h()
x dx
a a a
F()
x dx = a b c
l m n
12. SUMMATION OF DETERMINANTS
f() r g() r h()
r
Let D r = a b c
p q r
Then
n
Â
r = 1
D =
r
n n n
  Â
f() r g() r h()
r
r= 1 r= 1 r=
1
a b c
p q r
13. ADJOINT OF A MATRIX
The adjoint of a matrix is the transpose of the co-factors of
the corresponding elements of a given matrix.
If A be any square matrix, then
adj A = (C ij
) T
EXAMPLE 1: Let
then
EXAMPLE 2: Let
then
2 3
A = Ê ˆ
Á
Ë4 5 ˜
¯ ,
Ê 5 -4ˆ Ê 5 -3ˆ
adj ( A)
= Á =
Ë-3 2 ˜
¯
Á
Ë-4 2˜
¯ .
T
Ê1 2 3ˆ
B = Á2 Á
1 2 ˜,
˜
Ë3 2 4¯
Ê 1 2 2 2 2 1 ˆ
Á -
2 4 3 4 3 2 ˜
Á
˜
Á 2 3 1 3 1 2˜
adj ( B)
= Á- - ˜
Á 2 4 3 4 3 2˜
Á
2 3 1 3 1 2
˜
Á
-
Ë 1 2 2 2 2 1 ˜
¯
T
Ê 0 -2 1 ˆ
= Á-2 -5 4˜
Á
˜
Ë 1 4 -3¯
Ê 0 -2 1 ˆ
= Á-2 -5 4˜
Á
˜
Ë 1 4 -3¯
T
Theorem: If A be any square matrix, then
A◊ [adj ( A)] = | A| ◊ I = [adj ( A)]
◊ A
13.1 Properties of Adjoint of Matrix or Matrices
(i) If A be any square matrix of order n, then
1
|adj ( A)|
= | A| n-
Proof: We know that A◊ adj ( A) = | A| ◊In
fi | A◊ adj( A)| = ||A| ◊ I | = | A| n
fi | A||adj( A)| = || A| ◊ I | = | A| n
fi
| A|
|adj ( A)| = = | A|
| A|
n
n
n
n
n-1
( |AB| = |A||B|)
Hence, the result.
(ii) If |A| = 0, then |adj(A)| = 0
i.e. if A is singular, then adj(A) is also singular
n-1
(iii) adj ( kA) = k (adj ( A))
T
(iv) adj ( A ) = (adj A)
(v) adj ( AB) = (adj B)(adj A)
Proof: We have
( AB)adj( AB) = | AB|
In
fi (adj B)(adj A)( AB) adj ( AB)
= | AB|(adj B)(adj A)
I
T
fi (adj B) | A| I B(adj AB) = | AB|(adj B)(adj A)
n
fi | A| (adj B) B(adj AB) = | AB| (adj B)(adj A)
fi | ABI || | (adj AB) = |AB|(adj B)(adj A)
n
fi | A| | B| (adj AB) = | AB|(adj B)(adj A)
fi (adj AB) = (adj B)(adj A)
(vi) If A is a non-singular matrix, then
2
adj[adj ( A)]
| A| n -
= ◊ A
(vii)
Proof: We know that
A.adj(A) = |A|.I n
Replace A by adjA, we get
adj ( A) ◊ adj[adj ( A)] = |adj ( A)|
◊I
n-1
fi adj ( A) ◊ adj[adj ( A)] = | A|
I
fi
fi
fi
fi
fi
n
n
n-1
[ Aadj ( A)] ◊ adj (adj ( A)) = | A| ( AI n )
1
| A| I adj[adj ( A)] | A| n -
◊ = ◊ A
n
1
| A| adj[adj ( A)] | A| n -
= ◊ A
n - 1
◊
| A|
A
adj[adj ( A)]
=
| A|
2
adj[adj ( A)]
| A| n -
= ◊ A
2
( 1)
|adj[adj ( A)]|
= | A| n-
Proof: We have |[adj(A)]| = |A| n–1
n