Slide 1 : Minor of an Element :
Let |A| =|aij| be a determinant of order n, then
the determinant obtained by deleting the ith row and jth column in which the element aij lies is
called the minor of element aij and is denoted by Mij.
e.g. If D =
Slide 2 : is a determinant of order 3, then the minor of element a11 is obtained by deleting the first row and first column of D.
Thus
M11 =
Similarly minor of an element a32 is
M32 =
Slide 3 : Cofactor of an Element :
The co-factor of the element aij is (-1)i+j times its minor Mij. We shall denote the cofactor of an element by the corresponding capital letter.
Thus
Cofactor of aij = Aij = (-1)i+j Mij
Slide 4 : Consider the determinant
D =
The cofactor of an element a22 is obtained as follows:
First we find the minor M22 of a22 as:
M22 =
Slide 5 : Then to find the cofactor A22 of a22 , we
multiply the minor M22 by (-1)1 + 1 so that
A22 = (-1)1+1
=
Similarly, we can find the cofactors of other elements too.
Slide 6 : Note:-
If A = , then
det(A) = a11A11 + a12A12 + a13A13.
This is the first row determinant expression.
Similarly, we can express determinant in terms of any row or any column
e.g. det(A) = a11A11 + a21A21 + a31A31. i.e. det(A) is the sum of the products of the elements of any row
(or column) and the corresponding cofactors
Slide 7 : Adjoint of a Square Matrix:
The adjoint of a square matrix [aij] is defined as
the transpose of the matrix [Aij], where Aij are
the cofactors of the elements aij. Adjoint of A is
denoted by adj A.
Thus, if A = is a square matrix.
Then the matrix of cofactors is :
Slide 8 : Matrix of cofactors ( C ) =
Then the adjoint of matrix A is
adjA =
Slide 9 : Example 1: Find the adjoint of the matrix
A =
Solution: We find the cofactors of the elements of matrix A.
A11 = (-1)1 + 1 = (0 – 4) = - 4
Slide 10 : A12 = (-1)1 + 2
A13 = (-1)1 + 3
A21 = (-1)2+1
A22 = (-1)2+2 = -(15 + 12) = -27 = 3 = -12 = -1
Slide 11 : A23 = (-1)2 + 3
A31 = (-1)3 + 1
A32 = (-1)3 + 2
A33 = (-1)3 + 1 = -(1 + 6) = -7 = (8 + 0) = 8 = -(4 + 8) = -10 = (0 - 6) = -6
Slide 12 : ? Matrix of cofactors of A = = ? adj(A) = =
Slide 13 : Example 2: Find the adjoint of the matrix.
Solution :We find the cofactors of the elements of matrix A
Slide 14 :
Slide 15 : Inverse of a Square Matrix:
Let A be a square matrix of order n. If there exists a matrix B of order n such that AB = BA = I, where I is the identity matrix of order n, then the matrix A is said to be invertible and B is called the inverse (or reciprocal) of A.
The inverse of matrix A is denoted by A-1.
Thus B = A-1 or A = B-1.
With these notations, we can write
AA-1 = A-1A = I.
Slide 16 : Note 1:
The inverse is defined only for the square matrices.
Note 2:
From the definition, it is clear that if B is the inverse of A, then A is the inverse of B.
Slide 17 : Some Important Theorems:
Theorem 1: Prove that the inverse of a square matrix if it exists, is unique.
Proof: Let A be an invertible matrix.
And suppose that B and C be the two inverses of A.
Since B is the inverse of A , therefore
AB = BA = I ………….(1)
Again C is the inverse of A, therefore
AC = CA = I ………….(2)
Slide 18 : We want to show that B = C.
We have
B = B I (as I is the identity matrix)
= B(AC) (from equation (2))
= (BA) C (Associative property)
= I C (from equation (1))
= C
Thus, we have
B = C
i.e. the inverse of A is unique.
Slide 19 : Reversal law: The inverse of the product of two square matrices (If it exists) is equal to the product of their inverses taken in reverse order.
i.e.
If A and B are two invertible matrices of the same order, then (AB)-1 = B-1A-1.
Proof: Since A and B are two square matrices of same order, therefore AB is a square matrix.
By definition of inverse
(AB)(AB)-1 = I, (provided (AB)-1 exists)
Slide 20 : Pre-multiplying both sides by A-1
A-1 (AB)(AB)-1 = A-1 I
(A-1A) B (AB)-1 = A-1 (Since A-1 I = A-1)
IB(AB)-1 = A-1
B (AB)-1 = A-1
(B-1B)(AB)-1 =B-1A-1
I(AB)-1= B-1A-1
(AB)-1 = B-1A-1 Proved.
Slide 21 : Example 1:- Find the inverse of the following
Matrix : Solution: We first find the matrix of cofactors: A11 = = (-3 + 4) = 1
Slide 22 : A12 = = -(2 - 0) = -2 A13 = = (-2 + 0) = -2 A21 = = -(-3 + 4) = -1 A22 = = (3 - 0) = 3
Slide 23 : A23 = = -(-3 - 0) = 3 A31 = = (-12 + 12) = 0 A32 = = -(12 - 8) = -4 A33 = = (-9 + 6) = -3