Let A be a square matrix of order n. If there exists a square matrix B of order n such that
AB = BA = In
then the matrix B is called an inverse of A.
Note :
Let A be square matrix of order n. Then, A−1 exists if and only if A is non-singular.
Example 1 :
Find the inverse (if it exists) of the following:
Since |A| = 2 ≠ 0, it is non singular matrix. A-1 exists.
Example 2 :
Solution :
In order to find inverse of a matrix, first we have to find |A|.
|A| = 5(25 - 1) - 1(5 - 1) + 1(1 - 5)
= 5(24 ) - 1(4) + 1(-4)
= 120 - 4 - 4
= 112
Since |A| = 112 ≠ 0, it is non singular matrix. A-1 exists.
Example 3 :
Solution :
In order to find inverse of a matrix, first we have to find |A|.
|A| = 2(8 - 7) - 3(6 - 3) + 1(21 - 12)
= 2(1) - 3(3) + 1(9)
= 2 - 9 + 9
= 2
Since |A| = 2 ≠ 0, it is non singular matrix. A-1 exists.
Example 4 :
Solution :
Let A = F(α)
A-1 = (1/|A|) adj A
|A| = cos α [cos α - 0] - 0[0 - 0] + sin α[0 + sin α]
= cos2α + sin2α
|A| = 1
Hence proved.
Example 5 :
Solution :
= A2 - 3A - 7I2
Finding the value of A2 :
Finding the value of 3A :
Finding the value of 7I2 :
A2 - 3A - 7I2
Hence proved.
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