4.7 Inverse Matrix

4.7 Inverse Matrix
1. If A is a square matrix, B is another square matrix and A × B = B × A = I, then matrix A is the inverse matrix of matrix B and vice versa. Matrix A is called the inverse matrix of B for multiplication and vice versa.

2. The symbol A-1 denotes the inverse matrix of A.

3. Inverse matrices can only exist for square matrices but not all square matrices have inverse matrices.

4. If AB ≠ I or BA ≠ I, then A is not the inverse of B and B is not the inverse of A.

Example 1:
Determine whether matrix  A=( 2 9 1 5 )  is an inverse matrix of matrix B=( 5 9 1 2 ).

Solution:
AB=( 2 9 1 5 )( 5 9 1 2 ) =( 2×5+9×1 2×9+9×2 1×5+5×1 1×9+5×2 ) =( 10+( 9 ) 18+18 5+( 5 ) 9+10 ) =( 1 0 0 1 )=I AB=( 5 9 1 2 )( 2 9 1 5 ) =( 5×2+( 9 )×1 5×9+( 9 )×5 1×2+2×1 1×9+2×5 ) =( 10+( 9 ) 1818 2+2 9+10 ) =( 1 0 0 1 )=I AB=BA=I A is the inverse matrix of B and vice versa.


5. The inverse of a matrix may also be found using a formula.
If A=( a b c d ) , then the inverse matrix of A, A-1, is given by the formula below.
     A 1 = 1 adbc ( d b c a ), where adbc0    

6. ad – bc is known as the determinant of matrix A.

7. If the determinant, ad – bc = 0, then the inverse matrix of A does not exist.

Example 2:
Find the inverse matrix of A=( 6 1 9 1 )  using the formula.

Solution:
A=( 6 1 9 1 ) a=6, b=1, c=9, d=1 A 1 = 1 adbc ( d b c a ) A 1 = 1 6×1( 1×9 ) ( 1 1 9 6 ) A 1 = 1 6+9 ( 1 1 9 6 ) A 1 = 1 3 ( 1 1 9 6 )=( 1 3 1 3 3 2 )

Example 3:
The inverse matrix of ( 7 2 9 2 ) is r( 2 s 9 t ).  Find the value of r, of s and of t.

Solution:
Let A=( 7 2 9 2 ) A 1 = 1 7×2( 9 )×2 ( 2 2 9 7 ) A 1 = 1 4 ( 2 2 9 7 ) r( 2 s 9 t )= 1 4 ( 2 2 9 7 ) By comparison, r= 1 4 , s=2, t=7.