4.7 Inverse Matrix

4.7 Inverse Matrix
1. If A is a square matrix, is another square matrix and A × B = B × A = I, then matrix is the inverse matrix of matrix and vice versa. Matrix A is called the inverse matrix of 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 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:
A B = ( 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 A B = ( 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 ) 18 18 2 + 2 9 + 10 ) = ( 1 0 0 1 ) = I A B = B A = 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 a d b c ( d b c a ) , where a d b c 0
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 a d b c ( 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.

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