2.7 Inverse Matrix

2.7 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=(2915)A=(2915)  is an inverse matrix of matrix B=(5912).B=(5912).

Solution:
AB=(2915)(5912)=(2×5+9×12×9+9×21×5+5×11×9+5×2)=(10+(9)18+185+(5)9+10)=(1001)=IAB=(5912)(2915)=(5×2+(9)×15×9+(9)×51×2+2×11×9+2×5)=(10+(9)18182+29+10)=(1001)=IAB=BA=IA is the inverse matrix of B and vice versa.AB=(2915)(5912)=(2×5+9×12×9+9×21×5+5×11×9+5×2)=(10+(9)18+185+(5)9+10)=(1001)=IAB=(5912)(2915)=(5×2+(9)×15×9+(9)×51×2+2×11×9+2×5)=(10+(9)18182+29+10)=(1001)=IAB=BA=IA is the inverse matrix of B and vice versa.


5. The inverse of a matrix may also be found using a formula.
If A=(abcd)A=(abcd) , then the inverse matrix of A, A-1, is given by the formula below.
    A1=1adbc(dbca), 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=(6191)  using the formula.

Solution:

A=(6191)a=6,b=1,c=9,d=1A1=1adbc(dbca)A1=16×1(1×9)(1196)A1=16+9(1196)A1=13(1196)=(131332)


Example 3:
The inverse matrix of (7292)isr(2s9t).  Find the value of r, of s and of t.

Solution:
LetA=(7292)A1=17×2(9)×2(2297)A1=14(2297)r(2s9t)=14(2297)By comparison,r=14,s=2,t=7.

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