1. Group 7INVERSE MATRIX MEMBER’S NAME : Victoria Ros NoorAfidah Bt. MohdYatim Teh Ying Zhe Ng Kah Soon
2. Introduction The inverse of a matrix, A, is denoted by A⁻¹. The product of A x A⁻¹ is the identity matrix, I. Example: For matrix A = 3 5 , its inverse is A⁻¹ = 2 -5 1 2 -1 3 AA⁻¹ = 3 5 2 -5 = 1 0 and A⁻¹A = 2 -5 3 5 = 1 0 1 2 -1 3 0 1 -1 3 1 2 0 1
3. DETERMINING THE INVERSE OF A 2 x2 MATRIX A. Simultaneous Linear Equations Given, matrix A = 3 1 3 4 To find the inverse of matrix A, let A⁻¹ = a b c d A x A⁻¹ = I Then ; 3 1 a b = 1 0 3 4 c d 0 1 3a + c 3b + d = 1 0 < --------------( EQUAL MATRICES ) 3a + 4c 3b + 4d 0 1
4. 3a + c = 1 ------------ ① 3b + d = 0 ---------- ③ 3a + 4c = 0 ----------- ② 3b + 4d = 1 -------- ④ ① - ② : -3c = 1 ③ - ④ : -3d = -1 c =-1/3 d = 1/3 Substitute c = -1/3in equation ①. Substitute d = 1/3 in equation ③. 3a + (-1/3) = 1 3b + (1/3) = 0 -> a = 4/9 -> b = -1/9 Therefore, A⁻¹ = 4/9 -1/9 -1/3 1/3 To check the answer : AA⁻¹ = 3 1 4/9 -1/9 3 4 -1/3 1/3 = 1 0 = I (Identity matrix) 0 1
5. B. By Using Formula The inverse of a 2 x 2 matrix can also be obtained by using formula. In general, if A = a b , the inverse of matrix A is c d A⁻¹ = 1 d -b Change the positions of elements in the ad – bc -c a main diagonal and multiply the other elements by -1 = d -b ad – bc ad – bc -c a ad – bc ad - bc ad – bc is the determinant and written as | A |
6. Example 1 : 1. Find the inverse of the following matrices, by using the formula. (a) G = 4 3 (b) H = 5 2 2 2 10 4 SOLUTION (a) Determinant, |G| = ad-bc (b) Determinant, |H| = ad- bc = (4 x 2) – (3 x2 ) = (5 x 4) – ( 2 x 10) = 2 = 0 Therefore, G⁻¹ = ½ 2 -3 Therefore, H⁻¹ does not exist. -2 4 = 1 -3/2 -1 2
7. Example 2 : Determine whether matrix A is the inverse of matrix B. (a) A = 3 4 , B = 7 -4 5 7 -5 3 SOLUTION AB = 3 4 7 -4 5 7 -5 3 = 21+(-20) -12+12 35+(-35) -20+21 = 1 0 0 1 -> AB = I and BA = 1
8. EXERCISE 1. The inverse matrix of 3 2 is m 5 n . 6 5 -6 3 Find the values of m and n. SOLUTION The inverse matrix of 3 2 = 1 5 -2 6 5 3 x 5 – 2 x 6 -6 3 = 1/3 5 -2 compare with m 5 n -6 3 -6 3 Therefore, m = 1/3, n = -2
9. 2. Given the matrix B, find the inverse B⁻¹ by using the method of solving simultaneous B = 4 3 4 4 SOLUTION Let B⁻¹ = e f g h 4 3 e f = 1 0 4 4 g h 0 1
10. 4e + 3g 4f + 3h = 1 0 4e +4g 4f + 4h 0 1 4e + 3g = 1 ----------① 4f + 3h = 0 ----------③ 4e + 4g = 0 ----------② 4f + 4h = 1 ----------④ ② - ① : g = -1 ④ - ③ : h = 1 So, 4e + 3(-1) = 1 So, 4f + 3(1) = 0 e = 1 f = - 3 Therefore, B⁻¹ = 1 -3 -1 1
