The document discusses determining the inverse of matrices. It provides examples of finding the inverse of 2x2 matrices using simultaneous linear equations and formulas. The simultaneous equation method involves setting up equations where the product of the matrix and its inverse equals the identity matrix. The formula method uses the determinant of the matrix and the positions of the elements to determine the inverse. Exercises provide additional practice finding inverses of matrices using these methods.

1. Group 7INVERSE MATRIXMEMBER’S NAME : Victoria RosNoorAfidah Bt. MohdYatimTeh Ying Zhe Ng Kah Soon

2. IntroductionThe 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 EquationsGiven, matrix A = 3 13 4To find the inverse of matrix A, let A⁻¹ = a b c dA x A⁻¹ = IThen ; 3 1 a b = 1 03 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 -------- ④① - ② : -3c = 1 ③ - ④ : -3d = -1 c =-1/3 d = 1/3Substitute 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 1/3To 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 FormulaThe 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 dA⁻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 - bcad – 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 4SOLUTION (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 3SOLUTION 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. EXERCISE1. The inverse matrix of 3 2 is m 5 n . 6 5 -6 3Find the values of m and n.SOLUTIONThe 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 simultaneousB = 4 3 4 4SOLUTIONLet 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 14e + 3g = 1 ----------① 4f + 3h = 0 ----------③ 4e + 4g = 0 ----------② 4f + 4h = 1 ----------④② - ① : g = -1 ④ - ③ : h = 1So, 4e + 3(-1) = 1 So, 4f + 3(1) = 0 e = 1 f = - 3\4Therefore, B⁻¹ = 1 -3\4 -1 1

11. 3. A = 2 5 B = 8 5 3 8 3 2SOLUTIONBA = 7 -4 3 4 -5 3 5 7 = 21+ (-20) 28+(-28) -15 +15 -20+21 = 1 0 0 1

12. 4. Find the inverse A⁻¹,of A = 2 -3 4 -7SOLUTION 2 -3 = -14 + 12 4 -7 = -2  ad-bcA⁻¹ = 1 -7 3 |A| -4 2 = 1 -7 3 -2 -4 2 = 3.5 -1.5 2 -1

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