# Ex 4.6, 15

Last updated at March 11, 2017 by Teachoo

Last updated at March 11, 2017 by Teachoo

Transcript

Ex4.6, 15 If A = 2−3532−411−2, find A−1. Using A−1 solve the system of equations 2x – 3y + 5z = 11 3x + 2y – 4z = -5 x + y – 2z = -3 Step 1 Write equation as AX = B 2−3532−411−2 𝑥𝑦𝑧 = 11−5−3 i.e. AX = B Hence A = 2−3532−411−2 , X = 𝑥𝑦𝑧 & B = 11−5−3 Step 2 Calculate |A| |A|= 2−3532−411−2 = 2 (−4 + 4) − 3 (6 − 5) + 1 (12 − 10) = 2(0) − 3 (1) + 1(2) = −1 So, |A|≠ 0 ∴ The system of equation is consistent & has a unique solutions AX = B X = A-1 B Step 3 Calculating X = A-1 B Calculating A-1 Now, A-1 = 1|A| adj (A) adj A = A11A12A13A21A22A23A31A32A33′ = A11A21A31A12A22A32A13A23A33 A = 1−1234−52−13 𝐴11 = −4 + 4 = 0 𝐴12 = − −6−(−4) = − (−6 + 4) = −2 = 2 𝐴13 = 3 − 2 = 1 𝐴21 = (6 − 5) = −1 𝐴22 = −4 − 5 = −9 𝐴23 = −2−(−3 = −5 𝐴31 = 12−10=2 𝐴32 = −8−15=23 𝐴33 = 4− −9=13 Thus adj A = 0−122−9231−513 & |A| = –1 Now, A-1 = 1|A| adj A A-1 = 1−1 0−122−9231−513 = 01−2−29−23−15−13 Solving X = A-1 B 𝑥𝑦𝑧 = − 01−229−23−15−13 11−5−3 𝑥𝑦𝑧 = −5+6−22−45+69−11−25+39 𝑥𝑦𝑧 = 123 ∴ x = 1, y = 2 and z = 3

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.