Check sibling questions

Ex 4.6, 15 - If A = [2 -3 5 3 2 -4 1 1 -2], find A-1 and solve

Ex 4.6, 15 - Chapter 4 Class 12 Determinants - Part 2
Ex 4.6, 15 - Chapter 4 Class 12 Determinants - Part 3
Ex 4.6, 15 - Chapter 4 Class 12 Determinants - Part 4
Ex 4.6, 15 - Chapter 4 Class 12 Determinants - Part 5
Ex 4.6, 15 - Chapter 4 Class 12 Determinants - Part 6

Solve all your doubts with Teachoo Black (new monthly pack available now!)


Transcript

Ex4.6, 15 If A = [■8(2&−3&5@3&2&−4@1&1&−2)], find A−1. Using A−1 solve the system of equations 2x – 3y + 5z = 11 3x + 2y – 4z = -5 x + y – 2z = -3 Writing equation as AX = B [■8(2&−3&5@3&2&−4@1&1&−2)][■8(𝑥@𝑦@𝑧)] = [■8(11@−5@−3)] Ex4.6, 15 If A = [■8(2&−3&5@3&2&−4@1&1&−2)], find A−1. Using A−1 solve the system of equations 2x – 3y + 5z = 11 3x + 2y – 4z = -5 x + y – 2z = -3 Writing equation as AX = B [■8(2&−3&5@3&2&−4@1&1&−2)][■8(𝑥@𝑦@𝑧)] = [■8(11@−5@−3)] Ex4.6, 15 If A = [■8(2&−3&5@3&2&−4@1&1&−2)], find A−1. Using A−1 solve the system of equations 2x – 3y + 5z = 11 3x + 2y – 4z = -5 x + y – 2z = -3 Writing equation as AX = B [■8(2&−3&5@3&2&−4@1&1&−2)][■8(𝑥@𝑦@𝑧)] = [■8(11@−5@−3)] Hence A = [■8(2&−3&5@3&2&−4@1&1&−2)] , X = [■8(𝑥@𝑦@𝑧)] & B = [■8(11@−5@−3)] Calculating |A| |A|= |■8(2&−3&5@3&2&−4@1&1&−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 solution Now, AX = B X = A-1 B Calculating A-1 Now, A-1 = 1/(|A|) adj (A) adj A = [■8(A11&A12&A13@A21&A22&A23@A31&A32&A33)]^′ = [■8(A11&A21&A31@A12&A22&A32@A13&A23&A33)] A = [■8(1&−1&2@3&4&−5@2&−1&3)] Now, 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 = [■8(0&−1&2@2&−9&23@1&−5&13)] Now, A-1 = 1/(|A|) adj A A-1 = 1/(−1) [■8(0&−1&2@2&−9&23@1&−5&13)] = [■8(0&1&−2@−2&9&−23@−1&5&−13)] Solving X = A-1 B [■8(𝑥@𝑦@𝑧)] = [■8(0&1&−2@2&9&−23@−1&5&−13)] [■8(11@−5@−3)] " " [■8(𝑥@𝑦@𝑧)]" = " [█(0(11)+1(−5)+(−2)(−3)@2(11)+9(−5)+(−23)(−3)@−1(11)+5(−5)+(−13)(−3))] " " [■8(𝑥@𝑦@𝑧)]" = " [█(−5+6@−22−45+69@−11−25+39)] " " [■8(𝑥@𝑦@𝑧)]" = " [■8(1@2@3)] "∴ x = 1, y = 2 and z = 3 "

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.