CBSE Class 12 Sample Paper for 2023 Boards

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Question 35 - CBSE Class 12 Sample Paper for 2023 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Last updated at April 16, 2024 by

If A=[2&-3&5 3&2&-4 1&1&-2)], find A -1 . Use A -1 )to solve the following system of equations 2𝑥 − 3𝑦 + 5𝑧 = 11, 3x + 2y−4z, 𝑥 + 𝑦 − 2𝑧 = −3

 

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Transcript

Question 35 If A=[■8(2&−3&5@3&2&−4@1&1&−2)], find 𝐴^(−1). Use 𝐴^(−1)to solve the following system of equations 2𝑥 − 3𝑦 + 5𝑧 = 11, 3x + 2y − 4z = −5, 𝑥 + 𝑦 − 2𝑧 = −3The equations can be written as 2𝑥 − 3𝑦 + 5𝑧 = 11 3x + 2y − 4z = −5 𝑥 + 𝑦 − 2𝑧 = −3 So, the equation is in the form of [■8(2&−3&5@3&2&−4@1&1&−2)][■8(𝑥@𝑦@𝑧)] = [■8(11@−5@−3)] i.e. AX = B X = A–1 B Here, A = [■8(2&−3&5@3&2&−4@1&1&−2)] , X = [■8(𝑥@𝑦@𝑧)] & B = [■8(11@−5@−3)] Finding A–1 We know that A-1 = 1/(|A|) adj (A) Calculating |A| |A|= |■8(2&−3&5@3&2&−4@1&1&−2)| = 2(−4 + 4) + 3 (−6 + 4) + 5 (3 – 2) = 2(0) + 3(−2) + 5(1) = −1 Since |A|≠ 0 ∴ The system of equation is consistent & has a unique solution Now finding 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(2&−3&5@3&2&−4@1&1&−2)] 𝐴11 = −4 + 4 = 0 𝐴12 = −[−6+4] = 2 𝐴13 = 1 – 0 = 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)] & |A| = –1 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)] Now, X = A–1B [■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)]" =" [■8(0−5+6@−22−45+69@−11−25+39)] " " [■8(𝑥@𝑦@𝑧)]" =" [■8(1@2@3)] "∴ x = 1, y = 2 and z = 3"

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.