Question 24 (OR 1 st question)

If A = [3 1 2 3 2 -3 2 0 -1], find A –1

Hence, solve the system of equations:

3x + 3y + 2z = 1

x + 2y = 4

2x – 3y – z = 5

If A = [3 1 2 3 2 -3 2 0 -1], find A^-1. Hence solve system equations

Question 24 (Or 1st) - CBSE Class 12 Sample Paper for 2019 Boards - Part 2
Question 24 (Or 1st) - CBSE Class 12 Sample Paper for 2019 Boards - Part 3 Question 24 (Or 1st) - CBSE Class 12 Sample Paper for 2019 Boards - Part 4 Question 24 (Or 1st) - CBSE Class 12 Sample Paper for 2019 Boards - Part 5 Question 24 (Or 1st) - CBSE Class 12 Sample Paper for 2019 Boards - Part 6

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Question 24 (OR 1st question) If A = [■8(3&1&2@3&2&−3@2&0&−1)], find A–1 Hence, solve the system of equations: 3x + 3y + 2z = 1 x + 2y = 4 2x – 3y – z = 5 For our equation [■8(3&3&2@1&2&0@2&−3&−1)][■8(𝑥@𝑦@𝑧)] = [■8(1@4@5)] i.e. (𝐴^𝑇)X = B X = 〖(𝐴^𝑇)〗^(−1) 𝐵 X = 〖(𝐴^(−1))〗^𝑇 𝐵 (Because 〖(𝐴^𝑇)〗^(−1) = 〖(𝐴^𝑇)〗^(−1)) Here, A = [■8(3&1&2@3&2&−3@2&0&−1)] , X = [■8(𝑥@𝑦@𝑧)] & B = [■8(1@4@5)] Finding A–1 We know that A-1 = 1/(|A|) adj (A) Calculating |A|= |■8(3&1&2@3&2&−3@2&0&−1)| = 3(−2 + 0) − 1 (–3 + 6) + 2 (0 – 4) = –6 – 3 – 8 = −17 Since |A|≠ 0 ∴ The system of equation is consistent & has a unique solutions 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(3&1&2@3&2&−3@2&0&−1)] 𝐴11 = −2 + 0 = –2 𝐴12 = −[−3−(−6)] = − (−3+ 6) = −3 𝐴13 = 0 − 4 = – 4 𝐴21 = –[−1−0] = 1 𝐴22 = −3 – 4 = –7 𝐴23 = –[0−2] = 2 𝐴31 = −3−4= –7 𝐴32 = –[−9−6] = 15 𝐴33 = 6−3 = 3 Thus adj A = [■8(−2&1&−7@−3&−7&15@−4&2&3)] & |A| = –17 Now, A-1 = 1/(|A|) adj A A-1 = 1/(−17) [■8(−2&1&−7@−3&−7&15@−4&2&3)] = 1/17 [■8(2&−1&7@3&7&−15@4&−2&−3)] Now, X = 〖(𝐴^(−1))〗^𝑇 𝐵 [■8(𝑥@𝑦@𝑧)] = 1/17 [■8(2&−1&7@3&7&−15@4&−2&−3)]^′ [■8(1@4@5)] [■8(𝑥@𝑦@𝑧)] = 1/17 [■8(2&3&4@−1&7&−2@7&−15&−3)][■8(1@4@5)] " " [■8(𝑥@𝑦@𝑧)]" =" 1/17 [█(2(1)+3(4)+4(5)@−1(1)+7(4)+(−2)(5)@7(1)+(−15)(4)+(−3)(5))] " " [■8(𝑥@𝑦@𝑧)]" =" 1/17 [■8(2+12+20@−1+28−10@7−60−15)] " " [■8(𝑥@𝑦@𝑧)]" =" 1/17 [■8(34@17@−68)] " " [■8(𝑥@𝑦@𝑧)]" =" [■8(2@1@−4)] "∴ x = 2, y = 1 and z = "–4

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.