if A = [1 2 0  -2 1 0   0 -1 1], find A-1 . Hence Solve the system of equations;

𝑥 − 2𝑦 = 10

2𝑥 − 𝑦 − 𝑧 = 8

−2𝑦 + 𝑧 = 7

 

If A = [], find A^-1 . Hence, Solve the system of equations: x-2y=10

Question 36 (Choice 1) - CBSE Class 12 Sample Paper for 2021 Boards - Part 2
Question 36 (Choice 1) - CBSE Class 12 Sample Paper for 2021 Boards - Part 3
Question 36 (Choice 1) - CBSE Class 12 Sample Paper for 2021 Boards - Part 4
Question 36 (Choice 1) - CBSE Class 12 Sample Paper for 2021 Boards - Part 5 Question 36 (Choice 1) - CBSE Class 12 Sample Paper for 2021 Boards - Part 6

 

 

Note : This is similar to Question 33 of CBSE Sample Paper Maths Class 12 - 2020

Check the answer here https:// www.teachoo.com /10911/3128/Question-33-OR-2nd-Question/category/CBSE-Class-12-Sample-Paper-for-2020-Boards/

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Transcript

Question 36 (Choice 1) if A = [■8(1&2&0@−2&−1&−2@0&−1&1)], find A-1 . Hence Solve the system of equations; 𝑥 − 2𝑦 = 10 2𝑥 − 𝑦 − 𝑧 = 8 −2𝑦 + 𝑧 = 7 The equations are 𝑥 − 2𝑦 = 10 2𝑥 − 𝑦 − 𝑧 = 8 −2𝑦 + 𝑧 = 7 So, the equation is in the form of [■8(1&−2&0@2&−1&−1@0&−2&1)][■8(𝑥@𝑦@𝑧)] = [■8(10@8@7)] i.e. ATX = B X = (AT)–1 B X = (A−1)T B Here, A = [■8(1&2&0@−2&−1&−2@0&−1&1)], X = [■8(𝑥@𝑦@𝑧)] & B = [■8(10@8@7)] Finding A–1 We know that A-1 = 1/(|A|) adj (A) Calculating |A| |A|= |■8(1&2&0@−2&−1&−2@0&−1&1)| = 1(−1 − 2) − 2 (−2 – 0) + 0 (2 + 1) = –3 + 4 + 0 = 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(1&2&0@−2&−1&−2@0&−1&1)] 𝐴11 = −2 + 0 = –3 𝐴12 = −[2−0] = 2 𝐴13 = 1 – 0 = 2 𝐴21 = –[6−4] = –2 𝐴22 = 4 – 0 = 1 𝐴23 = –[2−0] = 1 𝐴31 = 0−(−4)= −4 𝐴32 = –[0−4] = 2 𝐴33 = −2−3 = 3 Thus adj A = [■8(−3&−2&−4@2&1&2@2&1&3)] & |A| = 1 Now, A-1 = 𝟏/(|𝐀|) adj A A-1 = 1/1 [■8(−3&−2&−4@2&1&2@2&1&3)] = [■8(−3&−2&−4@2&1&2@2&1&3)] Now, X = (A−1)T B [■8(𝑥@𝑦@𝑧)] = [■8(−3&−2&−4@2&1&2@2&1&3)]^𝑇 [■8(10@8@7)] [■8(𝑥@𝑦@𝑧)] =[■8(−3&2&2@−2&1&1@−4&2&3)][■8(10@8@7)] [■8(𝑥@𝑦@𝑧)]" =" [█(−3(10)+2(8)+2(7)@−2(10)+1(8)+1(7)@−4(10)+2(8)+3(7))] " " [■8(𝑥@𝑦@𝑧)]" =" [■8(−30+16+14@−20+8+7@−40+16+21)] " " [■8(𝑥@𝑦@𝑧)]" =" [■8(0@−5@−3)] "∴ x = 0, y = –5 and z = −3"

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Davneet Singh

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.