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Last updated at Jan. 23, 2020 by Teachoo

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Example 28 Solve the following system of equations by matrix method. 3x – 2y + 3z = 8 2x + y – z = 1 4x – 3y + 2z = 4 The system of equation is 3x – 2y + 3z = 8 2x + y – z = 1 4x – 3y + 2z = 4 Writing equation as AX = B [■8(3&−2&3@2&1&−1@4&−3&2)][■8(𝑥@𝑦@𝑧)] = [■8(8@1@4)] Hence A = [■8(3&−2&3@2&1&−1@4&−3&2)], 𝑥= [■8(𝑥@𝑦@𝑧)] & B = [■8(8@1@4)] Calculating |A| |A| = |■8(3&−2&3@2&1&−1@4&−3&2)| = 3 |■8(1&−1@−3&2)| – 1( – 2) |■8(2&−1@4&2)| + 3 |■8(2&1@4&−3)| = 3 (2 – 3) – 2 (4 + 4) + 3 (–6 – 4) = 3 (–1) + 2(8) + 3 (–10) = –3 + 16 – 30 = –17 Since, |A| ≠ 0 ∴ System of equation is consistent & has a unique solution Now, AX = B X = A-1 B Calculating A-1 A-1 = 1/(|A|) adj (A) adj (A) = [■8(A_11&A_12&A_13@A_21&A_22&A_23@A_31&A_32&A_33 )]^′ = [■8(A_11&A_21&A_13@A_12&A_22&A_23@A_13&A_32&A_33 )] A = [■8(3&−2&3@2&1&−1@4&−3&2)] M11 = |■8(1&−1@−3&2)| = 2 – 3 = –1 M12 = |■8(2&−1@4&2)| = 4 + 4 = 8 M13 = |■8(2&1@4&−3)| = –6 – 4 = –10 M21 = |■8(−2&3@−3&2)| = –4 + 9 = 5 M22 = |■8(3&3@4&2)| = 6 – 12 = – 6 M23 = |■8(3&−2@4&−3)| = –9 + 8 = –1 M31 = |■8(−2&3@1&−1)| = 1 – 2 = –1 M32 = |■8(3&3@2&−1)| = –3 – 6 = –9 M33 = |■8(3&−2@2&1)| = 3 + 4 = 7 Now, A11 = (–1)1+1 . M11 = (–1)2 . (–1) = –1 A12 = (–1)1+2 . M12 = (–1)3 . 8 = –8 A13 = (–1)1+3 . M13 = (–1)4 . ( –10) = –10 A21 = (–1)2+1 . M21 = (–1)3 . (5) = –5 A22 = (–1)2+2 . M22 = (–1)4 . (–6) = –6 A23 = (–1)2+3 . ( – 1) = (–1)5 . (–1) = 1 A31 = (–1)3+1 . M31 = (–1)4 . (–1) = –1 A32 = (–1)3+2 . M32 = (–1)5 . (–9) = 9 A33 = (–1)3+3 . M33 = (–1)6 . 7 = 7 Thus, adj (A) =[■8(−1&−5&−1@−8&−6&9@−10&1&7)] Now, A-1 = 1/(|A|) adj A Putting values = 1/(−17) [■8(−1&−5&−1@−8&−6&9@−10&1&7)] Also, X = A-1 B Putting values [█(■8(𝑥@𝑦)@𝑧)] = 1/(−17) [■8(−1&−5&−1@−8&−6&9@−10&1&7)][█(■8(8@1)@4)] [█(■8(𝑥@𝑦)@𝑧)] = 1/(−17) [■8(−1(8)+(−5)(1)+(−1)4@−8(8)+(−6)(1)+9(4)@−10(8)+1(1)+7(4) )] [█(■8(𝑥@𝑦)@𝑧)] = 1/(−17) [■8(−8&−5&−5@−64&−6&+36@−80&+1&+28)] [█(■8(𝑥@𝑦)@𝑧)] = 1/(−17) [█(■8(−17@−36)@−51)] [█(■8(𝑥@𝑦)@𝑧)] = [█(■8((−17)/(−17)@(−34)/(−(17)))@(−51)/(−17))] [█(■8(𝑥@𝑦)@𝑧)] = [█(■8(1@2)@3)] Hence x = 1, y = 𝟐 & z = 3

Examples

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Example 3

Example 4

Example 5

Example 6 Not in Syllabus - CBSE Exams 2021

Example 7 Not in Syllabus - CBSE Exams 2021

Example 8 Not in Syllabus - CBSE Exams 2021

Example 9 Important Not in Syllabus - CBSE Exams 2021

Example 10 Important Not in Syllabus - CBSE Exams 2021

Example 11 Not in Syllabus - CBSE Exams 2021

Example 12 Not in Syllabus - CBSE Exams 2021

Example 13 Not in Syllabus - CBSE Exams 2021

Example 14 Important Not in Syllabus - CBSE Exams 2021

Example 15 Important Not in Syllabus - CBSE Exams 2021

Example 16 Important Not in Syllabus - CBSE Exams 2021

Example 17

Example 18 Important

Example 19

Example 20

Example 21

Example 22

Example 23

Example 24 Important

Example 25

Example 26 Important

Example 27 Not in Syllabus - CBSE Exams 2021

Example 28 Important Not in Syllabus - CBSE Exams 2021 You are here

Example 29 Not in Syllabus - CBSE Exams 2021

Example 30

Example 31 Important Not in Syllabus - CBSE Exams 2021

Example 32 Important Not in Syllabus - CBSE Exams 2021

Example 33 Important Not in Syllabus - CBSE Exams 2021

Example 34 Important Not in Syllabus - CBSE Exams 2021

Chapter 4 Class 12 Determinants

Serial order wise

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.