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Example 28 - Solve by matrix method 3x-2y+3z=8 2x+y-z=1 - Examples

Example 28 - Chapter 4 Class 12 Determinants - Part 2
Example 28 - Chapter 4 Class 12 Determinants - Part 3
Example 28 - Chapter 4 Class 12 Determinants - Part 4
Example 28 - Chapter 4 Class 12 Determinants - Part 5
Example 28 - Chapter 4 Class 12 Determinants - Part 6
Example 28 - Chapter 4 Class 12 Determinants - Part 7

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Transcript

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

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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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.