# Example 28 - Chapter 4 Class 12 Determinants (Term 1)

Last updated at Jan. 23, 2020 by Teachoo

Last updated at Jan. 23, 2020 by Teachoo

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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Example 28 Important You are here

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Chapter 4 Class 12 Determinants (Term 1)

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.