# Example 28 - Chapter 4 Class 12 Determinants

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Hence A = 3−2321−14−32, 𝑥= 𝑥𝑦𝑧 & B = 814 Step 2 Calculate |A| |A| = 3−2321−14−32 = 3 1−1−32 – 1( – 2) 2−142 + 3 214−3 = 3 (2 – 3 ) – 2 (4 + 4) + 3 ( – 6 – 4) = 3 ( – 1) + 2(8) + 3 ( – 10) = – 3 + 16 – 30 = – 17 Thus, |A| ≠ 0 ∴ System of equation is consistent & has a unique solution Now, AX = B X = A-1 B Step 3 Calculate X = A-1 B A-1 = 1|A| adj (A) adj (A) = A11 A12 A13 A21 A22 A23 A31 A32 A33′ = A11 A21 A13 A12 A22 A23 A13 A32 A33 A = 3−2321−14−32 M11 = 1−1−32 = 2 – 3 = – 1 M12 = 2−142 = 4 + 4 = 8 M13 = 214−3 = – 6 – 4 = – 10 M21 = −23−32 = – 4 + 9 = 5 M22 = 3342 = 6 – 12 = – 6 M23 = 3−24−3 = – 9 + 8 = – 1 M31 = −231−1 = 1 – 2 = – 1 M32 = 332−1 = – 3 – 6 = – 9 M33 = 3−221 = 3 + 4 = 7 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) = −1−5−1−8−69−1017 Now, A-1 = 1|A| adj A Putting values = 1−17 −1−5−1−8−69−1017 Also, X = A-1 B Putting values 𝑥𝑦𝑧 = 1−17 −1−5−1−8−69−1017 814 = 1−17 −1 8+ −5 1+ −14−8 8+ −6 1+9 4−10 8+1 1+7 4 𝑥𝑦𝑧 = 1−17 −8−5−5−64−6+36−80+1+28 = 1−17 −17−36−51 = −17−17 −34−(17) −51−17 𝑥𝑦𝑧 = 123 Hence x = 1, y = 𝟐 & z = 3

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Example 28 You are here

Example 29

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Example 34 Important

Chapter 4 Class 12 Determinants

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.