# 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

Examples

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Example 2

Example 3

Example 4

Example 5

Example 6

Example 7

Example 8

Example 9

Example 10

Example 11

Example 12

Example 13

Example 14 Important

Example 15 Important

Example 16 Important

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

Example 28 You are here

Example 29

Example 30

Example 31

Example 32 Important

Example 33

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.