# Example 19 - Chapter 4 Class 12 Determinants

Last updated at April 16, 2024 by Teachoo

Examples

Example 1

Example 2

Example 3

Example 4

Example 5 Important

Example 6

Example 7 Important

Example 8

Example 9

Example 10

Example 11 Important

Example 12

Example 13 Important

Example 14

Example 15 Important

Example 16

Example 17 Important

Example 18

Example 19 Important You are here

Question 1 Deleted for CBSE Board 2025 Exams

Question 2 Deleted for CBSE Board 2025 Exams

Question 3 Deleted for CBSE Board 2025 Exams

Question 4 Important Deleted for CBSE Board 2025 Exams

Question 5 Important Deleted for CBSE Board 2025 Exams

Question 6 Deleted for CBSE Board 2025 Exams

Question 7 Deleted for CBSE Board 2025 Exams

Question 8 Deleted for CBSE Board 2025 Exams

Question 9 Important Deleted for CBSE Board 2025 Exams

Question 10 Important Deleted for CBSE Board 2025 Exams

Question 11 Important Deleted for CBSE Board 2025 Exams

Question 12 Deleted for CBSE Board 2025 Exams

Question 13 Important Deleted for CBSE Board 2025 Exams

Question 14 Important Deleted for CBSE Board 2025 Exams

Question 15 Important Deleted for CBSE Board 2025 Exams

Chapter 4 Class 12 Determinants

Serial order wise

Last updated at April 16, 2024 by Teachoo

Example 19 Use product [■8(1&−1&2@0&2&−3@3&−2&4)] [■8(−2&0&1@9&2&−3@6&1&−2)] to solve the system of equations x – y + 2z = 1 2y – 3z = 1 3x – 2y + 4z = 2 Consider the product [■8(1&−1&2@0&2&−3@3&−2&4)] [■8(−2&0&1@9&2&−3@6&1&−2)] =[■8(1(⤶7−2)+(⤶7−1)(9)+2(6)&1(0)+(−1)(2)+2(1)&1(1)+(−1)(−3)+2(−2)@0(−2)+2(9)+(−3)(6)&0(0)+2(2)+(−3)(1)&0(1)+0(−3)+(−3)(−2)@3(−2)+(−2)(9)+(6)&3(0)+(−2)(2)+4(1)&3(1)+(−2)(−3)+4(−2))] = [■8(−2−9+12&0−2+2&1+3−4@0+18−18&0+4−3&0−6+6@−6−18+24&0−4+4&3+6−8)] = [■8(1@0@0)" " ■8(0@1@0)" " ■8(0@0@1)] We know that AA-1 = I So [■8(−2&0&1@0&2&−3@6&1&−2)] is inverse of [■8(1&−1&2@0&2&−3@3&−2&4)] i.e. [■8(𝟏&−𝟏&𝟐@𝟎&𝟐&−𝟑@𝟑&−𝟐&𝟒)]^(−𝟏)= [■8(−𝟐&𝟎&𝟏@𝟗&𝟐&−𝟑@𝟔&𝟏&−𝟐)] Given equations are x – y + 2z = 1 2y – 3z = 1 3x – 2y + 4z = 2 Writing the equation as AX = B [■8(1&−1&2@0&2&−3@3&−2&4)] [■8(𝑥@𝑦@𝑧)] = [■8(1@1@2)] Here A = [■8(𝟏&−𝟏&𝟐@𝟎&𝟐&−𝟑@𝟑&−𝟐&𝟒)], X = [■8(𝒙@𝒚@𝒛)] & B = [■8(𝟏@𝟏@𝟐)] Now, AX = B X = A-1 B Here A = [■8(1&−1&2@0&2&−3@3&−2&4)] So, A-1 = [■8(1&−1&2@0&2&−3@3&−2&4)]^(−1)= [■8(−2&0&1@9&2&−3@6&1&−2)] X = A-1 B Putting values [■8(𝒙@𝒚@𝒛)] = [■8(−𝟐&𝟎&𝟏@𝟗&𝟐&−𝟑@𝟔&𝟏&−𝟐)] [■8(𝟏@𝟏@𝟐)] [■8(𝑥@𝑦@𝑧)] = [■8(−2(1)+0(1)+1(2)@9(1)+2(1)−3(2)@6(1)+1(1)−2(2))] =[■8(−2+0+2@9+2−6@6+1−4)] [■8(𝑥@𝑦@𝑧)] = [■8(0@5@3)] Hence x = 0 , y = 5 & z = 3