Check sibling questions

Example 33 - Use product to solve x-y+2z=1 2y-3z=1 3x-2y+4z=2

Example 33 - Chapter 4 Class 12 Determinants - Part 2
Example 33 - Chapter 4 Class 12 Determinants - Part 3
Example 33 - Chapter 4 Class 12 Determinants - Part 4
Example 33 - Chapter 4 Class 12 Determinants - Part 5

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Transcript

Example 33 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)] Now, X = A-1 B Putting Value [■8(𝑥@𝑦@𝑧)] = [■8(−2&0&1@9&2&−3@6&1&−2)] [■8(1@1@2)] = [■8(−2&0&1@9&2&−3@6&1&−2)] [■8(1@1@2)] (Calculated above) [■8(𝑥@𝑦@𝑧)] = [■8(−2(1)+0(1)+1(2)@9(1)+2(1)−3(2)@6(1)+1(1)−2(2))] [■8(𝑥@𝑦@𝑧)] =[■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

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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.