# Ex 4.6, 14 - Chapter 4 Class 12 Determinants

Last updated at Jan. 23, 2020 by Teachoo

Last updated at Jan. 23, 2020 by Teachoo

Transcript

Ex 4.6, 14 Solve system of linear equations, using matrix method. x − y + 2z = 7 3x + 4y − 5z = −5 2x − y + 3z = 12 The system of equations are x − y + 2z = 7 3x + 4y − 5z = −5 2x − y + 3z = 12 Writing equation as AX = B [■8(1&−1&2@3&4&−5@2&−1&3)] [■8(𝑥@𝑦@𝑧)] = [■8(7@−5@12)] Hence, A = [■8(1&−1&2@3&4&−5@2&−1&3)] , X = [■8(𝑥@𝑦@𝑧)] & B = [■8(7@−5@12)] Calculating |A| |A| = |■8(1&−1&2@3&4&−5@2&−1&3)| = 1 |■8(4&−5@−1&−3)| – ( –1) |■8(3&−5@2&3)| + 2 |■8(3&4@2&−1)| = 1 ( 12 – 5) + 1 (9 + 10) + 2 (–3 – 8) = 2 (7) + 1 (19) + 2 (–11) = 7 + 19 – 22 = 4 Since |A|≠ 0 ∴ The system of equation is consistent & has a unique solution Now, AX = B X = A-1 B Calculating A-1 Now, A-1 = 1/(|A|) adj (A) adj A = [■8(A11&A12&A13@A21&A22&A23@A31&A32&A33)]^′ = [■8(A11&A21&A31@A12&A22&A32@A13&A23&A33)] A = [■8(1&−1&2@3&4&−5@2&−1&3)] M11 = [■8(4&−5@−1&3)] = 12 – 5 = 7 M12 = |■8(3&−5@2&3)| = 9 + 10 = 19 M13 = |■8(3&4@2&−1)| = –3 – 8 = –11 M21 = |■8(−1&2@−1&3)| = −3 + 2 = –1 M22 = |■8(1&2@2&3)| = 3 – 4 = –1 M23 = |■8(1&−1@2&−1)| = –1 + 2 = 1 M31 = |■8(−1&2@4&−5)| = 5 – 8 = –3 M32 = |■8(1&2@3&−5)| = –5 – 6 = –11 M33 = |■8(1&−1@3&4)| = 4 + 3= 7 Now, A11 = 〖"(–1)" 〗^(1+1) M11 = (–1)2 . 7 = 7 A12 = 〖"(–1)" 〗^"1+2" M12 = 〖"(–1)" 〗^"3" . 19 = – 19 A13 = 〖(−1)〗^(1+3) . M13 = 〖(−1)〗^4 . – 11= –11 A21 = 〖(−1)〗^(2+1) M21 = 〖(−1)〗^3 . (–1) = + 1 A22 = 〖(−1)〗^(2+2) M22 = (–1)4 . (–1) = –1 A23 = 〖(−1)〗^(2+3). M23 = 〖(−1)〗^5. 1 = –1 A31 = 〖(−1)〗^(3+1). M31 = 〖(−1)〗^4 . (–3) = –3 A32 = 〖(−1)〗^(3+2) . M32 = 〖(−1)〗^5. (–11) = 11 A33 = 〖(−1)〗^(3+3) . M33 = (–1)6 . (–7) = 7 Thus, adj A = [■8(7&1&−3@−19&−1&11@−11&−1&7)] Now, A-1 = 1/(|A|) adj A A-1 = 1/4 [■8(7&1&−3@−19&−1&11@−11&−1&7)] Solving X = A-1 B [■8(𝑥@𝑦@𝑧)] = 1/4 [■8(7&1&−3@−19&−1&11@−11&−1&7)] [■8(7@−5@12)] [■8(𝑥@𝑦@𝑧)] = 1/4 [■8(7(7)+1(−5)+(−13)(12)@−19(7)+(−1)(−4)+11(12)@−11(7)+1(−5)+7(12))] [■8(𝑥@𝑦@𝑧)] = 1/4 [■8(49−5−36@−133+5+132@−77+5+84)] = 1/4 [■8(8@4@12)] [■8(𝑥@𝑦@𝑧)] = [■8(2@1@3)] Hence, x = 2 , y = 1, & z = 3

Ex 4.6

Ex 4.6, 1
Deleted for CBSE Board 2021 Exams only

Ex 4.6, 2 Deleted for CBSE Board 2021 Exams only

Ex 4.6, 3 Important Deleted for CBSE Board 2021 Exams only

Ex 4.6, 4 Deleted for CBSE Board 2021 Exams only

Ex 4.6, 5 Important Deleted for CBSE Board 2021 Exams only

Ex 4.6, 6 Deleted for CBSE Board 2021 Exams only

Ex 4.6, 7 Deleted for CBSE Board 2021 Exams only

Ex 4.6, 8 Deleted for CBSE Board 2021 Exams only

Ex 4.6, 9 Deleted for CBSE Board 2021 Exams only

Ex 4.6, 10 Important Deleted for CBSE Board 2021 Exams only

Ex 4.6, 11 Deleted for CBSE Board 2021 Exams only

Ex 4.6, 12 Deleted for CBSE Board 2021 Exams only

Ex 4.6, 13 Important Deleted for CBSE Board 2021 Exams only

Ex 4.6, 14 Important Deleted for CBSE Board 2021 Exams only You are here

Ex 4.6, 15 Important Deleted for CBSE Board 2021 Exams only

Ex 4.6, 16 Important Deleted for CBSE Board 2021 Exams only

Chapter 4 Class 12 Determinants

Serial order wise

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.