





Ex 4.6
Ex 4.6, 2 Deleted for CBSE Board 2022 Exams
Ex 4.6, 3 Important Deleted for CBSE Board 2022 Exams
Ex 4.6, 4 Deleted for CBSE Board 2022 Exams
Ex 4.6, 5 Important Deleted for CBSE Board 2022 Exams
Ex 4.6, 6 Deleted for CBSE Board 2022 Exams
Ex 4.6, 7
Ex 4.6, 8
Ex 4.6, 9
Ex 4.6, 10 Important
Ex 4.6, 11
Ex 4.6, 12
Ex 4.6, 13 Important
Ex 4.6, 14 Important You are here
Ex 4.6, 15 Important
Ex 4.6, 16 Important
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