Slide42.JPG

Slide43.JPG
Slide44.JPG Slide45.JPG Slide46.JPG Slide47.JPG Slide48.JPG Slide49.JPG

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Ex 4.5, 12 Solve system of linear equations, using matrix method. x − y + z = 4 2x + y − 3z = 0 x + y + z = 2 The system of equations is x − y + z = 4 2x + y − 3z = 0 x + y + z = 2 Step 1 Write equation as AX = B [■8(1&−1&1@2&1&−3@1&1&1)] [■8(𝑥@𝑦@𝑧)] = [■8(4@0@2)] Hence A = [■8(1&−1&1@2&1&−3@1&1&1)] , X = [■8(𝑥@𝑦@𝑧)] & B = [■8(4@0@2)] Step 2 Calculate |A| |A| = |■8(1&−1&1@2&1&−3@1&1&1)| = 1 |■8(1&−3@1&1)| – ( –1) |■8(2&−3@1&1)| + 1 |■8(2&1@1&1)| = ( 1 + 3) + 1 ( 2 + 3) + 1 (2 – 1) = 1 (4) + 1 (5) + 1 (1) = 4 + 5 + 1 = 10 Since |A|≠ 0 ∴ The system of equation is consistent & has a unique solution Now, AX = B X = A-1 B Step 3 Calculate 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&1@2&1&−3@1&1&1)] M11 = [■8(1&−3@1&1)] = 1 + 3 = 4 M12 = |■8(2&−3@1&1)| = 2 + 3 = 5 M13 = |■8(2&1@1&1)| = 2 – 1 = 1 M21 = |■8(−1&1@1&1)| = −1 – 1 = – 2 M22 = |■8(1&1@1&1)| = 1 – 1 = 0 M23 = |■8(1&−1@1&1)| = 1 + 1 = 2 M31 = |■8(−1&1@1&−3)| = 3 – 1 = 2 M32 = |■8(1&1@2&−3)| = –3 – 4 = – 5 M33 = |■8(1&−1@2&1)| = 3 + 2 = 3 A11 = 〖"(–1)" 〗^(1+1) M11= (–1)2 . 4 = 4 A12 = 〖"(–1)" 〗^"1+2" M12 = 〖"( –1)" 〗^"3" . 5 = – 5 A13 = 〖(−1)〗^(1+3) M13= 〖( −1)〗^4 . (1) = 1 A21 = (−1)^(2+1) M21= 〖( −1)〗^3 . (-2) = 2 A22 = 〖(−1)〗^(2+2) M22 = ( –1)4 . 0 = 0 A23 = 〖(−1)〗^(2+3). M23 = 〖(−1)〗^5. ( 2) = – 2 A31 = 〖(−1)〗^(3+1). M31 = 〖(−1)〗^4 . (2) = 2 A32 = 〖(−1)〗^(3+2) . M32 = 〖(−1)〗^5. ( – 5) = 5 A33 = 〖(−1)〗^(3+3) . M33 = ( –1)6 . 3 = 3 Thus , adj A = [■8(4&2&2@−5&0&5@1&−2&3)] & |A| = 10 So, A-1 = 1/(|A|) adj A A-1 = 1/10 [■8(4&2&2@−5&0&5@1&−2&3)] & B = [■8(4@0@2)] Now, solving X = A-1 B [■8(𝑥@𝑦@𝑧)] = 1/10 [■8(4&2&2@−5&0&5@1&1&3)] [■8(4@0@2)] = 1/10 [■8(4(4)+2(0)+2(2)@0(4)+0(0)+5(2)@2(4)+1(0)+3(2))] [■8(𝑥@𝑦@𝑧)] = 1/10 [■8(16+0+4@−20+0+10@4+0+6)] = 1/10 [■8(20@−10@10)] [■8(𝑥@𝑦@𝑧)] = [■8(2@−1@1)] Hence, x = 2 , y = –1, & z = 1

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.