     1. Chapter 4 Class 12 Determinants
2. Serial order wise
3. Ex 4.6

Transcript

Ex4.6, 15 If A = 2﷮−3﷮5﷮3﷮2﷮−4﷮1﷮1﷮−2﷯﷯, find A−1. Using A−1 solve the system of equations 2x – 3y + 5z = 11 3x + 2y – 4z = -5 x + y – 2z = -3 Step 1 Write equation as AX = B 2﷮−3﷮5﷮3﷮2﷮−4﷮1﷮1﷮−2﷯﷯ 𝑥﷮𝑦﷮𝑧﷯﷯ = 11﷮−5﷮−3﷯﷯ i.e. AX = B Hence A = 2﷮−3﷮5﷮3﷮2﷮−4﷮1﷮1﷮−2﷯﷯ , X = 𝑥﷮𝑦﷮𝑧﷯﷯ & B = 11﷮−5﷮−3﷯﷯ Step 2 Calculate |A| |A|= 2﷮−3﷮5﷮3﷮2﷮−4﷮1﷮1﷮−2﷯﷯ = 2 (−4 + 4) − 3 (6 − 5) + 1 (12 − 10) = 2(0) − 3 (1) + 1(2) = −1 So, |A|≠ 0 ∴ The system of equation is consistent & has a unique solutions AX = B X = A-1 B Step 3 Calculating X = A-1 B Calculating A-1 Now, A-1 = 1﷮|A|﷯ adj (A) adj A = A11﷮A12﷮A13﷮A21﷮A22﷮A23﷮A31﷮A32﷮A33﷯﷯﷮′﷯ = A11﷮A21﷮A31﷮A12﷮A22﷮A32﷮A13﷮A23﷮A33﷯﷯ A = 1﷮−1﷮2﷮3﷮4﷮−5﷮2﷮−1﷮3﷯﷯ 𝐴11 = −4 + 4 = 0 𝐴12 = − −6−(−4)﷯ = − (−6 + 4) = −2 = 2 𝐴13 = 3 − 2 = 1 𝐴21 = (6 − 5) = −1 𝐴22 = −4 − 5 = −9 𝐴23 = −2−(−3﷯ = −5 𝐴31 = 12−10=2 𝐴32 = −8−15﷯=23 𝐴33 = 4− −9﷯=13 Thus adj A = 0﷮−1﷮2﷮2﷮−9﷮23﷮1﷮−5﷮13﷯﷯ & |A| = –1 Now, A-1 = 1﷮|A|﷯ adj A A-1 = 1﷮−1﷯ 0﷮−1﷮2﷮2﷮−9﷮23﷮1﷮−5﷮13﷯﷯ = 0﷮1﷮−2﷮−2﷮9﷮−23﷮−1﷮5﷮−13﷯﷯ Solving X = A-1 B 𝑥﷮𝑦﷮𝑧﷯﷯ = − 0﷮1﷮−2﷮2﷮9﷮−23﷮−1﷮5﷮−13﷯﷯ 11﷮−5﷮−3﷯﷯ 𝑥﷮𝑦﷮𝑧﷯﷯ = −5+6﷮−22−45+69﷮−11−25+39﷯﷯ 𝑥﷮𝑦﷮𝑧﷯﷯ = 1﷮2﷮3﷯﷯ ∴ x = 1, y = 2 and z = 3

Ex 4.6 