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Misc 18 - If x, y, z are nonzero real numbers, then inverse - Finding Inverse of a matrix

Misc 18 - Chapter 4 Class 12 Determinants - Part 2
Misc 18 - Chapter 4 Class 12 Determinants - Part 3
Misc 18 - Chapter 4 Class 12 Determinants - Part 4
Misc 18 - Chapter 4 Class 12 Determinants - Part 5

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Misc 18 Choose the correct answer. If x, y, z are nonzero real numbers, then the inverse of matrix A = x﷮0﷮0﷮0﷮y﷮0﷮0﷮0﷮z﷯﷯ is A. 𝑥﷮−1﷯﷮0﷮0﷮0﷮ 𝑦﷮−1﷯﷮0﷮0﷮0﷮ 𝑧﷮−1﷯﷯﷯ B. xyz 𝑥﷮−1﷯﷮0﷮0﷮0﷮ 𝑦﷮−1﷯﷮0﷮0﷮0﷮ 𝑧﷮−1﷯﷯﷯ C. 1﷮xyz﷯ x﷮0﷮0﷮0﷮y﷮0﷮0﷮0﷮z﷯﷯ D. 1﷮xyz﷯ 1﷮0﷮0﷮0﷮1﷮0﷮0﷮0﷮1﷯﷯ Given A = x﷮0﷮0﷮0﷮y﷮0﷮0﷮0﷮z﷯﷯ We have to find A-1 We know that A-1 = 1﷮|A|﷯ adj (A) exists if |A| ≠ 0 Calculating |A| |A| = x﷮0﷮0﷮0﷮y﷮0﷮0﷮0﷮z﷯﷯ = x 𝑦﷮0﷮0﷮𝑧﷯﷯ – 0 0﷮0﷮0﷮𝑧﷯﷯ + 0 0﷮𝑦﷮0﷮0﷯﷯ = x (yz – 0) – 0 (0 – 0) + 0 (0 – 0) = x(yz) + 0 + 0 = xyz Since |A| ≠ 0 Thus, A-1 exist Now, adj A = A﷮11﷯﷮ A﷮21﷯﷮ A﷮31﷯﷮ A﷮12﷯﷮ A﷮22﷯﷮ A﷮32﷯﷮ A﷮13﷯﷮ A﷮23﷯﷮ A﷮33﷯﷯﷯ A = 𝑥﷮0﷮0﷮0﷮𝑦﷮0﷮0﷮0﷮𝑧﷯﷯ M11 = 𝑦﷮0﷮0﷮𝑧﷯﷯ = yz – 0 = yz M12 = 0﷮0﷮0﷮𝑧﷯﷯ = 0 – 0 = 0 M13 = 0﷮y﷮0﷮0﷯﷯ = 0 – 0 = – 0 M21 = 0﷮0﷮0﷮𝑧﷯﷯ = 0 – 0 = 0 M22 = x﷮0﷮0﷮𝑧﷯﷯ = xz – 0 = xz A11 = ( – 1)1 + 1 M11 = ( – 1)2 y z = yz A12 = ( – 1)1+2 M12 = ( – 1)3 0 = 0 A13 = ( – 1)1+3 M13 = ( – 1)4 0 = – 0 A21 = ( – 1)2+1 M21 = ( – 1)3. 0 = 0 A22 = ( – 1)2+2 M22 = ( – 1)4 x z = x z A23 = ( – 1)2+3 M23 = ( – 1)5 0 = 0 A31 = ( – 1)3+1 M31 = ( – 1)4 0 = 0 A32 = ( – 1)3+2 M32 = ( – 1)5 0 = 0 A33 = ( – 1)3+3 M33 = ( – 1)6 xy = xy Thus, adj (A) = A11﷮A21﷮A31﷮A12﷮A22﷮A32﷮A33﷮A23﷮A33﷯﷯ = 𝑥 𝑦﷮0﷮0﷮0﷮𝑧 𝑦﷮0﷮0﷮0﷮𝑥 𝑦﷯﷯ Now, A-1 = 1﷮|A|﷯ adj (A) = 1﷮𝑥𝑦𝑧﷯ 𝑥 𝑦﷮0﷮0﷮0﷮𝑧 𝑦﷮0﷮0﷮0﷮𝑥 𝑦﷯﷯ = 𝑦𝑧﷮𝑥𝑦𝑧﷯﷮0﷮0﷮0﷮ 𝑦𝑧﷮𝑥𝑦𝑧﷯﷮0﷮0﷮0﷮ 𝑦𝑧﷮𝑥𝑦𝑧﷯﷯﷯ = 1﷮𝑥﷯﷮0﷮0﷮0﷮ 1﷮𝑦﷯﷮0﷮0﷮0﷮ 1﷮𝑧﷯﷯﷯ = 𝑥﷮−1﷯﷮0﷮0﷮0﷮ 𝑦﷮−1﷯﷮0﷮0﷮0﷮ 𝑧﷮−1﷯﷯﷯ Thus, the correct option is A

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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.