Misc 18 (MCQ) - Chapter 4 Class 12 Determinants (Term 1)

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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