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Misc 8 Choose the correct answer. If x, y, z are nonzero real numbers, then the inverse of matrix A = [■8(x&0&0@0&y&0@0&0&z)] is A. [■8(𝑥^(−1)&0&0@0&𝑦^(−1)&0@0&0&𝑧^(−1) )] B. xyz[■8(𝑥^(−1)&0&0@0&𝑦^(−1)&0@0&0&𝑧^(−1) )] C. 1/xyz [■8(x&0&0@0&y&0@0&0&z)] D. 1/xyz [■8(1&0&0@0&1&0@0&0&1)] Given A = [■8(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| = |■8(x&0&0@0&y&0@0&0&z)| = x |■8(𝑦&0@0&𝑧)| – 0 |■8(0&0@0&𝑧)| + 0 |■8(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 = [■8(A_11&A_21&A_31@A_12&A_22&A_32@A_13&A_23&A_33 )] A = [■8(𝑥&0&0@0&𝑦&0@0&0&𝑧)] M11 = |■8(𝑦&0@0&𝑧)| = yz – 0 = yz M12 = |■8(0&0@0&𝑧)| = 0 – 0 = 0 M13 = |■8(0&y@0&0)| = 0 – 0 = – 0 M21 = |■8(0&0@0&𝑧)| = 0 – 0 = 0 M22 = |■8(x&0@0&𝑧)| = xz – 0 = xz M23 = |■8(x&0@0&0)| = 0 – 0 = 0 M31 = |■8(0&0@0&z)| = 0 – 0 = 0 M32 = |■8(x&0@0&0)| = 0 – 0 = 0 M33 = |■8(𝑥&0@0&y)| = x y – 0 = xy 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) = [■8(A11&A21&A31@A12&A22&A32@A33&A23&A33)] = [■8(𝑥 𝑦&0&0@0&𝑧 𝑦&0@0&0&𝑥 𝑦)] Now, A-1 = 1/(|A|) adj (A) = 𝟏/𝒙𝒚𝒛 [■8(𝒙 𝒚&𝟎&𝟎@𝟎&𝒛 𝒚&𝟎@𝟎&𝟎&𝒙 𝒚)] = [■8(𝑦𝑧/𝑥𝑦𝑧&0&0@0&𝑦𝑧/𝑥𝑦𝑧&0@0&0&𝑦𝑧/𝑥𝑦𝑧)] = [■8(𝟏/𝒙&𝟎&𝟎@𝟎&𝟏/𝒚&𝟎@𝟎&𝟎&𝟏/𝒛)] = [■8(𝑥^(−1)&0&0@0&𝑦^(−1)&0@0&0&𝑧^(−1) )] Thus, the correct option is A

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo