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Let M be a 3 × 3 invertible matrix with real entries and let I denote the 3 × 3 identity matrix. If M–1 = adj (adj M), then which of the following statement is/are ALWAYS TRUE ? (A) M = I (B) det M = 1 (C) M2 = I (D) (adj M)2 = I |
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Answer» \(\because\) M is invertible (given) \(\therefore|M|\neq0\) Now given that adj (adj M) \(=M^{-1}\) \(\Rightarrow\) |adj (adj M)| \(=|M|^{-1}\) (By taking determinant of both sides) \(\Rightarrow|adj\,M|^{3-1}=\frac{1}{|M|}\) \((\because|adj\,A|=|A|^{n-1}\) when n is order of matrix) \(\Rightarrow(|M|^{3-1})^2=\frac{1}{|M|}\) \((\because|adj\,M|=|M|^{3-1}=|M|^2)\) \(\Rightarrow|M|^4=\frac{1}{|M|}\) \(\Rightarrow|M|^5=1\) \(\Rightarrow\) |M| = 1 or |M| = 0 \(\Rightarrow\) |M| = 1 \((\because M^{-1}exists\Rightarrow|M|\neq0)\) Now, \(M^{-1}=adj\,(adj\,M)\) \(\Rightarrow\frac{adj\,M}{|M|}=|adj\,M|(adj\,M)^{-1}\) \((\because adj\,A=|A|A^{-1}\,and\,A^{-1}=\frac{adj\,A}{|A|})\) \(\Rightarrow\frac{adj\,M.adj\,M}{|M|}=|adj\,M|(adj\,M)^{-1}(adj\,M)\) (Multiplying both sides by adj M) \(\Rightarrow(adj\,M)^2=|M|^{3-1}|M|I\) \((\because A^{-1}A=I\) \(and\,|adj\,M|=|M|^{3-1}=|M|^2)\) \(=|M|^3I\) \(=I\;\;(\because|M|=1)\) \(\because(adj\,M)^2=I\) \(\Rightarrow(|M|M^{-1})^2=I\) \((\because adj\,M=|M|M^{-1})\) \(\Rightarrow(M^{-1})^2=I\) \((\because|M|=1)\) \(\Rightarrow M^{-2}=I\) \(\Rightarrow M^2M^{-2}=M^2I\) (Multiplying both sides by \(M^2)\) \(\Rightarrow I=M^2\) \((\because AI=A\,and\,AA^{-1}=I)\) \(\Rightarrow M^2=I\) |
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