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  • If A and B are invertible matrices, then which of the following is not correct? A. adj A = |A|. A-1 B. det (A)-1 = [det (A)]-1

If A and B are invertible matrices, then which of the following is not correct?
A. adj A = |A|. A^{-1}

B. det (A)^{-1} = [det (A)]^{-1}

C. (AB)^{-1} = B^{-1} A^{-1}
D. (A + B)^{-1} = B^{-1} + A^{-1}

Answers (1)

D)

We know, A and B are invertible matrices

\\ \vspace{\baselineskip} Consider (AB) B\textsuperscript{-1} A\textsuperscript{-1}\\ \\ \vspace{\baselineskip}$ \Rightarrow $ (AB) B\textsuperscript{-1} A\textsuperscript{-1} = A(BB\textsuperscript{-1}) A\textsuperscript{-1}\\ \\ \vspace{\baselineskip}= AIA\textsuperscript{-1} = (AI) A\textsuperscript{-1}\\ \\ \vspace{\baselineskip}= AA\textsuperscript{-1} = I\\ \\ \vspace{\baselineskip}$ \Rightarrow $ (AB)\textsuperscript{-1} = B\textsuperscript{-1} A\textsuperscript{-1} $ \ldots $ option (C)\\
\\ \\ \vspace{\baselineskip}Also AA\textsuperscript{-1} = I\\ \\ \vspace{\baselineskip}$ \Rightarrow $ $ \vert $ AA\textsuperscript{-1}$ \vert $ = $ \vert $ I$ \vert $ \\ \\ \vspace{\baselineskip}$ \Rightarrow $ $ \vert $ A$ \vert $ $ \vert $ A\textsuperscript{-1}$ \vert $ = 1\\

\\ \begin{aligned} &\Rightarrow|\mathrm{A}|^{-1}=\frac{1}{|\mathrm{~A}|}\\ &\therefore \operatorname{det}(A)^{-1}=[\operatorname{det}(A)]^{-1} \ldots(B)\\ &\text { We know that } \frac{|\mathrm{A}|^{-1}}{ }=\frac{\operatorname{adj} \mathrm{A}}{|\vec{A}|}\\ &\Rightarrow \text { adj } A=|A| \cdot A^{-1} \ldots \text { option }(A)\\ &\Rightarrow(\mathrm{A}+\mathrm{B})^{-1}=\frac{1}{|\mathrm{~A}+\mathrm{B}|} \operatorname{adj}(\mathrm{A}+\mathrm{B})\\ &{\text { But }}{\mathrm{B}}^{-1}+\mathrm{A}^{-1}=\frac{1}{|\mathrm{~B}|} \operatorname{adj} \mathrm{B}+\frac{1}{|\mathrm{~A}|} \text { adj } \mathrm{A} \end{aligned}

\therefore $ (A + B)\textsuperscript{-1} $ \neq $ B\textsuperscript{-1} + A\textsuperscript{-1}\\

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infoexpert22

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