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Exercise [13.37]

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Exercise [13.37]
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Shaun Culver Site Admin Joined: 25 Feb 2008, 13:32 Posts: 105 Location: Cape Town, South Africa	Exercise [13.37] Let $\text{A}$ and $\text{B}$ be non-singular square matrices. $(\text{AB})(\text{B}^{-1}\text{A}^{-1})=\text{A}(\text{BB}^{-1})\text{A}^{-1}=\text{A}I\text{A}^{-1}=\text{AA}^{-1}=I$ _________________ “Our imagination is stretched to the utmost, not, as in fiction, to imagine things which are not really there, but just to comprehend those things which are there.” Last edited by Shaun Culver on 08 Aug 2008, 18:51, edited 1 time in total.
14 Jul 2008, 01:11

Sameed Zahoor Joined: 12 Mar 2008, 10:57 Posts: 69 Location: India	Re: Exercise [13.37] Alternative solution: Let A be a non-singular square matrix such that $A=A_{1}A_{2}...A_{n}$ Let X be the required inverse of A.Then or $X(A_{1}A_{2}...A_{n})=I$ or $X(A_{1}A_{2}...A_{n})A^{-1}_{n}=I(A^{-1}_{n})$ (post-multiplying b/s by $A^{-1}_{n}$ ) or $X(A_{1}A_{2}...A_{n-1})(A_{n}A^{-1}_{n})=A^{-1}_{n}$ or $X(A_{1}A_{2}...A_{n-1})I=A^{-1}_{n}$ or $X(A_{1}A_{2}...A_{n-2})(A_{n-1}A^{-1}_{n-1})=A^{-1}_{n}A^{-1}_{n-1}$ (post-multiplying b/s by $A^{-1}_{n-1}$ ) or $X(A_{1}A_{2}...A_{n-2})I=A^{-1}_{n}A^{-1}_{n-1}$ Continuing in this fashion we will have, $X=A^{-1}_{n}A^{-1}_{n-1}A^{-1}_{n-2}...A^{-1}_{2}A^{-1}_{1}$
14 Jul 2008, 08:17

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