Linear Algebra and Its Applications, 4th Edition

Published by Brooks Cole
ISBN 10: 0030105676
ISBN 13: 978-0-03010-567-8

Chapter 1 - Section 1.6 - Inverses and Transposes - Problem Set - Page 52: 3


$$A^{-1}=BC^{-1}$$ $$A^{-1}=U^{-1}L^{-1}P$$

Work Step by Step

$$AB=C$$ Our goal is to find a formula for $A^{-1}$. Note that matrices are not mutually commutative, i.e. $a\cdot{b}\neq{b}\cdot{a}$. Thus, when multiplying both sides of the equation by any matrix, we must ensure that the inverse is always placed on the same side. Introduce $A^{-1}$ into the equation as a multiple on both sides. $$A^{-1}AB=A^{-1}C$$ Since $|A^{-1}\cdot{A}|=1$, $$B=A^{-1}C$$ We proceed to eliminate C from the RHS by multiplying C by its inverse, leaving $A^{-1}$. $$BC^{-1}=A^{-1}CC^{-1}=A^{-1}$$ We now aim to achieve this same objective with the equation $$PA=LU$$ As with earlier, introduce $A^{-1}$ to the equation as a multiple on both sides $$PAA^{-1}=LUA^{-1}$$ $$P=LUA^{-1}$$ $LU$ refers to the product $L\cdot{U}$. To isolate $A^{-1}$, we multiply both sides of the equation by the inverse of this product, $(LU)^{-1}$. $$(LU)^{-1}P=(LU)^{-1}LUA^{-1}=A^{-1}$$ Since $(LU)^{-1}=U^{-1}L^{-1}$, $$U^{-1}L^{-1}P=A^{-1}$$
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