-k= Xn–1 and Xn+1 = Xn-ơ• It follows from Eq.(1) that P=(A+B+C) Q+DP – (e d) (16) b Q= (A+B+C) P+ DQ (17) - (e – d) - subtracting (17) from (16), we get (P- Q) [(A+B+C+1) – D]=0, ce (A+B+C+1) – D +0, then P= Q. This is a
-k= Xn–1 and Xn+1 = Xn-ơ• It follows from Eq.(1) that P=(A+B+C) Q+DP – (e d) (16) b Q= (A+B+C) P+ DQ (17) - (e – d) - subtracting (17) from (16), we get (P- Q) [(A+B+C+1) – D]=0, ce (A+B+C+1) – D +0, then P= Q. This is a
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter2: Systems Of Linear Equations
Section2.4: Applications
Problem 32EQ
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