ABSTRACT
In Sec. 1.2 we showed that the problem a 11 b 1 a 21 b 1 ⋮ a p 1 b 1 + a 12 b 2 + a 22 b 2 ⋮ + a p 2 b 2 + ⋯ + ⋯ + ⋯ + a 1 p b p + a 2 p b p ⋮ + a p p b p = c 1 = c 2 ⋮ = c p https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203748923/a6b92606-1b5b-4bf2-bc51-ca8dfd20fd64/content/eq66.tif"/>
