Bn is still the same as An except for b11 = 1. So use linearity in the first row, where
Chapter 4, Problem 4.3.27(choose chapter or problem)
Bn is still the same as An except for b11 = 1. So use linearity in the first row, where [1 1 0] equals [2 1 0] minus [1 0 0]: |Bn| = 1 1 0 1 An1 0 = 2 1 0 1 An1 0 1 0 0 1 An1 0 . Linearity in row 1 gives |Bn| = |An| |An1| = .
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