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Linear Algebra Solutions 26

# Linear Algebra Solutions 26 - AB is thereFore singular as X...

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The associated normal equations are given by 5 10 30 10 30 100 30 100 354 a b c = 11 42 160 , which have the solution a = 1 / 5 , b = - 2 , c = 1. 24. Suppose that A is symmetric, i.e. A t = A and that AB is defined. Then ( B t AB ) t = B t A t ( B t ) t = B t AB, so B t AB is also symmetric. 25. Let A be m × n and B be n × m , where m > n . Then the homogeneous system BX = 0 has a non–trivial solution X 0 , as the number of unknowns is greater than the number of equations. Then ( AB ) X 0 = A ( BX 0 ) = A 0 = 0 and the m × m matrix
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Unformatted text preview: AB is thereFore singular, as X 6 = 0. 26. (i) Let B be a singular n × n matrix. Then BX = 0 For some non–zero column vector X . Then ( AB ) X = A ( BX ) = A 0 = 0 and hence AB is also singular. (ii) Suppose A is a singular n × n matrix. Then A t is also singular and hence by (i) so is B t A t = ( AB ) t . Consequently AB is also singular 29...
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