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# 8.2 - 1 10 0 1 Note the sum of the columns is the zero...

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4 ) A = - 1 1 0 - 1 0 1 0 - 1 1 . Note the sum of the columns is the zero vector so the rank of A is at most 2. It is in fact 2 as the first two columns are visibly dependent. Taking b to be the first column of A , we can solve Ax = b . A vector perpendicular to the first two columns will not be in the column space of A . Playing with numbers a bit (or taking the cross product of the first two columns) gives c = 1 - 1 1 as a vector for which there is no solution to Ax = c . For a given b , if we want a solution, b must lie in the plane spanned by the first two columns. this plane is perpendicular to c , so b must be perpendicular to c . 7 ) A T CA = - 1 - 1 0 1 0 - 1 0 1 1 1 0 0 0 2 0 0 0 2 - 1 1 0 - 1 0 1 0 - 1 1 = - 1 - 1 0 1 0 - 1 0 1 1 - 1 1 0 - 2 0 2 0 - 2 2 = 3 - 1 - 2 - 1 3 - 2 - 2 - 2 4 . The sum of the columns is zero so 1 1 1 is in the nullspace. The first two columns are visibly independent so the rank is 2. In A we have the loo along edge 1, then edge 3, then edge 2 backwards. This corresponds to Row 1 + Row 3 - Row 2 being the 0 Row.

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