# A the row reduced echelon form of the augmented

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Unformatted text preview: ow reduced echelon form of the augmented matrix is 2 3 1 0 ?1 ?2 0 60 1 2 3 07 6 7 40 0 0 0 15 00 0 00 Because the third row has its only nonzero entry in the last place, the system is inconsistent. (b) The row reduced echelon form of the augmented matrix is 3 2 1 0 ?1 ?2 7 60 1 2 3 ?4 7 7 6 40 0 0 0 05 00 0 0 0 Thus the general solution is 3 2 3 3 2 2 2 1 7 7 6 7 6 6 ?4 7 7 + s 6 ?2 7 + t 6 ?3 7 ; s; t 2 R x=6 4 05 4 15 4 05 1 0 0 The solution set is not a subspace since, for example, it doesn't contain the zero vector. The associated homogeneous system has the general solution given by 3 3 2 2 2 1 7 6 6 ?2 7 7 + t 6 ?3 7 ; s; t 2 R x = s6 4 05 4 15 1 0 MATA04Y page 3 The solution set is not a subspace since, again, it doesn't contain the zero vector. The solution set of the associated homogeneous system is x = tw where 3 2 1 6 ?2 7 7 w=6 4 15 0 and a basis for this space is the single vector w itself, since it clearly spans the solution set and, since it is non-zero, gives a basis. 2. Write B = (x; y; z) so that, since the \ABC = 90 0 = BA B C = ((x; y; z) ? (2; 1; ?1)) ((x; y; z) ? (2; 3; 3)) = (x ? 2; y ? 1; z + 1) (x ? 2; y ? 3; z ? 3) = (x ? 2)2 + (y ? 1)(y ? 3) + (z + 1)(z ? 3) Now the augmented matrix for the system of equations has reduced row echelon form 1 4 ?2 3 ?! 1 0 2 7 0 1 ?1 ?1 2 3 1 11 which has the form I so the column vectors form a basis of the associated 0 homogeneous system. (c) The reduced row echelon form of the augmented matrix is 2 3 1 0 ?1 0 6 60 1 5 2 0 ?2 7 6 7 6 17 40 0 0 1 ?2 5 00 00 0 so the general solution may be written as 3 2 3 2 6 1 6 ?5 7 6 ?2 7 7 6 7 t2R x = 6 2 7+t6 4 15; 4 05 0 ?1 2 and the two column vectors appearing here form a basis since, for example, row reduction gives 3 2 3 2 12 10 6 ?2 ?3 7 7 6 7 ?! 6 0 1 7 6 41 40 05 05 01 00 MATA04Y yielding the general solution page 4 x = 7 ? 2t y = ?1 + t z=t where t 2 R . Substituting thes...
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