Math 51 Linear Algebra Review (solutions)Jonathan LeeOctober 16, 2008Taking a Math 51 midterm in the near future? Then, this just might be for you!Computational things to knowParametrization• L =357+ t−23−7: t ∈ R(or others)• P =461+ s−400+ t0−60: s, t ∈ R(or others)Length and angle•√83,√65• 43/√83 · 65•3 5 71 8 0•pk~uk2+ k~vk2+ k~wk2Column space and NullspaceLetA =1 0 50 1 00 0 00 0 0B =1 0 50 1 00 0 10 0 0C =a b cd e fg h ij k l.• 2, plane, are not guaranteed (because a solution to Ax = b exists only when b is in this planeinside R4; if b is not in the plane, then no solution exists)• 1, line, infinitely many and non-unique (a line has infinitely many points on it!)1•adgj,behk(or others),−501(or others)•320+ span−501• { } or ∅ (don’t exist)• 3, 3-space, are not guaranteed (because a solution exists only if b is in this 3-dimensionalsubspace, which is smaller than R4; if b isn’t, then there is no solution!)• 0, point, unique•adgj,behk,cfil(or others),• { } or ∅•320•−221LetD =1 0 2 0 00 1 3 0 00 0 0 1 0E =a b c d ef g h i jk l m n o.• 3, 3-space, are guaranteed to (because a solution to Ax = b exists whenever b is in a certain3-dimensional subspace of R3; since the only such subspaces are all of R3, solutions mustalways exist)• 2, plane, infinitely many (and thus non-unique)•afk,bgl,din(and others),−2−3100,00001(and others)2•32000+ span−2−3100,00001•32010+ span−2−3100,00001Linear independenceLetA =1 0 2 0 00 1 3 0 00 0 0 1 0B =a b c d ef g h i jk l m n obe matrices such that B row reduces to A. For convenience, set~w =afk~x =bgl~y =chm~z =din.• dependent, b e cause 2 ~w + 3~x −~y = 0• independent, be cause the first, second and fourth columns of A are• independent, be cause the second, third and fourth columns of A are• {~w, ~x, ~z}, {~x, ~y, ~z}, {~w, ~y,
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