Exam 2 Review Sheet PDF

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Math 254 Exam 2 Review Sheet Effects of Row Operations on the Determinant: (1) Switching two rows of A changes the sign of det(A) (so we put a negative in front to compensate) (2) Multiplying a row by c causes multiplies the determinant by a factor of c (so we put 1/c in front to compensate) (3) Adding a scalar multiple of one row to another to replace the second row of this operation does NOT change the determinant. We can reduce a matrix to a triangular matrix to easily compute the determinant. Some Properties of the Determinant: (1) det(AB ) = det(A)det(B) *(2) det(cA) = cn det(A) (3) A is nonsingular iff det(A) 6= 0 (4) if A is nonsingular, det(A−1 ) = (5) det(AT ) = det(A).

1 det(A)

Cramer’s Rule: Consider the linear system corresponding to the matrix equation Ax = b. Let Ai be the matrix obtained by replacing the ith column of A by b. Then, if det(A) 6= 0, the system has a unique solution with components given by xi = |A|A|i| . Operations on Vectors in Rn : The standard operations for vector addition and scalar multiplication are performed componentwise. Linear Combinations: Vector w is a linear combination of vectors v1 , v2 , . . . , vk if there exist scalars c1 , c2 , . . . , ck such that w = c1 v1 + c2 v2 + · · · + ck vk . Given the vectors v1 , . . . , vk , w, we can solve for c1 , . . . , ck (if they exist) using a system of equations.

ii *Vector Space: A vector space is a nonempty set V with operations of vector addition and scalar multiplication such that all vectors u, v, w in V and all scalars c, d satisfy 1. u + v is in V ; 2. u + v = v + u; 3. u + (v + w) = (u + v) + w; 4. There is a zero vector 0 in V such that for all u in V , u + 0 = u; 5. For every u in V , there is an additive inverse −u ∈ V such that u + (−u) = 0; 6. cu is in V ; 7. c(u + v) = cu + cv; 8. (c + d)u = cu + du ; 9. c(du) = (cd)u; 10. 1u = u . Subspace: A nonempty subset W of a vector space V is s subspace of V if it is also a vector space. W can be shown to be a subspace if for every u, v ∈ W and scalar c, 1. u + v is in W ; 2. cu is in W . Span: Given set S = {v1 , v2 , . . . , vk }, the span of S, or span(S ), is the set of all linear combinations of the vectors in S. If set S spans vector space V , then for every vector v in V , there exist scalars c1 , c2 , . . . , ck such that c1 v1 + c2 v2 + · · · ck vk = v. Linear Independence: A set S = {v1 , v2 , . . . , vk } in vector space V is linearly independent if the equation c1 v1 + c2 v2 + · · · + ck vk = 0 has only the trivial solution c1 = c2 = · · · = ck = 0. Otherwise, the set is linearly dependent. Equivalently, if dependent, at least one vector can be written as a linear combination of (“depends on”) the others. Determining Span/Independence: To check if a given set of vectors spans a vector space or is independent, we attempt to solve the system resulting from the above definitions (for example, by using an augmented matrix). If the system produces a square coefficient matrix A, the set spans and is linearly independent if and only if det(A) 6= 0. Basis: A subset B of vector space V is a basis for V if it spans V and is linearly independent. If listing the vectors of a set S as column vectors forms a square matrix A, set S forms a basis if det(A) 6= 0.

iii Number of Vectors in a Basis/Dimension: Any basis for V must have the same number of vectors. This number is called the dimension of the vector space. Sets with the Wrong Number of Vectors: Suppose the dimension of vector space V is n. A set S1 with fewer than n vectors cannot span V , while a set S2 with more than n vectors must be linearly dependent. Basis for a Subspace: We can find a basis for a subspace W by separating the parameters to express the set as a linear combination; the resulting constant vectors form a basis for W .

* Indicates formula or properties will be provided if needed. **Note that other topics could appear (for example, proving a different property), but any necessary information would be provided....