Matrices and Linear Algebra Vector Spaces QUESTION 1 In order to find the vector
ID: 3112663 • Letter: M
Question
Matrices and Linear Algebra
Vector Spaces
QUESTION 1
In order to find the vector 2(u + 3v), you may either
(1) directly substitute the values given for vectors u and v, add vector u to 3v, then multiply the sum by the scalar 2 OR
(2) use the scalar distributive property and the associative property for scalar multiplication, then substitute the values for the given vectors u and v and add to get the desired vector.
True
False
QUESTION 2
The reasoning provided in the proof below of "If cv = 0, then c = 0 or v = 0," is valid.
Assuming cv = 0, then there are two choices to consider: either c is zero or c is NOT zero. If c is zero then part of the conclusion, namely c = 0, is met. Now, let's suppose that c is NOT zero. Since c is NOT zero, c-1 exists, and we have
c-1(cv) = c-10 a. Left multiplication property of equality
(c-1c)v = c-10 b. Associative property of scalar multiplication for vectors
1v = c-10 c. Inverse property of multiplication for the set of real nos.
v = c-10 d. Identity property of multiplication for the set of real nos.
v = 0 e. A scalar multiplication additive identity property
True
False
QUESTION 3
The additive identity of the vector space M5,1 is the matrix
True
False
QUESTION 4
Any matrix and its additive inverse in M1,4 may be expressed as the following respectively.
True
False
QUESTION 5
The set below with standard operations in R2 is not a vector space, since one may choose the real scalar -7 and (4,-8) from the given set so that -7(4,-8) = (-28,56) which does not belong to the set.
True
False
QUESTION 6
9(0, 7, 12) = (0, 63, 108)
So, clearly (0, 63, 108) is not in W.
True
False
QUESTION 7
The set of all n x n matrices with integer entries is not a subspace of Mn,n with the standard matrix operations. An example that substantiates this is
True
False
QUESTION 8
(u1, u2, u3) = c1(4, 7, 3) + c2(-1, 2, 6) + c3(2, -3, 5)
= (4c1 - c2 + 2c3, 7c1 +2c2 - 3c3, 3c1 +6c2 + 5c3)
produces a system of linear equations with a coefficient matrix that has a nonzero determinant of 228.
True
False
QUESTION 9
The set S = {(-4, -3, 4), (1, -2, 3), (6, 0, 0)} in R3 is linearly independent because the homogeneous linear system of equations produced based on this set does not row reduce to the I3 matrix.
True
False
QUESTION 10
In order to determine whether or not the set S = {x2 - 2x, x3 + 8, x3 - x2, x2 - 4} spans P3 one must try to find the solutions for
(c2 + c3)x3 + (c1 - c3 + c4)x2 - 2c1x + (8c2 - 4c4) = u1 + u2x + u3x2 + u4x3.
When this linear system is converted into an augmented matrix, the determinant of the coefficient matrix is nonzero. Thus, the set S spans P3.
True
False
QUESTION 11
The standard basis for the vector space M4,1 is
True
False
QUESTION 12
Given that S = {(0, 3, -2), (4, 0, 3), (-8, 15, -16)} for R3, since the standard basis for R3 already contains 3 vectors, it is only necessary to show that either S spans R3or S is a linearly independent set of vectors in R3.
True
False
QUESTION 13
The dimension of the vector space M3,2 is 9.
True
False
QUESTION 14
Rewriting the representative vector (2s - t, s, t, s) as shown below indicates that the spanning set for W = {(2s - t, s, t, s): s and t are real numbers} has two vectors in it.
(2s - t, s, t, s) = (2s, s, 0, s) + (-t, 0, t, 0) = s(2, 1, 0, 1) +t(-1, 0, 1, 0).
True
False
QUESTION 15
For the matrix given below, the rank is 3, since there are three rows in the matrix.
True
False
QUESTION 16
B1 = {(1, 2, -2, -4/3), (0, 0, 0, 1)} and B2 = {(1, 2, -2, 0), (0, 0, 0, 1)} are two possible bases for the row space of the given matrix below.
True
False
QUESTION 17
Since the rank of the matrix A below is 2 and the number of columns of the matrix A is 3, the nullity of the matrix A must be 1.
True
False
Explanation / Answer
I am really sorry but with the polices of chegg we can answer only 1 question, the first question
1) True because vectors have following properties=
Commutative (vector) P + Q = Q + P
Associative (vector) (P + Q) + R = P + (Q + R)
Additive identity There is a vector 0 such
that (P + 0) = P = (0 + P)
for all P
Additive inverse For any P there is a vector -P such that P + (-P) = 0
Distributive (vector) r(P + Q) = rP + rQ
Distributive (scalar) (r + s) P = rP + sP
Associative (scalar) r(sP) = (rs)P
Multiplicative identity For the real number 1,
1P = P for each P