Calculus Assignment Cross Product

Question 1
1. False (When we form a matrix from the vectors the determinant is 0, a set of vectors is
linearly independent only if the determinant is non-zero.)
2. True (the magnitude of two added vectors can be found by adding the magnitude of the
individual vectors.)
3. True (the area of the parallelogram is the magnitude of u*v)
4. True (vector space {0}=0-dimension)
5. True (If A · · → A 0 then Null(A) = Null(A’ ). Elementary row operations were originally
designed to preserve solutions of equations, so Ax = 0 if and only if A’x = 0.)
6 False (The rank of a matrix is also equal to the dimension of the column space)
Question 2
1. B (P1 (2,1,-3), P2 (2,6,4), V = P1P2=(2-2, 6-1, 4+3) = (0,5,7))
2. C (v1-v2, v1+v2)(2v2,2v1) = (2v1-2v2, 2v1+2v2)
3. A (kv = k(2, 1, -2) = (2k, k, -2k). magnitude kv = 9k2=122; k=4)
4. C (The rank of an m × n matrix is a nonnegative integer and cannot be greater than
either m or n)
5. A (rank A+ nullity A = n; where n=number of columns. Nullity = (8-3) = 5 )
6. C (two vectors are orthogonal if and only if their dot product is zero. (1,2, 0,−1).(-
2,3,1,4)=0)
Question 3
The cross-product of two vectors is found by;
[ i j k
a b c
d e f ] = (bf-ce)i – (af-cd)j + (ae-bd)k
Substituting;
[ i j k
1 1 1
−2 6 3 ] = (3-6)i – (3+2)j + (6+2)k
= -3i-5j+8k
= 0
Question 4
Span W(a, b ∈ R)
a + b ∈ W
The zero vector is in W because we take k=0

K(a+3b)
K(0)
K(a-3b)
All return zero as the answer since k=0.
The rest axioms holding for R are also true for W. So W is a vector space. Hence W is a
subspace of R3.
Question 5
To find the vector component of u along a;
= [ u .a
|a2
| ].a
¿(1 , 1, 6).(−1 , 0 ,3)
−12+02 +32 (−1, 0, 3)=
= −1 ,0 , 18
10 (-1, 0, 3)
= (1/10, 0, 54/10)
Question 6
w=k1u1+k2u2
(5, 8) = k1(1, 0) + k2(0, 4),
1k1 + 0k2 = 5
0k1 + 4k2 = 8
K1=5, K2=2
Therefore the coordinate vector of w is (5, 2)
Question 7
Transform the matrix into the reduced row echelon form:
[1 −3 1
2 −6 2
3 −9 3 ] = [1 −3 1
0 0 0
3 −9 3 ] = rref A = [1 −3 1
0 0 0
0 0 0 ]
rref A has two pivot columns,
dimW=(# columns of A)−(# pivot columns of rref A)=3−2=1
Question 8

Given any (x, y, z) ∈ R3 we have;
(x, y, z) = x(1, 0, 0) + y(0, 1, 0) + z(0, 0, 1).
So, any x, y, z) ∈ R3 is a linearly combinations of elements in S. So, R3 = Span(S).
Also, S is linearly independent: x(1, 0, 0)+y(0, 1, 0)+z(0, 0, 1) = (0, 0, 0) =⇒ x = y = z = 0.