Quiz 1 SolutionRayParkHyuk Park
Math 23a
September 27/2010
Quiz 1 Solution
1. Answer (E)
When the value of x is very close to the value of c, the value of f(x) is less than 0 since f(x) is continuous and f(c) b + 2c = 0
b 1
c...
Hyuk Park
Math 23a
September 27/2010
Quiz 1 Solution
1. Answer (E)
When the value of x is very close to the value of c, the value of f(x) is less than 0 since f(x) is continuous and f(c) < 0. Therefore c is not an upper bound of the set {x l f(x) is greater or equal to 0}.
2. Answer (B)
(Vector u) x (vector v) = matrix a
b
c
Matrix u1 u2 u3 = -u3 i + u3 j + (u1-u2) k = matrix –u3 = matrix a
1 1 0 u3 b
u1-u2 c
dot product of matrix a and matrix 0 = 0 + b + 2c -> b + 2c = 0
b 1
c 2
a = -u3, b = u3, and c = u1 – u2
u3 + 2u1 - 2u2 = 0
when u1 = 2, u2 = 3, and u3 = 2 vector u = 2
3
2+4+(-6) = 0 2
3. Answer (C)
Dot product of Vector a & Vector b = l vector a l l vector b l cos(T)
3+3 = root 10 * root 10 * cos(T)
6 = 10cos(T)
6/10 = cos (T)
3/5 = cos 53 -> 53 degrees
4. Answer
Matrix a b Matrix 1 = 2 (a – b) = 2 Do some algebra = a = 3
c d -1 -2 (c – d) = -2 b = 1
c = 2
Matrix a b Matrix 1 = 4 (a + b) = 4 d = 4
c d 1 6 (c + d) = 6
1/(ad-bc) * matrix d –b -> 1/10 matrix 4 -1 = Matrix S that represents T^-1
-c a -2 3
5. Answer
Rational number = countable
Rational number * Rational number = countable
Pick a point (x,y) from each disc such that x,y are rational number.
The set of these points is a subset of rational number * rational number so the set is countable.
6. Answer
Answer part a:
(Fast ball/Fast ball) = 0.5
(Curve ball/ Fast ball) = 0.5
(Curve ball/ Curve ball) = 0.3
(Fast ball/ Curve ball) = 0.7
Answer: A x A
[Matrix A]
0.5 0.7
0.5 0.3
Answer part b:
[Matrix A] [L] = [Matrix A][L]
0.5 0.7 7/12 7/12
0.5 0.3 5/12 5/12
Answer:
suppose that for any n.
A^n * L = L then A^n+1 * L = L
A^n * A^1 * L = L (A * L) = L
A^n * L = L
7. Answer
Step 1: T is onto and for any vector y form the vector [s] * vector y
T([s] * vector y) = [T] * [s] * vector y = [I] * vector y = vector y
Thus T is onto
Step 2: T is one to one
T(x1) = T(x2)
[s] * T(x1) = [s] * T(x2)
[I] * (x1) = [I] * T(x2)
(x1) = (x2)
Thus T is one to one
Step 3: If s is inverse of T, s 0 T is the identity transformation, so that the matrix for s 0 T is [s][T] = [I]
Thus the matrix s is the inverse of the matrix for T.
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