Please work out each of the given problems. Credit will be based on the steps that you show towards the final answer. Show your work.
Let
Find
The key to this is to find the coordinates of
(1,0) with respect to the basis
(1,4) and (0,2). This is given by
A
Multiplying, gives the coordinates (1,-2). Thus L(1,0) = L[1(1,4) - 2(0,2)] = L(1,4) - 2L(0,2) = (1,-1,3) + 2(2,1,4) = (5,1,11)
Suppose that you want to
change the graphic file from the one on the left to the one on the right.
What is the matrix needed to adjust the pixels appropriately?
Assume the center of the picture is the origin.
Solution This is a reflection across the y-axis. The corresponding matrix is
- {(
*x*,*y*,*z*) in R^{3}|*x*-*yz*= 0} - {(
*x*,*y*,*z*,*w*) in R^{4}|*x*+ 2*y*=*z*-*w*}
First, A. is not a subspace since (4,2,2) is in the set (4 - (2)(2) = 0), but 2(4,2,2) = (8,4,4) is not in the set (8 - (4)(4) = -8 is not 0). For B. we need to prove the two statements. Note that a vector in this
set is defined by 1. If
and the vector has the correct form 2. and this vector has the correct form. Therefore this is a subspace.
Answer the following true or false and
explain your reasoning. A.
If A and B
are n x n matrices and AB
= 0, then either A
= 0 or B = 0. B.
If A is a matrix with A C. If the columns of an D. If the nullity of an 8x11
matrix A. We have:Nullity + Rank = n 6 + Rank = 11 Rank = 5. E. If the second column of a 4x4
matrix A is not invertible.F. If (1,3,4,1)
Let
A.
Find the rank and the nullity of B.
Find a basis for the Null Space of C.
Find a basis for the Column Space of
v, ..., _{2}v}
be a set of linearly independent vectors and let_{n} v
be a vector in the span of S.
Prove that v
can uniquely be written as a
linear combination of elements of S.
That is that prove that if
_{2}v + ... + a_{2}_{n}v
and _{n }
v
= b_{1}v
+b_{1}_{2}v + ... + b_{2}_{n}v_{n }
then
a
Suppose that
_{2}v + ... + a_{2}_{n}v
and _{n }
v
= b_{1}v
+b_{1}_{2}v + ... + b_{2}_{n}v_{n }
Now subtract the two equations to get: v - v = a Combine like terms to get 0 = (a Now since a And the result follows.
Solution Let x_{1}v + x_{2}w =
bor that
This can be written as the matrix equation
This has a solution since rref(
Prove that if
r
We have
Since
As a consultant for the South Tahoe ski marketing group, your task is to make predictions on the ski pass that locals hold. You found that this year 55% of locals hold a pass for Heavenly, 30% hold a pass for Sierra-at-Tahoe, and 15% hold a pass for Kirkwood. Each year, 20% of the Heavenly pass holders defect to Sierra-at-Tahoe and 5% defect to Kirkwood. 10% of the Sierra-at-Tahoe pass holders defect to Heavenly and 20% defect to Kirkwood. 5% of the Kirkwood pass holders defect to Heavenly and 25% defect to Sierra-at-Tahoe. Assume that every pass holder remains a pass holder the following year and that everyone holds exactly one pass. Determine what the distribution of pass holders will be in five years.
Since we want to find the distribution after five years we take
So after 5 years, 29% will have Heavenly pass, 42% will have a Sierra-at-Tahoe pass, and 29% will have a Kirkwood pass.
Use the definition to find the determinant of the following matrix. Do not use your calculator.
Expand by the second row. Note that since the first and last entry of the second row are both 0, we only need the middle numbers.
Now expand each of these by their first rows to get
= 5[16 + 6] - 3[-24 -12] = 110 + 108 = 218.
(Any constructive remark will be worth full credit.) |