Rap:
Amy, Store B and charge it
(Here we go again I have; Store A. Better get it mail order).
Record store 20 different heterosexual couples, 72 mixed ones
I think this one is False---You still don't know all vectors of Rn. This is a single transformation.
I got this one down as True. The null space are those vectors [X] s.t. A[X]=0.
I got this one as True. If (A-I)^2=0 then A has a unity eigenvalue, which means |A|=|I|=1. Square matrices with nonzero determinants are invertable (e.g. A^-1 exists).
couples
interesting problem, the couples out of 9 thing that is
mixing them all together gives 9!/(9-2)!=9*8=72. But this equates to ab as different from ba, and in this case they ain't. So the free for all couples is 72/2=36
# of homosexual couples
Lesbians 5!/(5-2)!/2=5*4/2=10
Girly Boys 4!/(4-2)!/2=4*3/2=6
# of breeders
All-Lesbians-Girly Boys
Breeders 36-10-6=20
There is another way to come up with 20 breeders, but its too easy. :wink:
Mark wrote, "Try wrote: "715 ladybugs; 1430 dragonflies; 2860 moths; 1287 bees"
"What were you smoking in England?"
As a result of this slanderous statement being published on the web, I called an emergency meeting of the central committee to discuss the matter. The committee were unanimous in up holding Mark's comment.
Now, whilst I am not saying, ?'I was wrong' it would appear Mark is right.
Man this is good stuff.
Mark:
AMY
Assuming the cash discounts are applied after (as opposed to with) the original discount here's what Amy would pay as a percentage of the original price:
Store A
credit: 30%, cash: 27%
Store B
credit: 40%, cash: 32%
Store C
cash: 33.3%
The lowest price is at store A when paying with cash.
TAXES
Itemizing gives a $911 larger deduction.
The master speaks.
"Rap is starting to sound like Ahnold."
I thought he sounded like Homer Simpson
Doh!
Mark who may live in another dimension wrote:
"OK, I got out the linear algebra book"
What! Was it in one dimension? Or was it just long, narrow and of uniform breadth?
"(might as well be Greek)."
What! Do you not speak Greek?
The row space and the nullspace are orthogonal. As are the left nullspace and the column space.
(A-I)^2 = A^2 - 2A + I
Since (A-I)^2 = 0
2A - A^2 = I
A(2I-A) = I
Therefore, 2I-A is the inverse of A.
Let A be nxn. Attaching a basis of NS(A) to one of CS(A) gives a basis of R^n.
I'll guess that this is false, but that attaching a basis of NS(A) to one of RS(A) does give a basis of R^n.
True/False:
Let A be nxn. Attaching a basis of NS(A) to one of CS(A)
gives a basis of R^n.
Answer: FALSE.
Even though by the rank-nullity theorem the set consisting of
the v's and w's together has the right number of vectors to be
a basis, the set is usually not a basis, because it is usually a linearly dependent set.
An example is given by the matrix
(0 1
0 0)
for which NS=CS=span{(1 0)}, i.e. v_1=w_1=(1 0).
Clearly in this case the set {v_1 w_1} is not linearly
independent, hence it is not a basis.
True/False.
The row space is perpendicular to the null space.
Answer: TRUE.
The correct solution was: If v is in NS(A), then Av=0. Write A as
A= (r1 r2 r3 ... rn)^T (bad notation, but how else to type it?).
Then 0 = Av = (r1 v, r2 v, ..., rn v).
So the dot product of each row of A with v is 0.
If u is in the row space of A, then u can be written
as a linear combination of the rows of A: u=a1r1+a2r2+...+anrn. So,
u \cdot v= (a1r1+ ... +anrn) \cdot v= a1(r1 \cdot v) + ... +an(rn \cdot v)=a1(0) + ... +an(0)=0. So if u is in RS(A) and v is in CS(A), then u \cdot v =0.
True/False
If A is a square matrix such that (A-I)^2=0, then A is invertible.
Answer: True.
0=(A-I)^2=(A-I)(A-I)=A^2-2A+I, so 2A-A^2=I=A(2I-A)=(2I-A)A,
so 2I-A is the inverse of A.
It is now time to grab the old spice weasel and take it down a notch.
What does this mean
"O_ER_T_O_"
The product of three natural numbers is 24. How many different ways can this be done if the order of the 3 numbers does not matter
A number N divides each of 17 and 30 with the same remainder in each case. What is the largest value N can have
Consecutive numbers are whole numbers that follow in order such as 3, 4, 5. Find the smallest of the five consecutive numbers whose sum is 100.
Consider the counting numbers from I to 1000: 1, 2, 3, 4, ... , 1000. Which one of these numbers multiplied by itself, is closest to 1985
I build up castles.
I tear down mountains.
I make some men blind,
I help others to see.
What am I
Voiceless it cries,
Wingless flutters,
Toothless bites,
Mouthless mutters
I am
Glittering points
That downward thrust,
Sparkling spears
That never rust.
I am
Somewhere I saw the following, for which I now have an answer:
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