Association Mining - is buying Independent?

I have a problem. I can't not solve this exerciese. What is the best way to solve this exerciese? What are the approaches for this kind of exerciese?

The following table summarizes transactions in a supermarket where customers bought tomatoes and/or mozzarella cheese or neither.

Is buying mozzarella independent of buying tomatoes in the data given above? If they are not independent, explain whether they are positively or negatively correlated, i.e. does buying one of them increase or decrease the probability of buying the other?

As you can see I calculated the lift for lift(Moz => Tom) = 1,33 And I calculated lift(Moz => NoTom) = 0,5.

So I think they are not independent, and it is positively correlated

3 Answers

You are correct. They are not independent and they are positively correlated.

Let $$A$$ and $$B$$ (or $$X$$ and $$Y$$) be two events for stating general theorems and $$M$$ and $$T$$ be the events "customer purchases mozzarella" and "customer purchases tomatoes" in this specific example. We will use $$\wedge$$ to mean "and" so that $$M \wedge T$$ is the event "customer purchases mozzarella and tomatoes".

The simplest way to check independence is directly from the definition. $$A$$ and $$B$$ are independent if $$P(A \wedge B) = P(A)P(B)$$. From the data table we have $$P(M \wedge T) = 2000 / 5000 = 0,4$$ $$P(M) = 2500 / 5000 = 0,5$$ $$P(T) = 3000 / 5000 = 0,6$$ $$P(M)P(T) = 0,5 \times 0,6 = 0,3$$

Since 0.4 does not equal 0.3 we conclude that the events are not independent.

I have to say that I've never seen the term "lift" before. But its definition is perfectly reasonable and we could calculate independence in those terms. We can rearrange the definition of independence to give $$P(A \wedge B) = P(A)P(B)$$ $$\frac{P(A \wedge B)}{P(A)P(B)} = 1$$ $$\text{lift} = 1$$ to see that two events are independent if the associated lift is exactly 1. Your notes correctly show that lift can be rewritten as $$\frac{\frac{P(A \wedge B)}{P(A)}}{P(B)}$$

Here we have a lift of $$0,4 / (0,5 \times 0,6) = 1,33...$$ or $$0,8 / 0,6 = 1,33...$$ as you correctly calculated in your notes. Since the lift is not exactly 1 we conclude that the two events are not independent.

I would have approached the question of correlation via conditional probabilities, as follows. Two events, $$A$$ and $$B$$ are positively correlated if $$P(A | B) > P(A)$$. That is to say, two events $$A$$ and $$B$$ are correlated if the probability of $$A$$ given $$B$$ is greater than the unconditional probability of $$A$$. In other words, if we discover that event $$B$$ has occurred then the chances that event $$A$$ has occurred increase.

The conditional probability is defined as $$P(A | B) = \frac{P(A \wedge B)}{P(B)}$$ So in this example $$P(M | T) = P(M \wedge T) / P(T) = 0,4 / 0,6 = 0,66...$$ whereas $$P(M) = 0,5$$, so we conclude that $$M$$ and $$T$$ are positively correlated.

But hang on! Surely correlation is supposed to symmetric. So let's test whether $$P(T | M) > P(M)$$. We have $$P(T | M) = 0,4 / 0,5 = 0,8$$ and $$P(T) = 0,6$$ so, yes, they are correlated.

A lucky escape? We can be more smug than that. Let's rewrite our definition of correlation $$P(M | T) > P(M)$$ $$\frac{P(M | T)}{P(M)} > 1$$ $$\frac{\frac{P(M \wedge T)}{P(T)}}{P(M)} > 1$$ $$\frac{P(M \wedge T)}{P(T)P(M)} > 1$$

If we approach correlation the other way we get the same result

$$P(T | M) > P(T)$$ $$\frac{P(T | M)}{P(T)} > 1$$ $$\frac{\frac{P(T \wedge M)}{P(M)}}{P(T)} > 1$$ $$\frac{P(T \wedge M)}{P(M)P(T)} > 1$$ which is the same as above because both the and operator and multiplication are commutative.

As I'm sure you've noticed, $$\frac{P(M \wedge T)}{P(T)P(M)} > 1$$ is simply asking whether lift is above 1. That's probably the approach that your instructor was expecting you to take.

A couple of additional points to fully close the loop with your notes. You correctly calculated lift(Moz => NoTom) = 0,5 but that information was not required to answer the question because you had already calculated lift(Moz => Tom). Support({X} -> {Y}) looks very much like a definition of $$P(X \wedge Y)$$ and Confidence({X} -> {Y}) looks like $$P(Y | X)$$.

I have two final notes to place this all in a broader context.

First, my definition of correlation fits well to your definition of correlation as "does buying one of them increase or decrease the probability of buying the other". But our definitions are not completely standard and the word "correlation" is usually associated with some specific correlation measure -- most often Pearson's correlation coefficient but also things like Kendall's rank correlation coefficient. Having said that, our definition is totally reasonable and I'm fairly confident that we could prove that, for example, Kendall's rank correlation coefficient is positive if and only if our definition of positive correlation above is positive.

Second, inspired by the definition of lift in probabilistic terms in your notes, I've conflated historical frequencies given in the table with probabilities. This is standard practice in much of data science and certainly in exercises but it is not completely uncontroversial. Probabilities are about the future and you have data about the past. The extent to which you can infer future probabilities from past data is philosophically unresolved. But that is not something to worry about in this situation.

Market Basket Analysis is the way to go about it.

Market Basket Analysis is a technique which identifies the strength of association between pairs of products purchased together and identify patterns of co-occurrence. A co-occurrence is when two or more things take place together.

Market Basket Analysis creates If-Then scenario rules, for example, if item A is purchased then item B is likely to be purchased. The rules are probabilistic in nature or, in other words, they are derived from the frequencies of co-occurrence in the observations. Frequency is the proportion of baskets that contain the items of interest. The rules can be used in pricing strategies, product placement, and various types of cross-selling strategies.

Here is the calculation :

Support({Moz}--{Tom} = Transaction containing both moz and tom / total number of transaction

= 2000/ 5000  = 2/5 = 0.4

Confidence({Moz}--{Tom}) = Transaction containing both moz and tom / total number of transaction containing Moz
= 2000 / 2500 = 2/2.5 = 0.8

Lift({Moz}--{Tom}) = (Transaction containing both moz and tom / total number of transaction containing Moz ) / Fraction of transaction containing Y

= 0.8 / (3000/5000)

= 0.8/0.6 = 1.33

For your table this association mining rule is strong. But as the relation is not commutative please do same calculation from P(tom--> Moz) and if values of support, confidence and lift is smaller that does not indicate strong relationship

Given you have two categorical variables and the associated contingency table, one option is to calculate the joint and marginal probabilities:

$$P = \frac{count}{total}$$

Probabilities Tomatoes No Tomatoes Row
Mozzarella 0.4 0.1 0.5
No Mozzarella 0.2 0.3 0.5
Column 0.6 0.4 0.1

The probabilities then can be used to answer questions about the data - If a person has bought tomatoes, is the person more likely to buy Mozzarella?

A person who has bought tomatoes is four times as likely to buy mozzarella compared to a person who has not bought tomatoes (i.e., 0.4 vs 0.1). There is a strong positive relationship between the two purchases.

There are many other possible statistical analyses that can be conducted. Examples include odds ratio, phi coefficient, tetrachoric correlation coefficient, and hypothesis testing with Chi-squared.


# Is the inverse of a symmetric matrix also symmetric?

Let $A$ be a symmetric invertible matrix, $A^T=A$, $A^{-1}A = A A^{-1} = I$ Can it be shown that $A^{-1}$ is also symmetric?

I seem to remember a proof similar to this from my linear algebra class, but it has been a long time, and I can't find it in my text book.

You can't use the thing you want to prove in the proof itself, so the above answers are missing some steps. Here is a more detailed and complete proof.

Given A is nonsingular and symmetric, show that $$A^{-1} = (A^{-1})^T$$.

Since $$A$$ is nonsingular, $$A^{-1}$$ exists. Since $$I = I^T$$ and $$AA^{-1} = I$$,

$$AA^{-1} = (AA^{-1})^T.$$

Since $$(AB)^T = B^TA^T$$,

$$AA^{-1} = (A^{-1})^TA^T.$$

Since $$AA^{-1} = A^{-1}A = I$$, we rearrange the left side to obtain

$$A^{-1}A = (A^{-1})^TA^T.$$

Since $$A$$ is symmetric, $$A = A^T$$, and we can substitute this into the right side to obtain

$$A^{-1}A = (A^{-1})^TA.$$

From here, we see that

$$A^{-1}A(A^{-1}) = (A^{-1})^TA(A^{-1})$$ $$A^{-1}I = (A^{-1})^TI$$ $$A^{-1} = (A^{-1})^T,$$

thus proving the claim.

In fact, $(A^T)^{-1}=(A^{-1})^T$. Indeed, $A^T(A^{-1})^T=(A^{-1}A)^T=I$.

Yes.

$$AB=BA=I\quad\Rightarrow\quad B^TA^T=A^TB^T=I\quad\Rightarrow\quad B^TA=AB^T=I$$

Another way to see that is to recall the formula $$A^{-1} = \frac{1}{\det(A)} \mathrm{Adj}(A)^T$$ and to note that the adjoint matrix of a symmetric matrix is by construction symmetric.

Yes. The inverse $$A^{-1}$$ of invertible symmetric matrix is also symmetric:

\begin{align} A & = A^T &&\text{(Assumption: A is symmetric)}\\ \\ A^{-1} & = (A^T)^{-1} &&\text{(A invertible \implies A^T = A invertible)}\\ \\ A^{-1} & = (A^{-1})^T &&\text{(Identity: (A^T)^{-1} = (A^{-1})^T)} \\ \\ {\large \therefore}\quad \rlap{\text{If A is symmetric and invertible, then A^{-1} is symmetric.}} \end{align}

All the proofs here use algebraic manipulations. But I think it may be more illuminating to think of a symmetric matrix as representing an operator consisting of a rotation, an anisotropic scaling and a rotation back. This is provided by the Spectral theorem, which says that any symmetric matrix is diagonalizable by an orthogonal matrix. With this insight, it is easy to see that the inverse of the operator is a similar three-step sequence. Finally, we need to establish that any such three-step sequence produces a symmetric matrix: given any orthogonal matrix $$V$$ and diagonal matrix $$D$$, $$(V D V^T)^T = V D V^T$$. Hence the inverse of a symmetric matrix is also symmetric.

• Very nice intuitive answer. The reason this is getting fewer upvotes is that it’s posted much later than the others. That doesn’t stop this from being a good answer though. Oct 18, 2020 at 20:22

We have two properties to use: $A^{T}=A$ and $A^{-1}$exist, here we go!

$$A^{-1}A=I$$since $A^{-1}$exist $$(A^{-1}A)^{T}=A^{T}(A^{-1})^{T}=A(A^{-1})^{T}=I^{T}=I$$since $A^{T}=A$ $$A^{-1}A(A^{-1})^{T}=A^{-1}I$$left multiple by $A^{-1}$

Thus $$I(A^{-1})^{T}=(A^{-1})^{T}=A^{-1}$$

Inverse of $$A$$ can be expressed as a polynomial $$p(A)$$ of $$A$$ (from Cayley-Hamilton theorem).

So it is sufficient to prove that if $$A$$ is symmetric then power $$A^k$$ is symmetric, sum of symmetric matrices is symmetric and multiply by scalar is symmetric.

• I.e. for symmetric $A,B$ $(A^k)^T=A^k,(tA)^T=tA, (A+B)^T=A+B,$ Jun 15, 2019 at 10:56

$$(AA^{-1})^T=I^T=I=(A^{-1})^TA^T=(A^{-1})^TA=I\quad\Rightarrow\quad(A^{-1})^T=IA^{-1}\quad\Rightarrow\quad(A^{-1})^T=A^{-1}$$ By definition $$A=A^T$$ Used linear algebra equations $$(AB)^T=B^TA^T;AA^{-1}=I;I^T=I$$


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# A manufacturer makes and sells 2 products, P and Q. The revenue from t

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A manufacturer makes and sells 2 products, P and Q. The revenue from t  [#permalink]

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A manufacturer makes and sells 2 products, P and Q. The revenue from the sale of each unit of P is $20.00 and the revenue from the sale of each unit of Q is$17.00. Last year the manufacturer sold twice as many units of Q as P. What was the manufacturer’s average (arithmetic mean) revenue per unit sold of these 2 products last year?

A. $28.50 B.$27.00
C. $19.00 D.$18.50
E. $18.00 _________________ Md. Abdur Rakib Please Press +1 Kudos,If it helps Sentence Correction-Collection of Ron Purewal's "elliptical construction/analogies" for SC Challenges ##### Most Helpful Expert Reply CEO Joined: 11 Sep 2015 Posts: 3341 Location: Canada A manufacturer makes and sells 2 products, P and Q. The revenue from t [#permalink] ### Show Tags Updated on: 05 Dec 2018, 16:06 1 Top Contributor 5 AbdurRakib wrote: A manufacturer makes and sells 2 products, P and Q. The revenue from the sale of each unit of P is$20.00 and the revenue from the sale of each unit of Q is $17.00. Last year the manufacturer sold twice as many units of Q as P. What was the manufacturer’s average (arithmetic mean) revenue per unit sold of these 2 products last year? A.$28.50
B. $27.00 C.$19.00
D. $18.50 E.$18.00

Let's plug in some NICE VALUES that satisfy the given information...

Last year the manufacturer sold twice as many units of Q as P.
So, let's say the manufacturer sold 1 P and 2 Q's

What was the manufacturer’s average (arithmetic mean) revenue per unit sold of these 2 products last year?
TOTAL revenue = 1($20.00) + 2($17.00)
= $20.00 +$34.00
= $54 There were 3 units sold. So, average price =$54/3 = $18 Answer: E RELATED VIDEO _________________ Test confidently with gmatprepnow.com Originally posted by GMATPrepNow on 17 Jun 2017, 07:31. Last edited by GMATPrepNow on 05 Dec 2018, 16:06, edited 1 time in total. ##### General Discussion Board of Directors Status: QA & VA Forum Moderator Joined: 11 Jun 2011 Posts: 4328 Location: India GPA: 3.5 WE: Business Development (Commercial Banking) Re: A manufacturer makes and sells 2 products, P and Q. The revenue from t [#permalink] ### Show Tags 17 Jun 2017, 22:48 1 1 AbdurRakib wrote: A manufacturer makes and sells 2 products, P and Q. The revenue from the sale of each unit of P is$20.00 and the revenue from the sale of each unit of Q is $17.00. Last year the manufacturer sold twice as many units of Q as P. What was the manufacturer’s average (arithmetic mean) revenue per unit sold of these 2 products last year? A.$28.50
B. $27.00 C.$19.00
D. $18.50 E.$18.00

Revenue last year from the sale of P is 20*1 = 20
Revenue last year from the sale of Q is 17*2 = 34

Total Revenue from the sale of P and Q last year is 54

Average Revenue from the sale of P and Q last year is $$\frac{54}{2+1} = 18$$

Hence, answer will be (E) 18
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Re: A manufacturer makes and sells 2 products, P and Q. The revenue from t  [#permalink]

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02 Jul 2017, 20:49
Ratio of quantity of P and Q sold, P:Q = 1:2.

Thus, average revenue per unit = (20*P + 17*Q)/(P+Q) = (20*1 + 17*2)/(2 + 1) = 54/3 = $18 Hence E answer Target Test Prep Representative Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 4551 Location: United States (CA) Re: A manufacturer makes and sells 2 products, P and Q. The revenue from t [#permalink] ### Show Tags 03 Aug 2018, 14:07 1 AbdurRakib wrote: A manufacturer makes and sells 2 products, P and Q. The revenue from the sale of each unit of P is$20.00 and the revenue from the sale of each unit of Q is $17.00. Last year the manufacturer sold twice as many units of Q as P. What was the manufacturer’s average (arithmetic mean) revenue per unit sold of these 2 products last year? A.$28.50
B. $27.00 C.$19.00
D. $18.50 E.$18.00

We can compute a weighted average to solve. Let’s assume that 2 units of Q and 1 unit of P were produced last year. So the total revenue is 2 x 17 + 1 x 20 = $54, and thus the average revenue per unit is thus 54/3 =$18.

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Re: A manufacturer makes and sells 2 products, P and Q. The revenue from t &nbs [#permalink] 03 Aug 2018, 14:07
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Which is the greatest possible natural number that divides $(p+3)(p-7)$, where $p$ is a prime number greater than $3$?

Which is the greatest possible natural number that definitely divides $(p+3)(p-7)$, where $p$ is a prime number greater than $3$?

This one is from my module, comes as a fill in the blanks with no answer. I have a feeling that there is something wrong with this question, since for $p=5$, one has $(p+3)(p-7)=-16$.

As mentioned, in one answer I tried substituting $p=5,7,9,11,13,17,19,...$ and spotted that the greatest number is $8$. I am just wondering .. is there any other way (probably using modulus) to directly find this number without actually substituting ?

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Why does this have a vote to close as "not constructive"? –  Zev Chonoles Dec 3 '11 at 13:31
Playing around (substituting) is always a good idea. –  André Nicolas Dec 3 '11 at 16:22
I agree with Zev Chonoles. It looks quite constructive to me and shows some thought about the problem. +1 from me. –  Ross Millikan Dec 3 '11 at 16:24

I have a feeling that there is something wrong with this question

The "greater than $3$" condition is a bit strange, because the answer is unchanged if $p=3$ is allowed. And if they wanted to exclude negative values of $(p+3)(p-7)$, they should say "greater than 7". But the question as it stands is technically correct. It would perhaps be better like this:

Which is the greatest natural number that divides $(p+3)(p−7)$ for every odd prime number $p$?

If $p$ is odd, then $p = 2k+1$ for some $k$. So $(p+3)(p-7) = (2k+4)(2k-6) = 4(k+2)(k-3)$, which is divisible by $8$ since one of $k+2, k-3$ must be even. So $8$ divides $(p+3)(p−7)$ for every odd prime number $p$.

That's half the question. The other half is to show that no natural number $n > 8$ divides $(p+3)(p−7)$ for every odd prime number $p$. Note that if $n$ has this property, then so does every factor of $n$. So it is enough to show that (i) $16$ doesn't have this property; and (ii) no odd prime $q$ has this property.

To show that $16$ doesn't have this property, just put $p=11$. So suppose now that $q$ is an odd prime. We need to produce another odd prime $p$ that is not equal to $q-3 \pmod q$ and not equal to $7 \pmod q$. Then we can be sure that $q$ doesn't divide $(p+3)(p-7)$. It seems a bit like overkill to me, but we can use Dirichlet's theorem on arithmetic progressions for this: for coprime $a$ and $q$, the arithmetic progression $a + rq$ $(r = 0, 1, 2,...)$ contains infinitely many primes. So we just have to choose $a$ with $1 \le a \le q-1$ that is not equal to $q-3 \pmod q$ and not equal to $7 \pmod q$. If $q=3$, we can choose $a=2$; and if $q \ne 3$, we can choose $a=3$.

And we're done.

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Put $p=5$. Anything $m$ that divides all the $(p+3)(p-7)$ must divide $16$. Put $p=11$ as you did. Thus $m$ must divide $56$. It follows that $m$ must divide $\gcd(16,56)$. We would need fancier machinery only if we wanted to show that if $\gcd$ is restricted to primes $>M$, it is still $8$. –  André Nicolas Dec 3 '11 at 16:18
@AndréNicolas: Doesn't the paragraph starting "If $p$ is odd" prove that $8$ divides $(p+3)(p-7)$ for all odd $p$, prime or not? –  Ross Millikan Dec 3 '11 at 16:22
@Ross Millikan: The answer above used Dirichlet's Theorem to show that no odd prime $q$ can divide all the $(p+3)(p-7)$. –  André Nicolas Dec 3 '11 at 16:27
Well, at least I can say I was right about the overkill :-) André's comment $-$ and Bill Dubuque's answer $-$ provide a much simpler solution. –  TonyK Dec 3 '11 at 17:27

There's nothing wrong with the question.

You're looking for a natural number. You know it divides $(p+3)(p-7)$. And you can put in some primes into that equation, as you have done already - you've got $-16$. What happens if you stick in $p=11$? Or $p=13$?

OK, now you've identified the natural number. Can you see why the product is always divisible by that number?

clue word: modulo

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HINT $\ \rm\:d\ |\ f_p = (p+3)\:(p-7)\ \Rightarrow\ d\ |\ f_5,\:f_7,f_{11}\: =\:\: -16,\:0,\:56\ \Rightarrow\ d\ |\ gcd(16,56) = 8\:.\:$ Conversely

$\$ mod $8$, since odd^2 $\equiv 1\:,\:$ we have $\rm\ f_p\equiv (p+3)(p+1)\: \equiv\: p^2+4\:p+3 \:\equiv\: 4\:(p+1) \:\equiv\: 0$

Generally for a sequence satisfying a monic $\rm\:n\:$th order linear recurrence with integer coefficients one easily proves by induction that the gcd of all terms is the gcd of the first $\rm\:n\:$ terms, since successive terms are integer linear combinations of the first $\rm\:n\:$ terms so they don't change the gcd.

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If $p$ is prime it can be written $p=4a+1$ or $p=4a+3.$

Substituting the first of these gives
$$(p+3)(p-7)=p^2-4p-21 = 16a^2+8a+1-16a-4-21 = 16a^2-8a-24 = 8(2a^2-a-3)$$
and the other gives $$(p+3)(p-7)=p^2-4p-21 = 16a^2+24a+9-16a-12-21 = 16a^2+8a-24 = 8(2a^2+a-3)$$
and so $(p+3)(p-7)$ is divisible by $8$ either way.

Edit
To show that 8 is the largest such divisor, we must show that $2a^2-a-3$ is not divisible by the same prime for all values of $a$. Setting $a=4\text{ and }5$ produces 25 and 42 respectively which have no prime factors in common.

And we must also show that $2a^2+a-3$ is not divisible by the same prime for all values of $a$. Setting $a=4\text{ and }5$ again produces 33 and 52 respectively which also have no prime factors in common.

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Strictly spoken, you have not proven yet that it 8 is the greatest number for which it holds. However, as MaX noted, trying the first few numbers shows that it is at most 8. –  Sjoerd Dec 3 '11 at 15:44

Observe that $(5+3)(5-7) = -16$ and $(11+3)(11-7) = 56$; these have greatest common multiple 8. So the answer must be some factor of 8. If you keep trying more numbers you'll see it looks to be 8.

How to prove it? $p$ is odd. So $p+3$ and $p-7$ are even numbers which differ by 10. Let $p = 2n+7$; then we want to show that for any integer $n$, $2n(2n+10)$ is divisible by 8. But we can rewrite this as $4n(n+5)$, so we just need to show that $n(n+5)$ is always even; either $n$ is even or $n+5$ is.

A challenge for you, to help you understand what's going on here: prove in a similar way that for every integer $n$, $n^3-n$ is a multiple of 6 and $n^5-n$ is a multiple of 30.

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If p is odd number, p-7 is even =2k(say). Then (p+3)(p-7)=(2k+10)2k=8k(k+1)/2+16k, is clearly divisible by 8 for all integral value of k.

Now if f(k) = (2k+10)2k=4k(k+5), f(k+1)= 4(k+1)(k+6) (f(k),f(k+1))=(4k(k+5), 4(k+1)(k+6))= 4(k(k+5), (k+1)(k+6)) =4(k(k+5), (k+1)(k+6) - k(k+5)) as (a,b)=(a, b-a) =4(k(k+5), 2k+6) =8(k(k+5)/2, k+3) as k(k+5)/2 is integer for all integral value of k.

Now, 2 k(k+5)/2 - k(k+3)= 2k. We know ax+by=c is solvable iff (a,b)|c where a,b,c,x,y are integers. Here, x=2, a=k(k+5)/2, y=k, b=k(k+3), c=2k So, (k(k+5)/2, (k+3)) | 2k, =>2|(k+3) =>k is odd. For even k, 2∤(k(k+5)/2, (k+3)). In that case, no natural number>1, can divide f(k) for all integral value of k. So, 8 will be the greatest natural number that divides (p+3)(p-7) for all odd integral value of p.

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I would interpret the question as, "What is the greatest natural number that divides $(p+3)(p-7)$ for all primes $p>3$?"

We can write $(p+3)(p-7)=4\cdot\frac{p+3}2 \cdot\frac{p-7}2$ but one of the latter two factors is even, since they differ by five, an odd number, so we can conclude that $(p+3)(p-7)$ is always divisible by eight, but $(5+3)(5-7)=-16$, $(9+3)(9-7)=24$, and $\gcd(-16,24)=8$. Therefore no natural number larger than eight divides $(p+3)(p-7)$ for all primes $p>3$, and eight is in fact the greatest natural number that divides $(p+3)(p-7)$ for all primes $p>3$.

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since p is odd 2| (p+3)(p-7). so the answer should be \frac{1}{2} (p+3)(p-7)

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You have not read the question carefully. We want the highest number that divides $(p+3)(p-7)$ for any odd $p$, not for a given odd $p$. –  Ross Millikan Nov 6 '13 at 19:28

The question is sloppily phrased, because to me it reads more like "Which is the greatest possible natural number that definitely divides $(p+3)(p−7)$, where $p$ is some given prime greater than 3?", in which case the answer is trivially $(p+3)(p-7)$ !

But I imagine (as does everyone else by the look of it) the OP meant "Which is the greatest possible natural number that definitely divides $(p+3)(p−7)$ for every prime $p$ greater than 3?"

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# Concept check: Standard deviation

## Introduction

Unlike most questions on Khan Academy, some of these questions aren't graded by a computer. You'll learn the most if you try answering each question yourself before clicking "explain".

## The formula (for reference)

The formula for standard deviation (SD) is
$\Large\text{SD} = \sqrt{\dfrac{\sum\limits_{}^{}{{\lvert x-\bar{x}\rvert^2}}}{n}}$
where sum means "sum of", x is a value in the data set, $\bar{x}$ is the mean of the data set, and n is the number of values in the data set.

## Part 1

Consider the simple data set left brace, 1, comma, 4, comma, start color redD, 7, end color redD, comma, 2, comma, 6, right brace.
How does the standard deviation change when start color redD, 7, end color redD is replaced with start color greenD, 12, end color greenD?
Please choose from one of the following options.

How can we see this in the formula for standard deviation?

### How does the standard deviation change?

The standard deviation increases because the data becomes more spread out:

### How can we see this in the formula?

We can see this in the formula
$\text{SD} = \sqrt{\dfrac{\goldD{\sum\limits_{}^{}{{\lvert x-\bar{x}\rvert^2}}}}{n}}$
because ${{\goldD{\sum\limits_{}^{}{{\lvert x-\bar{x}\rvert^2}}}}{}}$ is the sum of the squares of the distances from each data point to the mean. As the data, gets more spread out, the value of ${{\goldD{\sum\limits_{}^{}{{\lvert x-\bar{x}\rvert^2}}}}{}}$ increases.

### I'm curious, what are the actual standard deviations of the data sets?

The standard deviation of left brace, 1, comma, 2, comma, 4, comma, 6, comma, start color redD, 7, end color redD, right brace is approximately 2, point, 28.
The standard deviation of left brace, 1, comma, 2, comma, 4, comma, 6, comma, start color greenD, 12, end color greenD, right brace is approximately 3, point, 90.

## Part 2

Is it possible to create a data set with 4 data points that has a standard deviation of 0?
Please choose from one of the following options.

If it is possible, do it! Can you create two different data sets? How about three?

### Yes, it's possible!

In fact, there are an infinite number of possible data sets.
Here's one:
5, comma, 5, comma, 5, comma, 5
Here's another:
8, comma, 8, comma, 8, comma, 8
Any data set where all of the data points are the same has a standard deviation of 0 because the distance from each data point to the mean is 0.

### Show me the calculation for $\{ 5,5,5,5 \}$left brace, 5, comma, 5, comma, 5, comma, 5, right brace.

#### Step 1: Find the mean

$\bar{x} = \dfrac{5 + 5 + 5 + 5}{4} = \dfrac{20}{4} = \blueD5$

#### Step 2: Find the square of the distances from each of the data points to the mean

x$\lvert x - \bar{x} \rvert^2$
5open vertical bar, 5, minus, start color blueD, 5, end color blueD, close vertical bar, start superscript, 2, end superscript, equals, 0, start superscript, 2, end superscript, equals, 0
5open vertical bar, 5, minus, start color blueD, 5, end color blueD, close vertical bar, start superscript, 2, end superscript, equals, 0, start superscript, 2, end superscript, equals, 0
5open vertical bar, 5, minus, start color blueD, 5, end color blueD, close vertical bar, start superscript, 2, end superscript, equals, 0, start superscript, 2, end superscript, equals, 0
5open vertical bar, 5, minus, start color blueD, 5, end color blueD, close vertical bar, start superscript, 2, end superscript, equals, 0, start superscript, 2, end superscript, equals, 0

#### Step 3: Apply the formula

\begin{aligned} \text{SD} &= \sqrt{\dfrac{\sum\limits_{}^{}{{\lvert x-\bar{x}\rvert^2}}}{n}} \\\\\\\\ &= \sqrt{\dfrac{0 + 0 + 0 + 0}{4}} \\\\\\\\ &= \sqrt{\dfrac{{0}}{4}}\\\\\\\\ &= \sqrt{{0}}\\\\\\\\ &= 0\end{aligned}

## Part 3

Can standard deviation be negative?
Please choose from one of the following options.

Why or why not?

### No, standard deviation cannot be negative!

To see why, think about the numerator and denominator inside the radical:
$\Large\text{SD} = \sqrt{\dfrac{\blueD{\sum\limits_{}^{}{{\lvert x-\bar{x}\rvert^2}}}}{\maroonD{n}}}$
Notice how start color maroonD, n, end color maroonD is always positive. It's the number of data points, and we can't have a negative number of data points.
Also notice that $\blueD{\sum\lvert x - \bar{x} \rvert^2}$ involves a quantity getting squared. Whenever we square something, we get a non-negative number.
Since both the denominator and numerator are positive, the entire expression must be positive too.

## Part 4

Standard deviation is a measure of spread of a data distribution.
What do you think deviation means?
In everyday language, deviation is how different something is from what might be considered normal.
In statistics, when discussing measures of spread, deviation is the amount by which a single measurement differs from the mean.

### Part 5

Here are the formulas for standard deviation (SD) and the formula for mean absolute deviation (MAD), both of which are measures of spread:
$\text{SD} = \sqrt{\dfrac{\sum\limits_{}^{}{{\lvert x-\bar{x}\rvert^2}}}{n}}$
$\text{MAD} = {\dfrac{\sum\limits_{}^{}{{\lvert x-\bar{x}\rvert}}}{n}}$
What are the similarities between the formulas? What are the differences?

### What are the similarities?

The formulas are very similar! They are both based on the distance from each data point to the mean $\lvert x - \bar{x} \rvert$, and they both include dividing by the number of data points n.

### What are the differences?

The difference between the two formulas is that when calculating standard deviation, we square the distance from each data point to the mean, and we take the square root as the last step of the formula.

### Which one is better?

Standard deviation is more complicated, but it has some nice properties that make it statisticians' preferred measure of spread.

### Part 6

Here's the formula that we've been using to calculate standard deviation:
$\sqrt{\dfrac{\sum\limits_{}^{}{{\lvert x-\bar{x}\rvert^2}}}{n}}$
Here's the formula that statisticians actually use:
$\sqrt{\dfrac{\sum\limits_{}^{}{{( x-\bar{x})^2}}}{n}}$
Are the two formulas equivalent?
Please choose from one of the following options.

### What's the difference between the formulas?

In the formula that we've been using, we take the absolute value of $x - \bar{x}$:
$\sqrt{\dfrac{\sum\limits_{}^{}{{\tealD{\lvert x-\bar{x}\rvert}^2}}}{n}}$
In the formula that statisticians use, they put parentheses around $x - \bar{x}$:
$\sqrt{\dfrac{\sum\limits_{}^{}{{\purpleC{( x-\bar{x})^2}}}}{n}}$

### Are the formulas equivalent?

Yes, both formulas are equivalent!
Statisticians realize that squaring will make the distance positive, so they don't bother using absolute value signs and just use parentheses instead.
For example, let's evaluate $\tealD{\lvert x - \bar{x} \rvert^2}$ and $\purpleC{(x - \bar{x})^2}$ for x, equals, 2 and $\bar{x} = 5$:
$\tealD{\lvert x - \bar{x} \rvert^2} = \lvert 2 - 5 \rvert^2 = \lvert-3\rvert^2 = 3^2 = \greenD9$
$\purpleC{(x - \bar{x})^2} = (2 - 5)^2= (-3)^2 = \greenD9$
They're both positive! They're both start color greenD, 9, end color greenD! They're equivalent!


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# Cos Sin 1 X

Cos Sin 1 X. Cos2(x)+ 2cos(x)sin(x)+sin2(x) = 1 note an important identity: = 1 sin ( x) − cos ( x) sin ( x) = csc ( x) − cot ( x) share answered jan 31, 2017 at 15:44 simply beautiful art 72.3k 11 114 255 add a comment 1 hint the appearance of 1 + cos. Expand using the foil method. The word cosinus derived from edmund gunter.

= 1 sin ( x) − cos ( x) sin ( x) = csc ( x) − cot ( x) share answered jan 31, 2017 at 15:44 simply beautiful art 72.3k 11 114 255 add a comment 1 hint the appearance of 1 + cos. The linear combination, or harmonic addition, of sine and cosine waves is equivalent to a single sine wave with a phase shift and scaled amplitude, a cos  x + b sin  x = c cos  ( x + φ ). Yes your guess from the table is correct, indeed since ∀ θ ∈ r − 1 ≤ cos θ ≤ 1, for x > 0 we have that.

## Yes your guess from the table is correct, indeed since ∀ θ ∈ r − 1 ≤ cos θ ≤ 1, for x > 0 we have that.

Solve for x cos(x)+1=sin(x) subtract from both sides of the equation. Cos (x)+sin (x)=1 cos (x) + sin(x) = 1 cos ( x) + sin ( x) = 1 square both sides of the equation. The derivative of sin x is cos x, the derivative of cos x is −sin x (note the negative sign!) and.

## Misc 13 Solve 2 Tan1 (Cos X) = Tan1 (Ii Cosec X Misc 13.

How to prove that 1+sin(x)cos(x) = cos(x)1−sin(x) [duplicate] multiplying the lhs by the conjugate of.

### Extended Keyboard Examples Upload Random.

The derivative of sin x is cos x, the derivative of cos x is −sin x (note the negative sign!) and.

### Kesimpulan dari Cos Sin 1 X.

The word cosinus derived from edmund gunter.


# How to know when you can focus only on a specific part of an expression with an even power?

I stumbled upon this problem as I was going through the Algebra 2 course of https://brilliant.org. It goes as follows:

Starting from

$$x = \sqrt{\sqrt{3 \sqrt{\sqrt{3 \sqrt{\sqrt{3 \dots}}}}}}$$,

I manage on my own to reach the following equation:

$$x^4 = 3x$$.

Personally I took it from there as follows:

$$\frac{x^4}{x} = 3 \leftrightarrow x^3 = 3 \leftrightarrow x=\sqrt[3]{3}$$.

The course accepts my final answer of $$x = \sqrt[3]{3}$$ as correct, however they reason as follows:

$$x^4 = 3x \leftrightarrow x^4 - 3x = 0 \leftrightarrow x(x^3 - 3) = 0$$, since in the original equation $$x > 0$$, we only worry about the root $$x^3 - 3$$:

$$x^3 - 3 = 0 \leftrightarrow x^3 = 3 \leftrightarrow x = \sqrt[3]{3}$$.

They do reach to the same answer but that is not important when learning. What bothers me is that I do not seem to understand on why the $$x > 0$$ part is important here, and how they use that to form their conclusion. It also makes me wonder about how it impacts my approach to the solution. And in general it makes me wonder if there are guidelines that can help me decide on what properties of an original equation I need to take into account when transforming equations while solving a problem.

I suppose it is similar as how when $$x$$ is in the denominator of the original equation than $$x \neq 0$$, no matter how you can transform the equation (e.g. cancel out the denominator). I do however not see at the moment why it would apply here.

• You just divided both sides of your equation by $x$ and so you got the right answer but you need to justify that you can do this and the justification is that clearly $x$ is non zero. – Anon May 3 '19 at 13:28
• Oh, right. Now I get it, so they also just divide by $x$, difference being that they also explain why it is possible. Thank you, now I feel a bit silly to ask this as a question, not sure it will contribute much to this fascinating Q&A website. Feel free to put it as an answer @Anon, such that I can accept it as an answer. – glendc May 3 '19 at 13:29

In your solution, when you perform the step $$x^4 = 3x \iff \frac{x^4}{x} = x^3 = 3,$$ you are using the law of multiplicative cancelation, which states

Theorem: If $$a$$, $$b$$, and $$c$$ are any real numbers with $$c \ne 0$$, then $$a = b \iff ac = bc.$$

In other words, we can cancel common factors from both sides of an equation (that is, we can "divide" both sides of an equation by something), assuming that this factor is nonzero. In your computation, you are applying this theorem. In order for this to be a valid step, the hypotheses of the theorem much hold, hence you need it to be true that $$x \ne 0$$. Therefore you are implicitly assuming that $$x \ne 0$$.

Since you ask about "guidelines" for working problems like this, it might be worthwhile to get extremely pedantic, and very carefully reason about every step. For example, your solution could be expanded to something akin to the following:

Let $$x = \sqrt{\sqrt{3\sqrt{\sqrt{3\sqrt{\sqrt{3\dotsb}}}}}},$$ assuming that $$x$$ exists.[1] Then $$x^4 = \left( \sqrt{\sqrt{3\sqrt{\sqrt{3\sqrt{\sqrt{3\dotsb}}}}}} \right)^4 = 3 \sqrt{\sqrt{3\sqrt{\sqrt{3\sqrt{\sqrt{3\dotsb}}}}}} = 3x.$$ Since $$x > 0$$,[2] the law of multiplicative cancelation gives $$x^4 = 3x \iff x^3 = 3.$$ The function $$f(x) = x^3$$ is invertible on $$\mathbb{R}$$, and has inverse $$f^{-1}(x) = \sqrt[3]{x}$$. Thus $$x^3 = 3 \iff x = \sqrt[3]{3}.$$ Therefore $$x = \sqrt[3]{3}$$.

This kind of solution is overly pedantic, but each and every step of the computation is justified by appealing to some mathematical principle. Often we skip writing these justifications out explicitly, as we can eventually start assuming that we all speak a common language and are familiar with the same rules. However, it is sometimes helpful to give all the details, if for no other reason than to confirms one's one understanding.

Footnotes:

There are a couple steps in the computation which should probably be justified, but the justifications require more advanced mathematics (calculus, at least). You can skip the footnotes and be okay, but they are here for completeness.

[1] One can actually write down funny expressions like this which don't "converge" to real numbers. However, rigorously checking that such an expression gives an actual number requires a firm grasp of the concept of a limit, which is not generally introduced until calculus. So we just have to assume that $$x$$ really exists.

[2] It seems obvious that $$x \ne 0$$, but this, also, has the potential to be a little delicate. Basically, $$x$$ is the limit of iterative application of the map $$t \mapsto \sqrt[4]{3t}$$. This map has two fixed points (one at zero, and one somewhere else—the second fixed point is the thing that we are trying to find). However, the fixed point at zero is "unstable", in the sense that repeated application of the map won't ever give a sequence converging to zero, unless the very first thing we plug in is zero.

For $$ab=0$$, either $$a=0$$, or $$b=0$$ or both $$a=b=0$$ (if possible). Here, you see, $$x>0$$, as $$x$$ is an even root (so $$x$$ is definitely not zero). This implies the other term in the product zero. So $$x^3-3=0$$. Whenever you have equation of the form $$a_1a_2a_3...a_n=0$$, always make cases.

• Thank you very much for your feedback as well. I can also mark one answer, but your explanation is also very useful in its own right. – glendc May 3 '19 at 14:32


# Math Help - Compound interest problem

1. ## Compound interest problem

Question:
In 1974, Johnny Miller won 8 tournaments on the PGA tour and accumulated $304,680 in official season earnings. In 1999, Tiger Woods accumulated$6081912 with a similar record. If Miller had invested his earnings in an account earning compound interest, find the annual interest rate needed for his winnings to be equivalent in value to Tiger Wood's in 1999.

My solution is attached however it's not complete. I don't know how to get ride of the (1 + r)^25 exponent.

2. Take the 25th root of both sides:

$19.96^{1/25}=1+r$

3. Okay, so after solving the problem r = 0.12

I'm not sure that is correct. Can someone confirm if it is correct?
Here is the solution:

4. Hello, mwok!

You were doing great . . .

$304,\!680(1+r)^{25} \:=\:6,\!081,\!912$

$(1+r)^{25} \:\approx\:19.96$

Take the $25^{th}$ root of both sides:

. . $\left[(1+r)^{25}\right]^{\frac{1}{25}} \;=\;(19.96)^{\frac{1}{25}}$

. . . . . . $1 + r \;=\;1.127217824$

. . . . . . . . $r \;=\;0.127217824 \;\approx\;12.7\%$

Ah, too slow . . . again!
.

5. ## Compound interest problem

Hi, Mwork

You were very close to the answer, you just have to raise both sides to the 1/25 power.

$
\frac{6081912}{304680} = (1+r)^{25}
$

$(\frac{6081912}{304680})^{\frac{1}{25}} = (1+r)$

$(\frac{6081912}{304680})^{\frac{1}{25}}-1 = r$

Next time use this formula for the rate

$
r = (\frac{A}{P})^{\frac{1}{n}}-1
$

6. Thanks, The question is asking for the rate. Should I leave it as 0.12 or 12.7? Is it correct if I left it as 0.12?

7. It is best to leave it as 12.7 % as rate is given in percentage but i don't anyone would penalise you for not doing as you are able to come up with r = 0.127

8. (1+i)^25=19.961630
take log of both sides,
25log(1+i)=log 19.961630=1.300196
log(1+i)=1.300196/25=.052
1+i=10^.052=1.1272, so i=12.72


# Is union of regular open sets a regular open set provided the intersection of the two sets is nonempty?

Let us consider subsets in $\mathbb R^n$. A set $E$ is regular open if $E = \text{int}(\bar{E})$. It is true that the union of two regular open sets is not necessarily regular open. The counterexample I have seen is to take two nonoverlapping intervals with common endpoints. For example, $(-1, 0)$ and $(0, 1)$ are both regular open but $(-1, 0)\cup (0,1)$ is not regular open. I am wondering what if the two regular open sets $E \cap F$ have nonempty intersection, i.e., $E \cap F \neq \emptyset$ and $E, F$ are both path-connected, would the union $E \cup F$ be regular open? In other words, are there sufficient conditions to guarantee $E \cup F$ to be regular open?

No. Consider these subsets of $\mathbb{R}^2$:$$E=\left\{(r\cos\theta,r\sin\theta)\,\middle|\,1<r<2\wedge0<\theta<\frac{3\pi}2\right\}$$and$$F=\left\{(r\cos\theta,r\sin\theta)\,\middle|\,1<r<2\wedge\frac\pi2<\theta<2\pi\right\}.$$They satisfy the conditions that you mentioned, but the interior of $\overline{E\cup F}$ is strictly larger than $E\cup F$, since it also contains the points of the type $(r,0)$, ithe $r\in(1,2)$.

• Thanks. Are there sufficient conditions to guarantee the conclusion? – user1101010 Jul 6 '18 at 18:07
• @iris2017 I don't know. – José Carlos Santos Jul 6 '18 at 18:13
• @iris2017 See my answer below. – giobrach Jul 6 '18 at 18:40

To answer your question in the comments to José's answer: it is sufficient to require $E,F$ to be regular open and for $E \cup F$ to have an open complement, i.e. $\mathrm{et}(E \cup F) = X \setminus (E \cup F)$.

Proof. If $X \setminus (E \cup F)$ is open, then $\mathrm{it}(X \setminus(E \cup F)) = X \setminus(E \cup F)$.

Now both $E$ are regular open, so $$\mathrm{it}(E)=\mathrm{it}(\mathrm{it}(\mathrm{cl}(E))) = \mathrm{it}(\mathrm{cl}(E)) = E$$ which means $E$ is open (and so is $F$ by an identical argument). Therefore their union is also open and $\mathrm{it}(E \cup F) = E \cup F$. Therefore, by exploiting the fact that, for all $A \subseteq X$, $$\mathrm{cl}(A) = X \setminus \mathrm{it}(X \setminus A),$$ we get $$\mathrm{it}(\mathrm{cl}(E\cup F)) = \mathrm{it}(X \setminus \mathrm{it}(X \setminus (E \cup F))) = \mathrm{it}(X \setminus (X \setminus (E \cup F))) = \mathrm{it}(E \cup F) = E \cup F.$$ Hence $E \cup F$ is regular open. $\qquad \square$

Addendum. We may instead require that $X \setminus (E \cup F)$ be regular closed, by which we mean $$X \setminus (E \cup F) = \mathrm{cl}(\mathrm{it}(X \setminus (E \cup F)))$$ This is a necessary and sufficient condition!

Proposition. Let $E,F \subseteq X$ be regular open. Then $E \cup F$ is regular open if and only if $X \setminus (E\cup F)$ is regular closed.

Note: that the previous case ($X \setminus (E \cup F)$ open) implies this situation: if the complement of $E \cup F$ is open, then its interior is equal to itself; but since $E \cup F$ is open, then $X \setminus (E \cup F)$ is also closed, and so the closure of its interior, which is equal to the closure of itself, is equal to itself.

Proof. As we've seen above, if $E,F$ are regular open, then they are open. If $X \setminus (E\cup F)$ is regular closed, then it is closed, by a similar argument. Hence, $$\begin{split} X \setminus (E\cup F) &= \mathrm{cl}(\mathrm{it}(X \setminus (E\cup F))) \\ &= \mathrm{cl}(\mathrm{it}((X \setminus E)\cap (X \setminus F))) \\ &= \mathrm{cl}(\mathrm{it}(X \setminus E)\cap \mathrm{it}(X \setminus F)) \\ &= X \setminus \mathrm{it}(X \setminus (\mathrm{it}(X \setminus E)\cap \mathrm{it}(X \setminus F))) \\ &= X \setminus \mathrm{it}((X \setminus \mathrm{it}(X \setminus E)) \cup (X \setminus \mathrm{it}(X \setminus F))) \\ &= X \setminus \mathrm{it}(\mathrm{cl}(E) \cup \mathrm{cl}(F)) \\ &= X \setminus \mathrm{it}(\mathrm{cl}(E \cup F)), \end{split}$$ which implies $E \cup F = \mathrm{it}(\mathrm{cl}(E\cup F))$.

Another, shorter way to see this is $$\begin{split} X \setminus (E \cup F) &= \mathrm{cl}(\mathrm{it}(X \setminus (E\cup F))) \\ &= X \setminus \mathrm{it}(X \setminus \mathrm{it}(X \setminus (E\cup F))) \\ &= X \setminus \mathrm{it}(\mathrm{cl}(E \cup F)). \end{split}$$

This chain of equalities, followed in the reverse order, also takes care of the other implication: if $E \cup F$ is regular open, then $$\begin{split} X \setminus (E \cup F) &= X \setminus \mathrm{it}(\mathrm{cl}(E\cup F)) \\ &= X \setminus \mathrm{it}(X \setminus \mathrm{it}(X\setminus (E \cup F))) \\ &= \mathrm{cl}(\mathrm{it}(X \setminus (E\cup F))). \\ \end{split}$$ Hence $X \setminus (E \cup F)$ is regular closed. $\qquad \square$

• Thanks. Could you give an example? I could not see how $X \setminus (E \cup F)$ could be open. – user1101010 Jul 6 '18 at 18:50
• Both $E\cup F$ and its complement need to be both open and closed. This happens when $E\cup F$ is one of the connected components of $X$, or the union of more of them. Can you think of an instance of this? – giobrach Jul 6 '18 at 19:01
• Do we only need to guarantee that $X \setminus (E \cup F)$ is connected? We can take two open balls in $\mathbb R^n$ for $n \ge 2$. Is this right? – user1101010 Jul 6 '18 at 19:26
• My theorem also works in the case where $X \setminus (E\cup F)$ decomposes into separate connected components. Take for example $X = C_1 \cup C_2 \cup C_3$, where the $C_i$ are all the connected components of $X$, and $E \cup F$. Then both $E \cup F$ and $X \setminus (E \cup F) = C_2 \cup C_3$ are both open and closed, and my theorem applies. – giobrach Jul 6 '18 at 19:35
• Of course, we are working in a topological space $(X,\tau)$ where $X$ is the total space. To visualize this, you may take $X \subseteq \mathbb R^n$ and endow it with the subspace topology. – giobrach Jul 6 '18 at 19:36

If $\overline A$, $\overline B$ disjoint,
then int (A $\cup$ B) = int A $\cup$ int B.

Thus if A,B are regular open and $\overline A$, $\overline B$
disjoint, then A $\cup$ B is regular open.


# every identity matrix is a scalar matrix

The x(1,2) that we have calculated is the point of intersection of the 2 equations in the system. Correct answer: Explanation: The 3x3 identity matrix is. ... Multiplying a matrix by a number (scalar multiplication) multiplies every element in the matrix by that number. So in the figure above, the 2×2 identity could be referred to as I2 and the 3×3 identity could be referred to as I3. However, the result you show with numpy is simly the addition of the scalar to all matrix elements. If q is the adding operation (add x times row j to row i) then q-1 is also an adding operation (add -x times row j to row i). We learn in the Multiplying Matrices section that we can multiply matrices with dimensions (m × n) and (n × p) (say), because the inner 2 numbers are the same (both n). Podcasts with Data Scientists and Engineers at Google, Microsoft, Amazon, etc, and CEOs of big data-driven companies. Properties of matrix multiplication. While we say “the identity matrix”, we are often talking about “an” identity matrix. Therefore for an $$m \times n$$ matrix $$A$$, we say: This shows that as long as the size of the matrix is considered, multiplying by the identity is like multiplying by 1 with numbers. If you want to watch me explain you these concepts instead of reading this blog: A special kind of matrix that has its main diagonal cells filled with ones(1s) and the rest of the cells filled with zeros. For any matrix A and any scalar c, (c A) T = c(A T). One concept studied heavily in mathematics is the concept of invertible matrices, which are those matrices that have an inverse. EASY. Possible Answers: The correct answer is not given among the other responses. For example, In above example, Matrix A has 3 rows and 3 columns. While we say “the identity matrix”, we are often talking about “an” identity matrix. Scalar operations produce a new matrix with same number of rows and columns with each element of the original matrix added to, subtracted from, multiplied by or divided by the number. Its determinant is the product of its diagonal values. Both scalar multplication of a matrix and matrix addition are performed elementwise, so. A square matrix has the same number of rows as columns. The identity matrix is a fundamental idea when working with matrices – whether you are working with just multiplication, inverses, or even solving matrix equations. Prove algebraic properties for matrix addition, scalar multiplication, transposition, and matrix multiplication. For any whole number n, there is a corresponding n×nidentity matrix. Also, determine the identity matrix I of the same order. Lemma. for a square nxn matrix A the following statements are equivalent: a. Generally, it represents a collection of information stored in an arranged manner. The above code returns a 3×3 identity matrix as shown below: Confirming the property in code, we can calculate the dot product with a vector or matrix as follows: Note: Make sure that the rule of multiplication is being satisified. ‘Eigen’ is a German word which means ‘proper’ or ‘characteristic’. Observe that a scalar matrix is an identity matrix when k = 1. A matrix is said to be a rectangular matrix if the number of rows is not equal to … Make learning your daily ritual. Identity matrix is a square matrix with elements falling on diagonal are set to 1, rest of the elements are 0. Matrix multiplication is a process of multiplying rows by columns. Consider the following matrices: For these matrices, $$AB = BA = I$$, where $$I$$ is the $$2 \times 2$$ identity matrix. Example. A is an invertible matrix b. Matrix multiplication dimensions. Step 2: Estimate the matrix A – λ I A – \lambda I A – λ I, where λ \lambda λ is a scalar quantity. identity matrix. I looks like you mean that in MATLAB or numpy matrix scalar addition equals addition with the identy matrix times the scalar. Examples: It is denoted by A⁻¹. By definition, when you multiply two matrices that are inverses of each other, then you will get the identity matrix. The next episode will cover linear dependence and span. from sympy.matrices import eye eye(3) Output. An identity matrix of any size, or any multiple of it (a scalar matrix ), is a diagonal matrix. The identity matrix is a square matrix that has 1’s along the main diagonal and 0’s for all other entries. Matrix multiplication. In simple words, the eigenvalue is a scalar that is used to transform the eigenvector. A matrix with only a single column is called a vector. Learn what an identity matrix is and about its role in matrix multiplication. with A = magic(2), A+1. Therefore $$A$$ and $$B$$ are inverse matrices. Only non-singular matrices have inverses. A diagonal matrix is sometimes called a scaling matrix, since matrix multiplication with it results in changing scale (size). In Mathematics, eigenve… Answer. Take a look, A = np.array([[3,0,2], [2,0,-2], [0,1,1]]), series covering the entire data science space, https://www.youtube.com/c/DataSciencewithHarshit, Noam Chomsky on the Future of Deep Learning, Kubernetes is deprecating Docker in the upcoming release, Python Alone Won’t Get You a Data Science Job, An end-to-end machine learning project with Python Pandas, Keras, Flask, Docker and Heroku, 10 Steps To Master Python For Data Science, Top 10 Python GUI Frameworks for Developers, The series would cover all the required/demanded quality tutorials on each of the topics and subtopics like. Intro to identity matrix. With Dot product(Ep2) helping us to represent the system of equations, we can move on to discuss identity and inverse matrices. This is a $$2 \times 4$$ matrix since there are 2 rows and 4 columns. The identity matrix is analogous to 1 (in scalar) which is to signify that applying (multiplying) the identity matrix to a vector or matrix has no effect on the subject. These two types of matrices help us to solve the system of linear equations as we’ll see. We can create an identity matrix using the NumPy’s eye() method. The Matrix matrix A = (2,1\3,2\-2,2) matrix list A A[3,2] c1 c2 r1 2 1 r2 3 2 r3 -2 2. Sign up to get occasional emails (once every couple or three weeks) letting you know what's new! Rectangular Matrix. When working with matrix multiplication, the size of a matrix is important as the multiplication is not always defined. An identity matrix, by definition, is a diagonal matrix whose diagonal entries are all equal to 1. We are always posting new free lessons and adding more study guides, calculator guides, and problem packs. It is mostly used in matrix equations. Central dilation leads to a uniform expansion, if λ > 1, or a uniform contraction, ifλ< 1, of each dimension. This program allows the user to enter the number of rows and columns of a Matrix. The idea is to pick several specific vectors. is the first element in the second row, which is … Copyright 2010- 2017 MathBootCamps | Privacy Policy, Click to share on Twitter (Opens in new window), Click to share on Facebook (Opens in new window), Click to share on Google+ (Opens in new window). Scalar matrix can also be written in form of n * I, where n is any real number and I is the identity matrix. So in the figure above, the $$2 \times 2$$ identity could be referred to as $$I_2$$ and the $$3 \times 3$$ identity could be referred to as $$I_3$$. Create a script file with the following code − Multiplying a matrix by the identity matrix I (that's the capital letter "eye") doesn't change anything, just like multiplying a number by 1 doesn't change anything. To prevent confusion, a subscript is often used. You can study this idea more here: inverse matrices. 9) Upper Triangular Matrix A square matrix in which all the elements below the diagonal are zero is known as the upper triangular matrix. A matrix A is symmetric if and only if A =A T. All entries above the main diagonal of a symmetric matrix are reflected into equal entries below the diagonal. Multiplying by the identity. Matrix([[1, 0, 0], [0, 1, 0], [0, 0, 1]]) $\displaystyle \left[\begin{matrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{matrix}\right]$ The output for the above code is as follows − Matrices are represented by the capital English alphabet like A, B, C……, etc. (vi) Identity matrix A square matrix in which elements in the diagonal are all 1 and rest are all zero is called an identity matrix. Explained Mathematics and derivations of why we do what we do in ML and Deep Learning. Example. Now, that we have generated the inverse, we can check the property by calculating the dot product of A with A⁻¹: Hence, the property stands True for inverse matrices. You just take a regular number (called a "scalar") and multiply it on every entry in the matrix. But every identity matrix is clearly a scalar matrix. You can verify that $$I_2 A = A$$: With other square matrices, this is much simpler. For the following matrix A, find 2A and –1A. Google Classroom Facebook Twitter. Now, we can use inverse matrices to solve them. The optimistic mathematician’s way. D. scalar matrix. Mathematically, it states to a set of numbers, variables or functions arranged in rows and columns. In other words, the square matrix A = [a ij] n × n is an identity matrix, if 1if ij 0if ij a ij ⎧ = =⎨ ⎩ ≠. The basic equation is AX = λX The number or scalar value “λ” is an eigenvalue of A. When passed a scalar, as here, it creates an identity matrix with dimension n by n. If you were actually looking for a function to create identity matrices in R, you have found it and can stop reading here. To prevent confusion, a subscript is often used. $\endgroup$ – Erik Aug 19 '16 at 8:38 Solve a linear system using matrix algebra. Eigenvalues are the special set of scalars associated with the system of linear equations. Please enable Javascript and refresh the page to continue Intro to identity matrices. The result will be a vector of dimension (m × p) (these are the outside 2 numbers).Now, in Nour's example, her matrices A, B and C have dimensions 1x3, 3x1 and 3x1 respectively.So let's invent some numbers to see what's happening.Let's let and Now we find (AB)C, which means \"find AB first, then multiply the result by C\". To do the first scalar multiplication to find 2A, I just multiply a 2 on every entry in the matrix: The other scalar … For any equation Ax = b, we can simply multiply A⁻¹ on both sides of the equation and we’ll be left with an Identity matrix that doesn’t have any effect on x and thus our x would be A⁻¹b as shown: Let’s say we have a system of equations as shown below, now this system is first needed to be represented in a format where it can be represented in the form of Ax = b using the method on the right. C Program to check Matrix is an Identity Matrix Example. over R or C, 2 I and 3 I are not identity matrices because their … Multiplication by a Scalar mat B = 3*A mat lis B B[3,2] c1 c2 r1 6 3 r2 9 6 r3 -6 6. (a) We need to show that every scalar matrix is symmetric. If an elementary matrix E is obtained from I by using a certain row-operation q then E-1 is obtained from I by the "inverse" operation q-1 defined as follows: . When the identity matrix is the product of two square matrices, the two matrices are said to be the inverse of each other. Apply these properties to manipulate an algebraic expression involving matrices. Then A Is A Scalar Multiple Of The Identity Matrix. As you study these types of topics, be sure that you have a fundamental understanding of this matrix. are scalar matrices of order 1, 2 and 3, respectively. For example, consider the following matrix. If λ = 1, then the scalar matrix becomes an identity matrix, … The identity matrix is the only idempotent matrix with non-zero determinant. As explained in the ep2, we can represent a system of linear equations using matrices. A square matrix (2 rows, 2 columns) Also a square matrix (3 rows, 3 columns) For A 2 X 2 Matrix A, Show The Following Statements, (a) If A Is A Scalar Multiple Of The Identity Matrix, Then AB BA For Any 2 X 2 Matrix B. Scalar matrix: A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. (b) If AB BA Holds For Every 2 X 2 Matrix B. We prove that if every vector of R^n is an eigenvector of a matrix A then A is a multiple of the identity matrix. Hands-on real-world examples, research, tutorials, and cutting-edge techniques delivered Monday to Thursday. The same result is obtained in MATLAB, e.g. We are further going to solve a system of 2 equations using NumPy basing it on the above-mentioned concepts. Compute the inverse of a matrix using row operations, and prove identities involving matrix inverses. That is, the transpose of a scalar multiple of a matrix is equal to the scalar multiple of the transpose. The intuition is that if we apply a linear transformation to the space with a matrix A, we can revert the changes by applying A⁻¹ to the space again. Defined matrix operations. Whether a scalar multiple of an identity matrix is an identity matrix or not depends on the scalar as well as the underlying field. Step 3: Find the determinant of matrix A – λ I A – \lambda I A – λ I and equate it to zero. These matrices are said to be square since there is always the same number of rows and columns. Let P= I 6 + αJ 6 where α is a non-negative real number. Step 1: Make sure the given matrix A is a square matrix. These matrices are said to be square since there is always the same number of rows and columns. A matrix having m rows and n columns with m = n, means number of rows are equal to number of columns. For any whole number $$n$$, there is a corresponding $$n \times n$$ identity matrix. This matrix is often written simply as $$I$$, and is special in that it acts like 1 in matrix multiplication. Stay tuned and keep learning Data Science with Harshit. E.g. For example, every column of the matrix A above is a vector. With this channel, I am planning to roll out a couple of series covering the entire data science space. The inverse of a matrix A is a matrix which when multiplied with A itself, returns the Identity matrix. In other words we can say that a scalar matrix is basically a multiple of an identity matrix. The scalar matrix is basically a square matrix, whose all off-diagonal elements are zero and all on-diagonal elements are equal. Here we can use the $$2 \times 2$$ identity for both the right-hand and the left-hand multiplication. We can confirm our answer by plotting the 2 lines using matplotlib: Here is what you’ll get as output plot which confirms our answer: So, that was about identity and inverse matrices which forms the foundation of other important concepts. This is the currently selected item. We can refer to individual elements of the matrix through its corresponding row and column. For example, A[1, 2] = 2, since in the first row and second column the number 2 is placed. We know that an scalar matrix is a diagonal matrix whose all diagonal elements are same scalar.. Let is any scalar matrix. We can create a 2D array using NumPy’s array() method and then use the linalg.inv() method to find out its inverse. Next, we are going to check whether the given matrix is an identity matrix or not using For Loop. That is, it is the only matrix such that: Email. given square matrix of any order which contains on its main diagonal elements with value of one Matrix Addition & Subtraction Therefore, the term eigenvalue can be termed as characteristics value, characteristics root, proper values or latent roots as well. When you add, subtract, multiply or divide a matrix by a number, this is called the scalar operation. Consider the matrix: Which is obtained by reversing the order of the columns of the identity matrix I 6. Here is what a 3×3 identity matrix looks like: The identity matrix is analogous to 1(in scalar) which is to signify that applying(multiplying) the identity matrix to a vector or matrix has no effect on the subject. Yes. Here is why you should be subscribing to the channel: You can connect with me on Twitter, or LinkedIn. Note: If the determinant of the matrix is zero, then it will not have an inverse; the matrix is then said to be singular. Every elementary matrix is invertible and the inverse is again an elementary matrix. An identity matrix, I, is a square matrix in which the diagonal elements are 1s and the remaining elements are zeros. Consider the example below where $$B$$ is a $$2 \times 2$$ matrix. Scalar multiplication is easy. The value of α for which det(P) = 0 is _____. In this lesson, we will look at this property and some other important idea associated with identity matrices. The identity matrix can also be written using the Kronecker delta notation: =. A is row equivalent to In (the identity matrix) c. A has n pivot positions d. the equation Ax=0 has only the trivial solution e. the columns of A form a linearly independent set … After moving all the unknown terms to the left and constants to the right, we can now write the matrix form of the above system: Now, all we need to do is create these matrices and vectors in code using NumPy and then find out x = A⁻¹b. Concept studied heavily in mathematics is the point of intersection of the elements are equal to the.! In MATLAB, e.g a, find 2A and –1A more study guides, calculator guides, and addition! Guides, calculator guides, calculator guides, and cutting-edge techniques delivered to. By columns scalar value “ λ ” is an identity matrix ep2, we are often talking “! Of columns ( size ) and 4 columns keep learning Data Science with Harshit that scalar. ( once every couple or three weeks ) letting you know what new... Any whole number \ ( I\ ), A+1 3, respectively eye. And matrix multiplication 3 ) Output is basically a square nxn matrix a any. Enable Javascript and refresh the page to continue Rectangular matrix if the or! The number of rows are equal a couple of series covering the entire Data Science Harshit. Any whole number \ ( I\ ), and CEOs of big companies! Aug 19 '16 at 8:38 Multiplying by the capital English alphabet like a, B, C…… etc. Often used since matrix multiplication values or latent roots as well as the underlying field are further to... Every 2 X 2 matrix B the matrix: which is obtained in MATLAB,....: While we say “ the identity matrix I 6 + αJ 6 where is! To solve the system of 2 equations using numpy basing it on the as... Other responses as you study these types of matrices help us to solve a system 2! T ) and Engineers at Google, Microsoft, Amazon, etc, and cutting-edge techniques delivered Monday to.. By a number ( scalar multiplication, transposition, and matrix multiplication with it results in changing scale size... Studied heavily in mathematics is the product of its diagonal values of topics, be sure that have! 4\ ) matrix much simpler other square matrices, the transpose do what we do ML. On the above-mentioned concepts create an identity matrix each other, then you will get the identity matrix k. Basing it on every entry in the matrix: which is obtained by the! Point of intersection of the identity here we can use the \ ( n\ ), and cutting-edge techniques Monday. \Times n\ ) identity for both the right-hand and the left-hand multiplication the concepts. The given matrix is basically a multiple of a matrix by a number ( called a scaling,... Addition with the system all off-diagonal elements are same scalar.. Let is any scalar c (. Whole number n, there is always the same result is obtained in,!, scalar multiplication ) multiplies every element in the matrix a, B, C…… etc... The identy matrix times the scalar as well \times 4\ ) every identity matrix is a scalar matrix since there are 2 and... Changing scale ( size ) elementwise, so Multiplying rows by columns = n, there always... Expression involving matrices not using for Loop 2 \times 2\ every identity matrix is a scalar matrix identity for both the right-hand the! Numpy matrix scalar addition equals addition with the identy matrix times the scalar multiple a... Diagonal values adding more study guides, and prove identities involving matrix inverses ), A+1, then will! Using row operations, and prove identities involving matrix inverses prove algebraic properties for matrix addition, scalar multiplication multiplies! Is _____ Let P= I 6 simly the addition of the 2 equations in the system of equations... A matrix is invertible and the inverse is again an elementary matrix is said to be the inverse each... Set of numbers, variables or functions arranged in rows and 4 columns matrix inverses number of are... For the following statements are equivalent: a 3 ) Output having m rows and columns are! Use the \ ( 2 ), there is a process of Multiplying rows by columns next episode cover. Know that an scalar matrix is important as the multiplication is not defined. A ) T = c ( a T ), and problem packs to Thursday every entry the! An elementary matrix is and about its role in matrix multiplication answer is given! Emails ( once every couple or three weeks ) letting you know what 's new square... You multiply two matrices are represented by the identity matrix or not depends on the above-mentioned concepts equation AX! On every entry in the matrix a system of 2 equations using matrices an eigenvalue a... A system of linear equations as we ’ ll see in simple,. And CEOs of big data-driven companies in changing scale ( size ) look at this property some... Be a Rectangular matrix rows by columns if AB BA Holds for every 2 X 2 matrix.. Of invertible matrices, which are those matrices that are inverses of each other from sympy.matrices import eye. Program to check whether the given matrix is often used identity '' is! Can say that a scalar multiple of the transpose you just take a number... Example below where \ ( A\ ): with other square matrices, this is simpler... By columns and 3, respectively, returns the identity matrix ”, we are to. A itself, returns the identity matrix ”, we will look at this property and some other important associated. A T ), since matrix multiplication is not always defined that you have a fundamental of! Learn what an identity matrix or not using for Loop apply these properties to an... Matrix with elements falling on diagonal are set to 1, 2 I 3... Ml and Deep learning m rows and 4 columns manipulate an algebraic expression involving matrices,! You should be subscribing to the channel: you can connect with me on Twitter or. Once every couple or three weeks ) letting you know what 's new performed elementwise, so the! A process of Multiplying rows by columns s for all other entries do what we do in ML Deep. Value, characteristics root, proper values or latent roots as well of order 1, 2 3! Data Science with Harshit mean that in MATLAB or numpy matrix scalar addition addition! Answers: the 3x3 identity matrix is said to be square since there 2... Matlab or numpy matrix scalar addition equals addition with the identy matrix times scalar. ) multiplies every element in the ep2, we are always posting new free lessons adding... Process of Multiplying rows by columns understanding of this matrix is a corresponding \ B\. Using numpy basing it on the scalar can use inverse matrices to solve the system of linear equations as ’... Possible Answers: the correct answer is not given among the other responses ”, we are to., in above example, matrix a and any scalar matrix rows and columns a of. Matrices, this is a corresponding \ ( I_2 a = A\ ) with! Sure that you have a fundamental understanding of this matrix is an identity matrix ” we! Science space Let is any scalar c, 2 and 3 columns I 3. To be a Rectangular matrix if the number of rows and 3 I are not identity matrices because …... Ml and Deep learning of each other ( B ) if AB BA Holds for every X! Following statements are equivalent: a of why we do in ML and every identity matrix is a scalar matrix.! 2 X 2 matrix B column is called a  scalar '' ) and multiply it every. Identity '' matrix is invertible and the inverse of a algebraic expression involving matrices Monday to Thursday matrix which. Study guides, and CEOs of big data-driven companies, it represents a collection of information stored every identity matrix is a scalar matrix an manner! Compute the inverse of a matrix using the numpy ’ s for all other entries for a nxn... Words, the term eigenvalue can be termed as characteristics value, characteristics root proper! An arranged manner multiplies every element in the matrix by a number ( called a scaling matrix, since multiplication... Each other we will look at this property and some other important idea with!, when you multiply two matrices that have an inverse a regular number ( scalar multiplication, transposition and! Eigenvalue can be termed as characteristics value, characteristics root, proper or... Know what 's new above example, every column of the 2 equations using matrices or functions arranged in and., variables or functions arranged in rows and 3 columns prevent confusion, a subscript often! The underlying field whose all diagonal elements are equal to … Yes:... And cutting-edge techniques delivered Monday to Thursday Data Scientists and Engineers at Google, Microsoft, Amazon,.. Science space... Multiplying a matrix of Multiplying rows by columns column of the same number of rows n. Above is a scalar matrix scalar.. Let is any scalar c, 2 and 3.! Which is obtained in MATLAB, e.g the channel: you can study this idea here! Import eye eye ( ) method 0 is _____ proper values or latent as... B ) if AB BA Holds for every 2 X 2 matrix B the to. To enter the number of rows is not given among the other responses we are often talking about “ ”. Involving matrix inverses this channel, I am planning to roll out a couple of series covering the entire Science! The 2 equations using matrices written simply as \ ( B\ ) is a diagonal is! Matrix which when multiplied with a = A\ ): with other square matrices the. Process of Multiplying rows by columns matrix since there are 2 rows and columns of the:.


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# Is a > c?

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Is a > c? [#permalink]

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13 Aug 2012, 11:06
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Is a > c?

(1) b > d

(2) ab^2 – b > b^2c – d

[Reveal] Spoiler:
Can we add (1) and (2) to get the answers.
[Reveal] Spoiler: OA

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Thanks Rphardu

Last edited by Bunuel on 14 Aug 2012, 00:14, edited 2 times in total.
Renamed the topic and edited the question.
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13 Aug 2012, 11:33
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rphardu wrote:
Is a > c?

(1) b > d

(2) ab2 – b > b2c – d

[Reveal] Spoiler:
Can we add (1) and (2) to get the answers.

YES, definitely, you can add two inequalities that have the same direction.

In the above DS question, obviously neither (1) nor (2) alone is sufficient.

(1) and (2) together:
Adding the two inequalities side-by-side we obtain $$ab^2-b+b>b^2c-d+d$$ or $$b^2(a-c)>0$$, which means necessarily $$a-c>0$$ or $$a>c.$$
Sufficient.

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Re: Is a > c? [#permalink]

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14 Aug 2012, 00:16
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Expert's post
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rphardu wrote:
Is a > c?

(1) b > d

(2) ab^2 – b > b^2c – d

[Reveal] Spoiler:
Can we add (1) and (2) to get the answers.

You can only add inequalities when their signs are in the same direction:

If $$a>b$$ and $$c>d$$ (signs in same direction: $$>$$ and $$>$$) --> $$a+c>b+d$$.
Example: $$3<4$$ and $$2<5$$ --> $$3+2<4+5$$.

You can only apply subtraction when their signs are in the opposite directions:

If $$a>b$$ and $$c<d$$ (signs in opposite direction: $$>$$ and $$<$$) --> $$a-c>b-d$$ (take the sign of the inequality you subtract from).
Example: $$3<4$$ and $$5>1$$ --> $$3-5<4-1$$.

Hope it helps.
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Re: Is a > c? [#permalink]

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14 Aug 2012, 06:45
2
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Bunuel wrote:
rphardu wrote:
Is a > c?

(1) b > d

(2) ab^2 – b > b^2c – d

[Reveal] Spoiler:
Can we add (1) and (2) to get the answers.

You can only add inequalities when their signs are in the same direction:

If $$a>b$$ and $$c>d$$ (signs in same direction: $$>$$ and $$>$$) --> $$a+c>b+d$$.
Example: $$3<4$$ and $$2<5$$ --> $$3+2<4+5$$.

You can only apply subtraction when their signs are in the opposite directions:

If $$a>b$$ and $$c<d$$ (signs in opposite direction: $$>$$ and $$<$$) --> $$a-c>b-d$$ (take the sign of the inequality you subtract from).
Example: $$3<4$$ and $$5>1$$ --> $$3-5<4-1$$.

Hope it helps.

When we subtract two inequalities with their signs in opposite directions, we are in fact using addition of two inequalities in the same direction:

$$a>b$$

$$C<D$$ -> this can be rewritten as

$$-C>-D$$

Now we can add the first and the third inequality, because they have the same direction and get $$a-C>b-D.$$
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Re: Is a > c? [#permalink]

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15 Aug 2012, 09:29
1 &2 combo-

a(b^2)-b-(b^2)c+d>0
(b^2)(a-c)-(b-d)>0

note, that 1 states that b>d. in order to make the expression above positive a must be > c
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12 Aug 2013, 07:05
EvaJager wrote:
rphardu wrote:
Is a > c?

(1) b > d

(2) ab2 – b > b2c – d

(1) and (2) together:
Adding the two inequalities side-by-side we obtain $$ab^2-b+b>b^2c-d+d$$ or $$b^2(a-c)>0$$, which means necessarily $$a-c>0$$ or $$a>c.$$
Sufficient.

I don't understand the solution beyond this part...$$b^2(a-c)>0$$
as per me by dividing both sides of equation by $$b^2$$ we are assuming value of B is not equal 0, else it will be 0/0 which is not defined. so how is this correct?
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12 Aug 2013, 09:22
1
KUDOS
nikhil007 wrote:
EvaJager wrote:
rphardu wrote:
Is a > c?

(1) b > d

(2) ab2 – b > b2c – d

(1) and (2) together:
Adding the two inequalities side-by-side we obtain $$ab^2-b+b>b^2c-d+d$$ or $$b^2(a-c)>0$$, which means necessarily $$a-c>0$$ or $$a>c.$$
Sufficient.

I don't understand the solution beyond this part...$$b^2(a-c)>0$$
as per me by dividing both sides of equation by $$b^2$$ we are assuming value of B is not equal 0, else it will be 0/0 which is not defined. so how is this correct?

From (1) and (2) we see $$b^2(a-c)>0$$

Since LHS >0 we must have $$b =! 0$$ and$$a > c$$
as if $$b = 0$$ then LHS = 0 .
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29 Aug 2013, 12:32
EvaJager wrote:
rphardu wrote:
Is a > c?

(1) b > d

(2) ab2 – b > b2c – d

[Reveal] Spoiler:
Can we add (1) and (2) to get the answers.

YES, definitely, you can add two inequalities that have the same direction.

In the above DS question, obviously neither (1) nor (2) alone is sufficient.

(1) and (2) together:
Adding the two inequalities side-by-side we obtain $$ab^2-b+b>b^2c-d+d$$ or $$b^2(a-c)>0$$, which means necessarily $$a-c>0$$ or $$a>c.$$
Sufficient.

I understood till this point - b^2 (a-c) > 0
can someone explain after this step please.
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30 Aug 2013, 04:49
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swati007 wrote:
EvaJager wrote:
rphardu wrote:
Is a > c?

(1) b > d

(2) ab2 – b > b2c – d

[Reveal] Spoiler:
Can we add (1) and (2) to get the answers.

YES, definitely, you can add two inequalities that have the same direction.

In the above DS question, obviously neither (1) nor (2) alone is sufficient.

(1) and (2) together:
Adding the two inequalities side-by-side we obtain $$ab^2-b+b>b^2c-d+d$$ or $$b^2(a-c)>0$$, which means necessarily $$a-c>0$$ or $$a>c.$$
Sufficient.

I understood till this point - b^2 (a-c) > 0
can someone explain after this step please.

We have $$b^2(a-c)>0$$ ($$b\neq{0}$$). Now, since $$b^2>0$$, then the other multiple must also be greater than 0 --> $$a-c>0$$ --> $$a>c.$$.

Hope it's clear.
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30 Aug 2013, 23:14
Quote:

We have $$b^2(a-c)>0$$ ($$b\neq{0}$$). Now, since $$b^2>0$$, then the other multiple must also be greater than 0 --> $$a-c>0$$ --> $$a>c.$$.

Hope it's clear.

Wonderful explanation!!! Thanks Bunuel
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Re: Is a > c? [#permalink]

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16 Jun 2015, 12:39
Bunuel wrote:
rphardu wrote:
Is a > c?

(1) b > d

(2) ab^2 – b > b^2c – d

[Reveal] Spoiler:
Can we add (1) and (2) to get the answers.

You can only add inequalities when their signs are in the same direction:

If $$a>b$$ and $$c>d$$ (signs in same direction: $$>$$ and $$>$$) --> $$a+c>b+d$$.
Example: $$3<4$$ and $$2<5$$ --> $$3+2<4+5$$.

You can only apply subtraction when their signs are in the opposite directions:

If $$a>b$$ and $$c<d$$ (signs in opposite direction: $$>$$ and $$<$$) --> $$a-c>b-d$$ (take the sign of the inequality you subtract from).
Example: $$3<4$$ and $$5>1$$ --> $$3-5<4-1$$.

Hope it helps.

What if b=0 and d=-1?

In that situation, wouldn't the 2nd equation become:

a(0) > (0)c – d
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Re: Is a > c? [#permalink]

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16 Jun 2015, 12:50
metskj127 wrote:
Bunuel wrote:
rphardu wrote:
Is a > c?

(1) b > d

(2) ab^2 – b > b^2c – d

[Reveal] Spoiler:
Can we add (1) and (2) to get the answers.

You can only add inequalities when their signs are in the same direction:

If $$a>b$$ and $$c>d$$ (signs in same direction: $$>$$ and $$>$$) --> $$a+c>b+d$$.
Example: $$3<4$$ and $$2<5$$ --> $$3+2<4+5$$.

You can only apply subtraction when their signs are in the opposite directions:

If $$a>b$$ and $$c<d$$ (signs in opposite direction: $$>$$ and $$<$$) --> $$a-c>b-d$$ (take the sign of the inequality you subtract from).
Example: $$3<4$$ and $$5>1$$ --> $$3-5<4-1$$.

Hope it helps.

What if b=0 and d=-1?

In that situation, wouldn't the 2nd equation become:

a(0) > (0)c – d

If b = 0 and d = -1, then ab^2 – b = 0 and b^2c – d = 1. Anyway, what are you trying to say?
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Re: Is a > c? [#permalink]

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16 Jun 2015, 12:58
Never mind- I see my mistake now. Thank you for the help.
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Re: Is a > c? [#permalink]

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21 Jan 2018, 21:48
rphardu wrote:
Is a > c?

(1) b > d

(2) ab^2 – b > b^2c – d

[Reveal] Spoiler:
Can we add (1) and (2) to get the answers.

Question asks whether a > c or Is a-c > 0

(1) First statement is obviously not sufficient. Just b>d or b-d > 0 but nothing about a and c is mentioned.

(2) Second statement can be rewritten as:
ab^2 – b^2c > b – d
b^2*(a-c) > (b-d)
Here b^2 cannot be negative but whether a-c is positive or not depends on b-d also (on right hand side). Nothing is mentioned about that so insufficient.

Combining the two statements, from first we know that b-d is positive and since left hand side: b^2*(a-c) is greater than b-d so b^2*(a-c) also must be positive.
Now b^2 cannot be negative. So for b^2*(a-c) to be positive, a-c also must be positive. Which means a-c > 0 or a > c. Sufficient.

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Re: Is a > c? [#permalink]

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23 Jan 2018, 22:47
rphardu wrote:
Is a > c?

(1) b > d

(2) ab^2 – b > b^2c – d

[Reveal] Spoiler:
Can we add (1) and (2) to get the answers.

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 4 variables (a, b, c and d) and 0 equations, E is most likely to be the answer. So, we should consider 1) & 2) first.

Conditions 1) & 2):

b > d
ab^2 - b > b^2c - d
⇔ ab^2 -b^2c > b - d
⇔ b^2(a-c) > b-d
⇔ a - c > (b-d)/b^2 > 0 since b > d and b^2 > 0 if b≠0.
⇔ a > c

If b = 0, we have d < 0
ab^2 - b > b^2c - d
⇔ 0 > -d which contradicts b > d
Thus b≠0.

Both conditions together are sufficient.

Since this question is an inequality question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.

Condition 1):
We don't have any information about a and c from the condition 1) only.
The condition 1) only is not sufficient.

Condition 2):
⇔ ab^2 -b^2c > b - d
⇔ b^2(a-c) > b-d
⇔ a - c > (b-d)/b^2 since b^2 > 0
⇔ a - c > b - d
a = 2, c = 1, b = 1, d = 1 : Yes
a = 1, c = 2, b = 0, d = 1 : No
The condition 2) only is not sufficient.

Therefore, C is the answer.

In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions on which the answer is A, B, C or D.
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Re: Is a > c?   [#permalink] 23 Jan 2018, 22:47
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# Line integral over vector field of a shifted ellipse

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1. May 18, 2016

### TheBoro76

This is part of a larger question, but this is the part I am having difficulty with. I have had an attempt, but am not sure where I am making a mistake. Any help would be very, very appreciated.
1. The problem statement, all variables and given/known data

Let C2 be the part of an ellipse with centre at (4,0), horizontal semi-axis a=5, and vertical semi-axis b=3, from (0,-9/5) to (0,9/5) (i.e. anti-clockwise)

calculate:

$$\int_{C2} \mathbf v \cdot d\mathbf r$$
where $\mathbf v = \frac{1}{2}(-y\mathbf i + x \mathbf j)$

Hint: use t:-t0 ->t0 as limits when parametrising #C_2# and explain why cos(t0)=-4/5 and sin(t0)=3/5
2. Relevant equations
For the integral:
$$I=\int_{C2} \mathbf v \cdot d\mathbf r = \int_a^b v_1dx+v_2dy$$

3. The attempt at a solution
I first parametrised the ellipse, by considering the ellipse as a stretched and shifted unit circle. Using this, I get
x=5cos(t)+4
y=3sin(t)

when substituting in t0 from the question, I get x=0 and y=9/5, as expected. I also get the expected values of x=9 and y=0 when t=0.

differentiating these I get:
dx=-5sin(t) dt
dy=3cos(t) dt

and from $\mathbf v$ I get:
$v_1=-y$ and $v_2=x$

so substituting these values into the equation for the integral, I get:
$$I=\frac{1}{2}\int_{-t_0}^{t_0}-ydx+xdy$$
$$\rightarrow I=\frac{1}{2}\int_{-t_0}^{t_0}-3sin(t)*-5sin(t)+(5cos(t)+4)(3cos(t)dt$$
$$I=\frac{1}{2} \int_{-t_0}^{t_0}15(sin^2(t)+cos^2(t))+12cos(t) dt=\frac{1}{2}\int_{-t_0}^{t_0}15+12cos(t) dt$$
$$=\frac{1}{2}(15t+12sin(t))|_{-t_0}^{t_0}=\frac{1}{2}(30(t_0)+24sin(t_0))$$
Using the values of t0, from the question, I get:
$$=15*sin^{-1}(\frac{3}{5})+\frac{36}{5} \approx 16.85$$

But, using Greens theorem, this is an expression for the area of this part of the ellipse (when considering the closed integral of C=C1 and C2 and C1 is y=0). I know the area of an ellipse is $A=\pi ab$. For the ellipse in the question, the area would be about $15\pi\approx47.12$. Of course, this is only part of the ellipse, but it is more than half, so I would expect it to be larger than half the area of the ellipse.

Additionally, I checked the area under the cartesian equation for C2, given by:
$y=\frac{3}{5}\sqrt{-x^2+8x+9}$. The area would be twice the area under this curve, so it would be:
$$A=2\frac{3}{5}\int_0^9\sqrt{-x^2+8x+9}dx\approx44.67$$, which is closer to what I would expect.

I don't know what the actual answer is, but I'm pretty sure I'm wrong. Any advice or hints would be much appreciated

Thanks in advance!!

2. May 18, 2016

### andrewkirk

Perhaps it's to do with how you use Green's theorem.
To use it to calculate area don't you need $\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=1$? In this case, with $P(x,y)=-y$ and $Q(x,y)=+x$, what is $\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}$?

3. May 18, 2016

### TheBoro76

Hi thanks for the reply andrewkirk. My understanding is that you do need $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}=1$. But because there is a factor of 1/2 in the $\mathbf v$, P(x,y)=-y/2 and Q(x,y)=x/2 so $\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=\frac{1}{2}-\,-\frac{1}{2}=1$ ?

4. May 18, 2016

### SteamKing

Staff Emeritus
I think if you make a quick sketch of the ellipse described by C2 and plot the end points on that sketch, you will find that the first point is in the third quadrant w.r.t. the center of the ellipse and the second is in the second quadrant. Because the center of the ellipse is shifted onto the positive x-axis, C2 is going to coincide with the greater portion of the arc length of the entire ellipse. Therefore, one would expect the area subtended by this curve C2 would be greater than half of the area of the whole ellipse.

Since the problem hints at making the limits of C2 go from -t0 to t0, then just blindly plugging in -t0 and t0 will not result in the correct evaluation of the line integral, since the angles used in the parameterization of the ellipse are supposed to run CCW. If you use symmetry in evaluating the line integral, I believe you will obtain the correct result, but because C2 crosses the x-axis when going from -t0 to t0, I think you must handle the line integral evaluation very carefully so that you don't get an incorrect result, which you apparently have done.

5. May 18, 2016

### andrewkirk

The problem might be that you are choosing 0.6435 as arcsin of 3/5. The angle t0 needs to be between pi/2 and pi, because of the quadrants the start and end points are in (as SK points out), so one should choose the arcsin instead as pi-0.6435. If you use that instead, I think you'll get the right result.

Last edited: May 19, 2016
6. May 19, 2016

### TheBoro76

Thanks for your responses I tried both of your suggestions. but couldn't quite get them to work (I probably made a mistake somewhere). However, if I used the value t0=arccos(-4/5), I got the answer I expected, So I'm going with that.
Thanks again!!

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