QUESTIONS - CDOG - MATRICES

QUESTIONS - ANSWERS - CDOG - MATRICES


cdog has left a new comment on your post "MATHS - 8 - MATHS FOR ALL - PROBLEM SOLVING TECHNI...":

Solve the matrix equation {[3,a][c,d]}={[b,a-c][b+d,4a]} same thing as Matrix 3 a = Matrix b a-c c d b+d 4a Thanks again!

LHS =
3,A
C,D
RHS =
B,A-C
B+D,4A
EQUATING CORRESPONDING ELEMENTS..
B=3……………….1
A-C=A……C=0………………2
B+D=C=0…..FROM EQN.2…HENCE
B+D=….FROM. EQN.1.
3+D=0
D=-3…………………………………3
4A=D = -3 FROM EQN.3.
A = -3/4……………………………..4
HENCE
A=-3/4
B=3
C=0
D=-3

QUESTIONS - CDOG - MATRICES

QUESTIONS - ANSWERS - CDOG - MATRICES

cdog has left a new comment on your post "MATHS - 8 - MATHS FOR ALL - PROBLEM SOLVING TECHNI...":

Hello! I need help with this basic linear algebra question. Find the condition the b's must satisfy for the system to be CONSISTENT. x1+4x2-3x3=b1 2x1+6x2+5x3=b2 x1+6x2-14x3=b3 Thank you! Cdog
DETERMINANT OF COEFFICIENT MATRIX =C =
1 4 -3
2 6 5
1 6 -14
R2=R2-2*R1; R3=R3-R1
1 4 -3
0 -2 11
0 2 -11
R3=R3+R2
1 4 -3
0 -2 11
0 0 0
DETERMINANT = 0
HENCE THE EQNS. ARE LINEARLY DEPENDENT.
HENCE THE CONSTANTS SHALL BE IN THE SAME LINEAR COMBINATION AS THE
COEFFICIENTS ARE.
FROM ABOVE TRANSFORMATIONS , WE NOTE THAT …USING E=EQN.1…ETC
X1+6X2-14X3 = 3(X1+4X2-3X3)-(2X1+6X1+5X3)….HENCE
E3=3*E1-E2
HENCE B3=3*B1-B2 IS THE RELATION NEEDED BETWEEN B1,B2,B3.

MATHS - 8 - MATHS FOR ALL - PROBLEM SOLVING TECHNIQUES

MATHS - 8 - MATHS FOR ALL - PROBLEM SOLVING TECHNIQUES
Let us make our start in this programme with “Problem Solving Methods”. At the outset two points are to be mentioned . First point is that that they are general methods that could be used any where , not only in maths but in all situations in life. Our power of imagination and application is the only limitation for their adoption in arts or science or in life. The second point is that the word “problem” as is commonly used and understood, need not necessarily mean a difficulty. It is used here to denote a task or job or work or even a project to be carried out. To illustrate the first point ,as we go along ,we shall give few examples to show how they could be applied to solution of problems beyond mathematics.

This journey as indicated by the title appears limit less….there are no boundaries drawn…. So let us mark our boundaries, a start line and a finish line or a goal to reach. Let it be maths from high school to under-graduate or intermediate level. Now we are ready to make our journey.
The first rule as you have noted above is to know where we stand and then to know where we want to go. Few tips to make this journey a happy one are …..(we all want to be happy !! Is it not ? otherwise we do not undertake such a journey!!)……..
One is to be interested in making the journey.
Next is that the path of our travel should be a happy one.
Next is to derive happiness with each step forward , as we have already traveled a distance from the start line and if we keep moving we are bound to reach the finish line .
Never look at the goal with a despair at its distance , instead be happy that you are nearer to it now.
To sum up , the GOLDEN RULE for a happy journey is -

The travel along the path itself should make us happy and each step or progress along the path, however small it may be , should make us more happy as we are nearer to our goal by that extent.

Methods No.1,2 & 3 :
1.Start from what is given (Hypothesis) and go towards what is to be proved .(Conclusion) and reach there.
2. Start from what is to be proved always mentioning “To prove that”(TPT) and travel backwards towards what is given and when you get there the problem is solved ,because ,the point you reached to prove is already given to us as a starting point.
3. Start from what is given as in 1. and go towards what is to be proved ..you may find that you are not able to go beyond a point …stop there ..now go to what is to be proved and travel back wards as in 2, always mentioning TPT till you reach the point where you broke off in the forward journey…If you reach there , then also the problem stands solved as that point is already obtained from what is given to us .

A GOOD PRACTICE WHILE USING THESE METHODS IS TO WRITE CLEARLY WHAT IS TO BE PROVED AT A PROMINENT PLACE, LIKE THE UPPER RIGHT HAND CORNER SO THAT YOU KNOW WHERE TO GO AND WHEN TO STOP.

All these methods you have used many times in algebra ,trigonometry etc..,to prove certain equations where one of the sides of the equation usually the left hand side (LHS) is taken as the hypothesis and the right hand side (RHS) as the conclusion.
In the above context , it is to be emphasized ,that in memorizing various formulae ,it is as important to know the LHS as the RHS and also the conditions or basic assumptions made in deriving a formula. Very often we find that these are missed in practice by several students.
In real life situations , this could be interpreted as …if we have to meet then …I can start from my place and travel to your place or you can start from your place and travel to my place or both of us can start from our respective places and travel to a common place of meeting .
This is all very nice to say but , how do we travel the path you may ask ? Please note that , now , we are only outlining the different methods available to choose to solve a problem , The path in algebra or trigonometry or any particular topic will be taken up at appropriate time as we make our journey. However during this part also ,we shall indicate the detailed path in a few specific examples wherever needed to illustrate a point , as you shall see below .
Another difficulty that may arise in adopting these 3 methods is that some times ,it is not clear what the starting point is , or more so , where to stop.
Problem in “locus” is one such case. Many are not clear about this term and more so.. what is meant by or what is to be done to find a locus. So let us try to understand this term.
Locus is the path traced by a point which moves as per a given condition. A simple example is the locus of a point which moves such that its distance from a fixed point is always constant… This problem can come up in geometry or in coordinate geometry.
In geometry ,we are expected to describe the path traveled as a geometric figure, which in this case is a circle with the fixed point as center and the constant distance as the radius.
In coordinate geometry , the difficulty is more felt, both in understanding what to do and to know when to stop. As you know in coordinate geometry , we use x and y coordinates to denote a point and an equation connecting x and y like x + y = 5 to determine the path traced by the point. We trace the path by taking various values for x and finding the corresponding values of y from the equation and plotting them on a graph. So to find the locus ,we shall start by assuming any point P with coordinates (x,y) as a point on the locus. Now we apply the given condition to connect and find a link between x and y .The moment we get such a link as an equation ,that is called the locus and the problem is over .In this case to apply the given condition let us take the fixed point C with coordinates (h,k) as the center and the fixed distance radius as r .
So the given condition means CP = r . or

This represents an equation linking x and y and so we call this as the locus and stop here .That is, the problem is over.

Method No.4 :

REDCUTIO - AD – ABSURDUM : This is a method of proving a theorem by contradiction. We start by assuming that the conclusion (to be proved) is wrong. Then we end up with a contradiction to a known truth or the hypothesis (given data). Hence we conclude that our assumption is wrong….that is the theorem is correct.
This is a very powerful method with a wide variety of applications. You are aware of this method in solving theorems in geometry and in proving irrational numbers etc. Now let us see its application to a seemingly tough and not usually heard of problem but well within our scope.
You know a rational number as that which can be expressed as p / q where p and q are integers. You also know that square root of 2 is about 1.414….etc and it is an irrational number because it can not be expressed as p / q where p and q are integers. Now the question is can an irrational number raised to an irrational number power be a rational number ? or the theorem to be proved is that an irrational number raised to an irrational number power can be a rational number .
At the out set it looks baffling as we do not even know how to raise an irrational number to an irrational number without using logarithms or a calculator.
But let us remember that we have a powerful tool in REDCUTIO - AD – ABSURDUM . Let us put it to use .
Let us assume that the theorem is wrong .That is an irrational number raised to an irrational number power can not be a rational number .
We know square root of 2 is an irrational number. Let us raise it to again square root of 2 power
= {SQRT(2)}^{SQRT(2)}
As per our assumption it is an irrational number. Now let us raise this irrational number to square root of 2 power
=[{SQRT(2)}^{SQRT(2)}]^{SQRT(2)}={SQRT(2)}^2=2
which should once again be an irrational number as per our assumption.
Look now what we get …
We get 2 as the answer .
It is a perfectly rational number….even an integer….so our assumption is wrong . So the theorem that an irrational number raised to an irrational number power can be a rational number is correct. Got It ? How does it look ?
Now , one question about your understanding of this proof .What if that our first step , that is , square root of 2 raised to square root of 2 power is an irrational number is not correct ?
Well my Dear Watson as Sherlock Homes used to say , then there is no problem …the theorem is already proved at the first stage itself !!!!.
ses.
Well, there are many more methods like induction, analogy, generalisation, pattern search , iteration ,recurrence etc which we shall deal with next .

QUESTIONS - SHERILL - INTEGRATION

QUESTIONS - ANSWERS - SHERILL - INTEGRATION


sherill has left a new comment on your post "QUESTIONS - CDOG - VECTORS":

Please find the area between the two curves f(x)=1/(x+2) and g(x)=e^x over the interval [0,1]

AREA = INTEGRAL 0 TO 1 [ E^X - {1/(X+2)}]=[E^X - LN(X+2)] FROM 0 TO 1
=E-LN(3)-1+LN(2)=E-1-LN(1.5)

Posted by sherill to Maths,management,counselling at 6:38 AM

QUESTIONS - CDOG - VECTORS

QUESTIONS - ANSWERS - CDOG - VECTORS
cdog has left a new comment on your post "QUESTIONS - CDOG - VECTORS

Let u=(-1,1,0,2). Determine whether u is orthogonal to the subspace spanned by the vectors w1=(0,0,0,0), w2=(1,-1,3,0), w3=(4,0,9,2).
WE HAVE TO SHOW THAT
U.[AW1+BW2+CW3]=0…. FOR ANY VALUE OF A,B AND C
IT WILL BE SO IF
U.W1=U.W2=U.W3=0
HERE WE HAVE
U.W1=-1*0+1*0+0*0+2*0=0…OK
U.W2=-1*1+1*-1+0*3+2*0=-2…..NOT ZERO HENCE NOT ORTHOGONAL

HENCE U IS NOT ORTHOGONAL TO THE SUBSPACE SPANNED BY …W1,W2,W3.

QUESTIONS - CDOG -VECTORS

QUESTIONS - ANSWERS - CDOG - VECTORS
cdog has left a new comment on your post "QUESTIONS - CDOG - VECTORS":

Hello, again. Can you please demonstrate how to solve the following two problems? Thanks. 1)Do there exist scalars k,l
HOPE YOU MEAN K AND L( YOU TYPED 1 OR L).PLEASE TYPE TO GIVE CLARITY AND LEAVE NO AMBIGUITY.
such that the vectors u=(2,k,6), v=(l,5,3)
IS THIS (L,5,3) OR (1,5,3)….TAKING AS (L,5,3)
and w=(1,2,3)
IS THIS (L,2,3) OR (1,2,3)?….TAKING AS (1,2,3)
are mutually orthogonal with respect to the Euclidean inner product?
HOPE YOU MEAN
U=(2,K,6)
V=(L,5,3) AND
W=(1,2,3).ASSUMING SO
U.V=2L+5K+18=0……………………….1
U.W=2+2K+18=0……K=-10………….2
V.W=L+10+9=0…………L=-19…..………3….
FROM 1 …..2L=-5K-18=50-18=32……….L=16…………….4
EQN.3 GIVES L= -19 WHERE AS EQN.4 GIVES L=16.
HENCE NOT POSSIBLE.

PLEASE TYPE CLEARLY DISTINGUISHING L AND 1.
THEN ONLY THIS COULD BE ANSWERED PROPERLY
ANY WAY THE METHOD AS GIVEN ABOVE IS SIMPLE.
JUST CHECK THE 3 EQNS. FOR A CONSISTENT SOLUTION.THEN OK.OTHERWISE NOT POSSIBLE.

QUESTIONS - CDOG - VECTORS

QUESTIONS - ANSWERS - CDOG - VECTORS

cdog has left a new comment on your post "QUESTIONS - HOPE - FUNCTIONS":

Can someone please help me with this problem? (u,v) is the inner product general notation. Let u=(u1,u2) and v=(v1,v2) Show that (u,v)=4u1v1+u2v1+u1v2+4u2v2 is an inner product on R^2 by verifying that the four inner product axioms hold. Here are the four axioms: axiom 1: (u,v)=(v,u) axiom 2: (u+v,z)=(u,z)+(v,z) axiom 3: (ku,v)=k(u,v) axiom 4: (v,v)>= 0 and (v,v)=0 if and only if v=0.

THE DEFINITION TO QUALIFY AS AN INNER PRODUCT IS SPECIFIED BY 4 AXIOMS ABOVE AND WE NEED TO SHOW THAT THEY ARE SATISFIED FOR THE GIVEN EXAMPLE.
GIVEN
U = (U1,U2)….AND….V=(V1,V2)…..AND…..
(U,V) = 4U1V1+U2V1+U1V2+4U2V2
TO SHOW THAT
(U,V) IS THE INNER PRODUCT AS DEFINED BY THE 4 AXIOMS.
1.(U,V)=(V,U)
LHS = (U,V)=4U1V1+U2V1+U1V2+4U2V2
= 4V1U1+V2U1+V1U2+4V2U2…BY COMMUTATIVE AND ASSOCIATIVE PROPERTIES OF THE SCALAR COMPONENTS OF VECTORS =(V,U)=RHS

SIMILARLY APPLYING THE DISTRIBUTIVE ,COMMUTATIVE ,ASSOCIATIVE AND SCALAR MULTIPLICATION PROPERTIES OF SCALAR COMPONENTS OF THE VECTORS WE CAN PROVE THE REST AS FOLLOWS.
2.(U+V,Z)=(U,Z)+(V,Z)
LHS = 4(U1+V1)Z1+(U2+V2)Z1+(U1+V1)Z2+4(U2+V2)Z2
=[4U1Z1+U2Z1+U1Z2+4U2Z2]+[4V1Z1+V2Z1+V1Z2+4V2Z2]
=(U,Z)+(V,Z)=RHS
3.(KU,V)=K(U,V)
LHS = 4KU1V1+KU2V1+KU1V2+4KU2V2
=K[4U1V1+U2V1+U1V2+4U2V2]=K(U,V)
4.(V,V)>=0……
(V,V)=4V1^2+V2V1+V1V2+4V2^2
=[V1^2+2V1V2+V2^2]+3[V1^2+V2^2]
=(V1+V2)^2+3(V1^2+V2^2)
SUM OF 2 PERFECT SQUARES WHICH IS ALWAYS>=0

AND…..(V,V)=0
AS GOT ABOVE IF (V,V)=0,WE SHOULD HAVE
V1+V2=0….AND…
V1^2+V2^2=0…WHICH BEING AGAIN A SUM OF 2 PERFECT SQUARES, IS ONLY POSSIBLE IF AND ONLY IF V1=V2=0….THAT IS V=(V1,V2)=(0,0)=0
AS REGARDS THE REQUIREMENT OF IF AND ONLY IF,WE SEE THAT IF
V=0…THEN V1=V2=0…THEN (V,V)=0…ETC…

QUESTIONS - ANGELA - LOGS AND EXPONENTS

QUESTIONS - ANSWERS - ANGELA - LOGS AND EXPONENTS


Angela has left a new comment on your post "QUESTIONS - HOPE - FUNCTIONS":

Hello. Do you mind telling me if I got these 2 correct? Thanks! Solve the following for x = -3 and tell what kind of function it is. 0.1^x -.001. This is an exponential function. Solve the following for x = 3 and tell why ‘e’ is an irrational number. e^x 20.0855 This is irrational because the number doesn’t terminate or repeat.



DO YOU MEAN
0.1^X=0.001…IF SO
X=3 IS CORRECT ANSWER AS
0.1^3 = 0.1*0.1*0.1=0.001
IF YOU MEANT
0.1^(-X)= 0.001…THEN
(1/10)^(-X)= 0.001
10^X = 0.001
X=-3…..SINCE 10^(-3) = 1/10^3=1/1000=0.001
YES IT IS AN EXPONENTIAL FUNCTION.

E^3 = 20.0855 IS CORRECT.
YOUR EXPLANATION ON ‘E’ IS ALSO OK.

QUESTIONS - ANGELA - LOGARITHMS

QUESTIONS - ANSWERS -ANGELA - LOGARITHMS

Angela has left a new comment on your post "QUESTIONS - POLYNOMIALS":

1) Determine the value of the following function if a = 10 and x = 100, and tell which number is the base. log a^x 2) Determine the value of the following function if a = 2 and x = 1024. log a^x 3) Express log 2 10y as the sum or difference of two logarithms. 4) The definition of a google is the number 10^100. What is the base 10 logarithm of a google? Thank you!

AS YOU KNOW IF
B^P = N …WE SAY THAT
LOGARITHM OF ‘N’ WITH RESPECT TO BASE ‘B’ IS ‘P’.
GENERALLY WE ADOPT TWO BASE VALUES TO TABULATE LOGARITHMS.THEY ARE BASE 10 AND BASE ‘E’….AN IRRATIONAL NUMBER OFTEN ENCOUNTERED IN MATHEMATICS.
THE NORMAL CONVENTION IS TO USE
LOG(X)..TO INDICATE LOGARITHMS TO BASE 10 .....
LN(X) ..TO INDICATE LOGARITHMS TO BASE ‘E’. AND
IF ANY OTHER BASE IS ADOPTED,THEN TO SPECIFY IT AS A SUFFIX AFTER LOG TO BE FOLLOWED BY THE NUMBER LATER.THAT IS WE HAVE TO WRITE LOG TO BASE B OF N.

NOW ON TO YOUR PROBLEMS.HAVE YOU TYPED THEM PROPERLY.I THINK YOU MAY BE MEANING
LOG(X) TO BASE ‘A’[‘A’ IS PUT AS A SUFFIX AFTER LOG THEN FOLLOWED BY X].ANY WAY FOR THE FIRST QUESTION ,I AM GIVING ANSWERS TO 2 VARIANTS…AS YOU TYPED AND AS I THINK IT MAY BE.FOR THE OTHER PROBLEMS , I AM GIVING THE ANSWER ASSUMING MY INTERPRETATION OF YOUR TYPING.

1.AS YOU TYPED.
HENCE GOING BY THIS CONVENTION WE GET..
LOG(A^X)=X*LOG(A)=100LOG(10)=100*1=100
SINCE LOG(10) IS UNDERSTOOD TO BE TO BASE 10
BY THE ABOVE NOMENCLATURE AND AS WE KNOW
10^1 =10 AND HENCE LOG(10) TO BASE 10 =1

AS PER MY INTERPRETATION OF YOUR TYPING
1. LOG(100) TO BASE 10 =2…SINCE
10^2 =100
2. LOG (1024) TO BASE 2 = 10……SINCE
2^10 = 1024
3. LOG (10Y) TO BASE 2 = LOG(10) TO BASE 2 + LOG(Y) TO BASE 2.

QUESTIONS - HOPE - FUNCTIONS

QUESTIONS - ANSWERS - FUNCTIONS
On 12/1/06, Hope wrote:
> Hope has left a new comment on your post "QUESTIONS- HOPE -
> INEQUALITIES,RATIO-PROPORTION":
>
> If two relations in x and y are inverses of one another, then across what
Ø line are they reflections of one another?
For the function
Y=f(x) , we consider x as independent variable and y as dependent variable.
The inverse function is derived from the same relation as above
Treating y as independent variable and x as dependent variable.
Hence they represent reflections along the line y=x
==============================================
Determine f^-1(x) for the
following function. f(x) = 10 (x – 1)
1.first let y = f(x)
let y=f(x) =10(x-1)=10x-10
2. next transpose and find x in terms of y.
10x = y+10
x = (y+10)/10
3.now the resultant equation is the inverse function…that is the
equation obtained for x on RHS is the f^(-1) function.
To put it in standard form replace y with x on RHS to get
F^(-1)[x]
hence
f^(-1) [x] = (x+10)/10
==========================================
Determine the solution to the
following composite function. (f^-1 ◦ f) (y)
to find
f^(-1)[f(y)]
let z = f(y)…….from f(x)=10x-10
z = 10y-10
f^(-1)[x]=(x+10)/10
f^(-1)[z] = (z+10)/10
but
z=10y-10…..hence
f^(-1)[f(y)]= (z+10)/10 = (10y-10+10)/10 = y

> Posted by Hope to Maths,management,counselling at 7:15 PM