thanks!

sometimes i find it difficult to do back the maths and physics question that i've left in incompleted... (hard to think and recall the concept)

Beware! That's a typical LAZY student syndrome. All fundamental principles are stated in your textbooks. What were the reasons you took Math and Physics in STPM? Do you really want to become a Mathematician or a Physicist?

This post has been edited by Critical_Fallacy: Dec 28 2013, 01:42 AM

Refer to post #470 by Critical_Fallacy. The method to compute a 4X4 matrix is depicted very clearly.

Nevertheless, the formula you stated here is correct.

Yes, these are all rather advanced for a 1st year university student but given time, you will get used to it.

Hi Flame Haze, RED-HAIR-SHANKS, maximR, crazywing26 and delsoo,

Most chemical equations in STPM/A-level are sufficiently simple that they could have been balanced by trial and error, but for more complicated chemical equations we will need a systematic method. There are various methods that can be used, but I will give one that uses Matrix theory.

To balance a chemical equation, we let x1, x2, x3, and x4 be positive integers that balance the equation



Equating the number of atoms of each type on the two sides yields

Hydrogen (H) :: 1 x1 = 3 x3
Chloride (Cl) :: 1 x1 = 1 x4
Sodium (Na) :: 3 x2 = 1 x4
Phosphorus (P) :: 1 x2 = 1 x3
Oxygen (O) :: 4 x2 = 4 x3

from which we obtain the homogeneous linear system



The augmented matrix for this system is



I leave it for you to show that the reduced row echelon form of the augmented matrix for this system is



from which we conclude that the general solution of the system is

x1 = t, x2 = t/3, x3 = t/3, x4 = t

where t is arbitrary. To obtain the smallest positive integers that balance the equation, we let t = 3, in which case we obtain x1 = 3, x2 = 1, x3 = 1, and x4 = 3. Substituting these values in the original problem produces the balanced equation

I'm simply amazed by the application of matrices in solving Chemistry-related question like this one, thanks for it. I'm going to note this down for sure, in case I'm going to need to use this method in the near future. But, knowing that the time given for answering a question in STPM would be really scarce, do you reckon that we're able to balance the said equation under the short amount of time using Gauss-Jordan Elimination? But, I guess with given appropriate space and time to practice, we can really master it.

Why the total charge of two spheres of.two.isolated conducting spheres with different radii AnD charges but connected by a wire is.the same?

I've applied the Gauss-Jordan Elimination to convert it into reduced row-echelon form. Sorry for my messy workings. Am I correct? If there is any other simpler workings, please post it up, thanks.

Finally . Thank you ! When Susskind mentioned 'bright' high school student , he did really mean 'bright' because this is still beyond me .

Hi Flame Haze, RED-HAIR-SHANKS, maximR, crazywing26 and ystiang,

Another important application of Linear Algebra & Matrix Theory is to find a polynomial whose graph passes through a specified set of points in the plane. And this process is called polynomial interpolation. Now let us consider the more general problem of finding a polynomial whose graph passes through n points with distinct x-coordinates



THEOREM :: Given any n points in the xy-plane that have distinct x-coordinates, there is a unique polynomial of degree n − 1 or less whose graph passes through those points.

Let us now consider how we might go about finding the interpolating polynomial whose graph passes through the n points. According to the THEOREM, the graph of this polynomial is the graph of the equation



It follows that the coordinates of the points must satisfy



In these equations the values of x’s and y’s are assumed to be known, so we can view this as a linear system in the unknowns . From this point of view the augmented matrix for the system is



and hence the interpolating polynomial can be found either by reducing this matrix to reduced row echelon form (Gauss–Jordan elimination), or by solving the linear system using Cramer’s Rule or .

P.S. MATRIX is not designed to be solved by hand, but very efficient in a numerical computing environment, which allows matrix manipulations, plotting of functions, and implementation of algorithms. Learn how to solve Linear Equations with MINVERSE and MMULT functions in Microsoft Excel.

Hi Flame Haze, RED-HAIR-SHANKS, maximR, crazywing26 and ystiang,

EXAMPLE :: Find a cubic polynomial whose graph passes through the points



Solution :: Since there are four points, we will use an interpolating polynomial of degree n = 3. Denote this polynomial by



and denote the x- and y-coordinates of the given points by

x1 = 1, x2 = 2, x3 = 3, x4 = 4 ... and ... y1 = 3, y2 = −2, y3 = −5, y4 = 0

Thus, the augmented matrix for the linear system in the unknowns a0, a1, a2, and a3 is



I leave it for you to solve by hand (if you are interested), and here is the computation using MS Excel:



from which it follows that a0 = 4, a1 = 3, a2 = −5, a3 = 1. Thus, the interpolating polynomial is



The graph of this interpolating polynomial is shown below:

Sorry but I've forgotten every single thing about matrixes , A-level maths doesn't seem to deal with them, and I hope I won't be encountering them in Further Maths. Will take some time to catch up, busy spring cleaning the house for CNY , and new semester is around the corner.

House cleaning is important and you are an amazing character!

I suppose that you will encounter Matrices and Transformations in AS MODULE – Further Pure 1, and Matrix Algebra and Solution of Linear Equations in A2 MODULE - Further Pure 4.

come someone explain this? What is the use for this?

Hi danny88888,

Some textbooks fail to explain the essential applications of Matrix Algebra. To find out the important application of the rank of a matrix, please read the confession of a Control & Instrument Engineer:

To a Systems Designer/Engineer, whether they are Mechanical Systems & Electrical Systems, or Fluid Systems & Thermal Systems, the concepts of controllability and observability play an important role in the design of control systems in state space (Matrix form):



where

x = n-state vector
u = r-control input vector
y = m-output vector
A = n × n state matrix
B = n × r control input matrix
C = m × n output matrix

A system is said to be controllable, if an input u to a system can be found that takes every state variable from a desired initial state to a desired final state ; otherwise the system is said to be uncontrollable. In other words, an nth-order plant is completely controllable if the matrix



is of rank n, where P is called the controllability matrix.

A system is said to be observable, if it is possible to determine the initial-state vector from the observation of the output y over a finite interval of time from t0; otherwise the system is said to be unobservable. Thus, an nth-order plant is completely observable if the matrix



is of rank n, where Q is called the observability matrix.

Remark: The conditions of controllability and observability may govern the existence of a complete solution to the control system design problem. The solution to this problem may not exist if the system considered is not controllable. Although most physical systems are controllable and observable, corresponding mathematical models may not possess the property of controllability and observability.Then it is necessary to know the conditions under which a system is controllable and observable.

Quick question once again:



First part was cleared. The solution of the second part was blunt and depressing.



What logic is this? Please, enlighten me.

In the meantime I'm off to study Abelian groups etc etc.

ZERO,

This is purely algebraic manipulations. It is pretty obvious that



Next, we divide both sides by y²:



Then, we denote roots of the cubic function as



Now we compare Eq.(1) with the properties of Sum and Product of Cubic Roots:



From Eq.(3), we can detrmine



With the results from (4) and (5), we can solve



This post has been edited by Critical_Fallacy: Dec 29 2013, 05:46 PM

I understood every part of your explanation except this. Why would Sn be equivalent to the original roots? To my understanding, when the sum of two numbers is raised to any power, they give rise to intermediate terms like those in binomial theorem. Correct me if I'm wrong.

ZERO,

It was originated from here:



Therefore, when we add up the three equations below:



This post has been edited by Critical_Fallacy: Dec 29 2013, 06:51 PM

Critical_Fallacy, I do wonder how can you think in such brilliant ways of solving mathematical problems? In ways that usually people would not thought of yet effective. Is it with sufficient practise or critical thinking? Just asking only