The aim of today's class was to find the inverse of a matrix and then use it in order to solve simultaneous equations. We started by recalling how to find the inverse of a matrix as we had learnt it in a previous lesson.
We must first find the determinant. This is done by multiplying a by d and b by c and then subtracting the two results:

If the determinant is equal to zero, then there is no point in going on to find the inverse because there isn't one. In all other case, we then apply this formula to find the inverse:

To solve simultaneous equations with matrices by hand, this is what we have to do:
Let's take as an example the equations


We have to insert the variables into matrices, like this:

x=
b=
so that Ax=b
It is very important not to do xA=b , because the order of operations matters and it wouldn't work.
We then multiply both sides by the inverse of A:
Ax=b
Multiplying A by its inverse gives us the Identity matrix, which is like multiplying by one, so:
x=b
If we apply this to our example, the inverse is:







Finally, we can multiply b by the inverse to find the values of x and y:





so x=2 and y=2.
For the rest of the lesson we practised solving these simultaneous equations on a worksheet.

Posted by marzzi.
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This Friday we will be looking at the effect of transforming a vector using multiplication by a matrix.

You will need to download this file and extract it to a folder, then double-click on index.html.

Download file

Actually this morning I decided to convert this JavaScript to Flash. You can go here to use it.

Posted by DWe.
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When multiplying matrices, it is not as straightforward as multiplying numbers in brackets:

For example, when a 2 by 2 matrix is multiplied by another 2 by 2 matrix it will look something like:
Basically the columns of the second matrix are multiplied into the rows of the first matrix.



Extra notes:
The identity matrix times any matrix ( lets say A ) will equal to A, the identity matrix looks like:

If you want to know more on how to multiply more complicated matrices you can also visit this website:

http://www.analyzemath.com/matrices/matrices.html

Posted by erihiu.
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How do we find the determinant and inverse of a matrix?
To find the determinant of a matrix you need to multiply the opposite corners and subtract them from each other.
For Example:





To find the inverse, first the determinant must be found. Then to find the inverse the following is done:


If detA=0 then Ax=B has no solutions.

Proof:








(identity matrix)

Posted by sarger.
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If you want a 2 by 2 matrix use:

\left[\begin{array}{cc}a&b\\c&d\end{array}\right]

which produces:



You can also put expressions into the various positions of the matrix. Example:

\left[\begin{array}{cc}cos(\theta)&-sin(\theta)\\sin(\theta)&cos(\theta)\end{array}\right]

produces:



Don't forget, if you want line breaks, use <br /> at the end of a line, or wrap your text with <p> and </p>.

Posted by DWe.
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Vectors can be used to determine a particular point on a segment. Take, for an example, a segment which can be expressed in the form of y=1x+4 and place a specific point with coordinates of (2; 6) on it.

This point can be expressed as a position vector, which must be figured. This position vector can be determined by adding several other already known vectors together. The constants in the diagram below are the y-intercept (c) and the slope of the examined line. The value of “c” can also be expressed in terms of a position vector, the value of which can be written as or in the case of the particular example --> .

The observed segment can be expressed in terms of many position vectors, since a vector must have a magnitude, and this segment is endless. One can choose a specific magnitude for the vector and utilize it. In this example, it is (upper 2=x-value, bottom 2= y-value.).

So… we must add vector “c” and “vector 1” to obtain the purple vector. The formula to find the purple position vector is a= tb+c, where “t” represents the number of times the chosen magnitude is taken to obtain the desired point. In the case of this example the formula would be equivalent to: a= 1 + .

Note: If the point coordinates of the point were (4; 8), “t” would be 2.

Posted by eleova.
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Here is a sample picture of what we were working on with vectors before.

Here is a useful link to a tutorial on the simple matrix algebra we have been working on.

http://www.maths.surrey.ac.uk/interactivemaths/emmaspages/option1.html

Posted by DWe.
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Hello Class,

Macyen has the privilege of being the first person to post. At the end of his/her post, (s)he will choose the next person.

Hope this works out!

Mr. Wees

Posted by DWe.
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Hey folks,

If you were involved in the ISMTF Junior Competition, I am requesting some submissions about the event in the words of the students if at all possible. You are not required to write anything, but whatever you write has a chance to end up in the school yearbook!

Mr. Wees

Posted by DWe.
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