Revision:Matrices - The Student Room
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Revision:Matrices

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Matrices are tables of numbers. The numbers are put inside big brackets. Matrices are given 'orders', which basically describe the size of the matrices. The order is the number of rows 'by' the number of columns. So a 2 by 3 matrix has 2 rows and 3 columns.

Contents

Adding and Subtracting

Adding and subtracting matrices is fairly straight-forward. Adding and subtracting of matrices, however, can only occur if the matrices have the same order.

For example:

\left( \begin{array}{ccc}1&2&4\\11&3&10 \end{array}\right) + \left(\begin{array}{ccc}1&1&1\\1&1&1\end{array}\right)

is defined as, and equal to:

\left( \begin{array}{ccc}1+1&2+1&4+1\\11+1&3+1&10+1\end{array}\right) = \left( \begin{array}{ccc}2&3&5\\12&4&11\end{array}\right)

Multiplying

Matrix multiplication is non-commutative. This means that for matrices A, B:

AB != BA

ie. it matters which way around you multiply. This should be obvious enough from the method, as should the other rule - which is that the second matrix must have as many rows as the first has columns.

The method is as follows, for a 2x2 matrix, the top left term in the first matrix is multiplied by the top left of the second, and added to the top right of the first multiplied by the bottom left term in the second to give the top left term in resultant matrix. This pattern of 'along the first; down the second' is followed for all terms until a complete result matrix, C, is achieved - and works for any order of matrix.

So, for example:

\left( \begin{array}{ccc}1&2\\3&4 \end{array}\right)  \left(\begin{array}{ccc}5&6\\7&8\end{array}\right) = \left( \begin{array}{ccc}1*5+2*7&1*6+2*8\\3*5+4*7&3*6+4*8 \end{array}\right) = \left(\begin{array}{ccc}19&22\\43&50\end{array}\right)

Dividing, Inversion, and Determinant

Conventional division does not work with matrices, just like multiplication's also a bit different. It's still logical though, once some thought is applied:

So say we have a question, for matrices A,B,X - find X where AX = B. Of course what we want to do is to divide B by A, that's basic algebra that you've been doing since Y6 SATs. However there is of course a twist.

We can't just divide each term (just like we couldn't multiply each term), so how do we divide two things, with only additive and multiplicative rules? Well, we can multiply each side* by what we want to divide raised to the power -1! (*The other side, A, when multiplied by A^-1 gives the Identity matrix I which by definition leaves just X when multiplied by it - just like the multiplicative identity of conventional number system 1 doesn't alter any number when multiplied)

This is known as inverting the matrix, and written A^-1. The inverse is found by multiplying the matrix (with the leading diagonal numbers swapped, and non-leading diagonals changed to the negative of themselves), by 1/determinant. Where the determinant is the product of numbers in leading diagonal minus the product of the others.

ie:

For A = \left(\begin{array}{ccc}a&b\\c&d\end{array}\right), A^-1 = 1/(ad-bc) \left(\begin{array}{ccc}d&-b\\-c&a\end{array}\right)

Transposing

Transposing matrices, written A^T or A' for matrix A, can be pictured as a 'flip' about an axis running from top-left to bottom-right. A more proper definition would be that the nth row becomes the nth column, and vice versa.

An example may well be more clear:

For A = \left(\begin{array}{ccc}1\\2\end{array}\right), A^T =  \left(\begin{array}{ccc}1&2\\\end{array}\right)

For B = \left(\begin{array}{ccc}1&2\\3&4\end{array}\right), B^T =  \left(\begin{array}{ccc}1&3\\2&4\end{array}\right)

Comments

This article does not currently say much about anything.

Added example to multiplication, added basis for transposing, inverses, and division - work still needs to be done. //FO12DY

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