http://upload.wikimedia.org/math/7/b...5aea486b5c.png

Ok, I understand how to get it this far. But how are they getting this?

http://upload.wikimedia.org/math/d/3...b2f3136bae.png

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- Feb 13th 2007, 09:13 AMGruntThe coefficients-vectors method
http://upload.wikimedia.org/math/7/b...5aea486b5c.png

Ok, I understand how to get it this far. But how are they getting this?

http://upload.wikimedia.org/math/d/3...b2f3136bae.png - Feb 13th 2007, 09:20 AMRichard Rahl
Hello there!

I'm not sure what you are trying to do, multiplication or dot product (the single dot usually indicates dot product).

But the matrix below is the result of multiplying those two matricies. Is that what you are having a problem with? - Feb 13th 2007, 09:20 AMGrunt
nm, i just needed to sit back and look at it more.

- Feb 13th 2007, 11:29 AMThePerfectHacker
It makes no difference which method you use to multiply matrices, whichever one you understand the best. The important thing is that you know how to multiply them. The one I use is the standard dot product between the rows and coloums.

- Feb 18th 2007, 01:16 PMSoroban
Hello, Grunt!

Someone taught you a**very**messy way of multiplying matrices.

. . Shame on them!

ThePerfectHacker referred to a "standard dot product".

. . I thought*everyone*used it.

Let's multiply "a row times a column".Code:`| 4 |`

[ 1 2 3 ] · | 5 | = (1)(4) + (2)(5) + (3)(6)

| 6 |

= 4 + 10 + 18 = 32

Your problem looks like this:Code:`| 3 1 |`

| 1 0 2 | | |

| | · | 2 1 |

| -1 3 1 | | |

| 1 0 |

. . by each column in the second matrix.

Here we go . . .

Row-1 times Column-1Code:`| 3 |`

| 1 0 2 | · | 2 | = (1)(3) + (0)(2) + (2)(1) = 5

| 1 |

In Row-1, Column-1 of the product matrix (upper left).

Row-1 times Column-2Code:`| 1 |`

| 1 0 2 | · | 1 | = (1)(1) + (0)(1) + (2)(0) = 1

| 0 |

Row-2 times Column-1Code:`| 3 |`

|-1 3 1 | · | 2 | = (-1)(3) + (3)(2) + (1)(1) = 4

| 1 |

Row-2 times Column-2Code:`| 1 |`

|-1 3 1 | · | 1 | = (-1)(1) + (3)(1) + (1)(0) = 2

| 0 |

Answer:Code:`| 5 1 |`

| 4 2 |