Order of Matrix Multiplication

**bugatti79** · October 17th 2012, 10:24 AM

Folks,

What is the order of matrix multiplication for the following problem

$[K^e]=[T^e]^T[\overline K^e][T^e]$

where $[\overline K^e]$ and $[T^e]$ are 4*4 matrices.

I would like to know the order before I tackle the problem..

Thanks

**johnsomeone** · October 17th 2012, 12:34 PM

Matrix multiplication is associative - it doesn't make a difference "when" you multiply two matricies together (although you have to keep the order straight since it's not commutative.).

If A, B, and C are any matricies such that the dimensions make their product meaningful, then (AB)C = A(BC).
That means:
ABC can be calculated by first multiplying AB, then mulitplying the result on the right by C, or
ABC can be calculated by first multiplying BC, then mulitplying the result on the left by A,
and the result will be the same either way.
i.e. (AB)C = A(BC).

("God does not care about our mathematical difficulties. He integrates empirically." - Einstein. That's a wonderful quote!)

**bugatti79** · October 17th 2012, 01:11 PM

Originally Posted by johnsomeone

Matrix multiplication is associative - it doesn't make a difference "when" you multiply two matricies together (although you have to keep the order straight since it's not commutative.).

If A, B, and C are any matricies such that the dimensions make their product meaningful, then (AB)C = A(BC).
That means:
ABC can be calculated by first multiplying AB, then mulitplying the result on the right by C, or
ABC can be calculated by first multiplying BC, then mulitplying the result on the left by A,
and the result will be the same either way.
i.e. (AB)C = A(BC).

("God does not care about our mathematical difficulties. He integrates empirically." - Einstein. That's a wonderful quote!)

OK thanks.

I am a little puzzled how they calculated

$[K^e]=\left [ \begin{matrix} \cos^2 \theta & 0.5 \sin 2\theta & - \cos^2 \theta & -0.5 \sin 2 \theta \\ 0.5sin 2 \theta & sin^2 \theta &-0.5 \sin 2 \theta & - \sin^2 \theta\\- \cos^2 \theta &-0.5 \sin 2 \theta & \cos^2 \theta & 0.5 \sin 2 \theta \\-0.5 \sin 2 \theta & - \sin^2 \theta & 0.5 \sin 2 \theta & \sin^2 \theta \end{matrix} \right]$ given

$[\overline K^e]=\left [ \begin{matrix} 1 & 0 & -1 & 0 \\ 0 & 0 &0 & 0\\-1&0 & 1 & 0 \\0 & 0 & 0 & 0 \end{matrix} \right]$ and $[T^e]=\left [ \begin{matrix} \cos \theta & \sin \theta & 0 & 0 \\ -\sin \theta & \cos \theta &0 & 0\\ 0 &0 & \cos \theta & \sin \theta \\0 & 0 & - \sin \theta & \cos \theta \end{matrix} \right]$

because if you look at the second row for $[\overline K^e]$ , they are all 0's so how can there be non zero terms in $K^e$ ...?

**johnsomeone** · October 17th 2012, 02:24 PM

I just now multiplied the matricies out, and got the result you just posted. $[K^e]=[T^e]^T[\overline K^e][T^e]$ .

(The frequent appearance there of $0.5\sin(2\theta)$ is just another way to write $\cos(\theta)\sin(\theta)$ .)

**bugatti79** · October 18th 2012, 05:00 AM

Originally Posted by bugatti79

OK thanks.

I am a little puzzled how they calculated

$[K^e]=\left [ \begin{matrix} \cos^2 \theta & 0.5 \sin 2\theta & - \cos^2 \theta & -0.5 \sin 2 \theta \\ 0.5sin 2 \theta & sin^2 \theta &-0.5 \sin 2 \theta & - \sin^2 \theta\\- \cos^2 \theta &-0.5 \sin 2 \theta & \cos^2 \theta & 0.5 \sin 2 \theta \\-0.5 \sin 2 \theta & - \sin^2 \theta & 0.5 \sin 2 \theta & \sin^2 \theta \end{matrix} \right]$ given

$[\overline K^e]=\left [ \begin{matrix} 1 & 0 & -1 & 0 \\ 0 & 0 &0 & 0\\-1&0 & 1 & 0 \\0 & 0 & 0 & 0 \end{matrix} \right]$ and $[T^e]=\left [ \begin{matrix} \cos \theta & \sin \theta & 0 & 0 \\ -\sin \theta & \cos \theta &0 & 0\\ 0 &0 & \cos \theta & \sin \theta \\0 & 0 & - \sin \theta & \cos \theta \end{matrix} \right]$

because if you look at the second row for $[\overline K^e]$ , they are all 0's so how can there be non zero terms in $K^e$ ...?

Starting in steps I calculate $[\overline K^e] [T^e]=\left [ \begin{matrix} \cos \theta & \sin \theta & -\cos \theta & -\sin \theta \\ 0 & 0 &0 & 0\\- \cos \theta &\sin \theta & \cos \theta & -\sin \theta \\0 & 0 & 0 & 0 \end{matrix} \right]$ ....?

**johnsomeone** · October 18th 2012, 07:04 AM

Rather than my trying to explain anything, let me suggest that you just work it out.

Multiply that on the left by $[T^e]^t$ . All will become clear.

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