matrix multiplication proof

**duoc12** · April 8th 2010, 11:35 PM

can anyone please direct me to or give me a proof of why matrices are multiply the way they do with the rows of the first matrix multiply by the corresponding columns of the second matrix?

**Sudharaka** · April 9th 2010, 12:12 AM

Originally Posted by duoc12

can anyone please direct me to or give me a proof of why matrices are multiply the way they do with the rows of the first matrix multiply by the corresponding columns of the second matrix?

Dear duoc,

Matrix multiplication is defined that way. Therefore there is no proof to why matrices are multiplied the way they do. If you need to clarify the definition of matrix multiplication please refer, Multiplication of Matrices

Hope this will help you.

**Prove It** · April 9th 2010, 01:15 AM

In addition, the REASON matrix multiplication has been defined that way is to create a shorthand method of writing systems of equations and shorthand methods of solving them.

I.e.

$a_{11}x_1 + a_{12}x_2 + a_{13}x_3 + \dots + a_{1n}x_n = b_1$

$a_{21}x_1 + a_{22}x_2 + a_{23}x_3 + \dots + a_{2n}x_n = b_2$

$\vdots$

$a_{n1}x_1 + a_{n2}x_2 + a_{n3}x_3 + \dots + a_{nn}x_n = b_n$

Can be written as

$\left[\begin{matrix}a_{11} & a_{12} & \dots & a_{1n} \\ a_{21} & a_{22} & \dots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \dots & a_{nn}\end{matrix}\right]\left[\begin{matrix}x_1 \\ x_2 \\ \vdots \\ x_n\end{matrix}\right] = \left[\begin{matrix}b_1 \\ b_2 \\ \vdots \\ b_n \end{matrix}\right]$

Or in even shorterhand...

$A\mathbf{x} = \mathbf{b}$ ...

And then by using some matrix algebra

$A^{-1}A\mathbf{x} = A^{-1}\mathbf{b}$

$I\mathbf{x} = A^{-1}\mathbf{b}$

$\mathbf{x} = A^{-1}\mathbf{b}$ .

If $A^{-1}$ can be calculated, that means that you can easily find $\mathbf{x}$ .

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