# Matrix Multiplication

• Feb 3rd 2010, 01:45 PM
millerst
Matrix Multiplication
Let A = (aij) be an m x n matrix. Let r and s be between or equal to 1 to m. Let Irs be the matrix whose rs-composition is 1 and such that all other components are equal to 0.

(a) What is IrsA.

I'm totally confused, can someone explain where to begin.
• Feb 3rd 2010, 11:58 PM
Hello millerst
Quote:

Originally Posted by millerst
Let A = (aij) be an m x n matrix. Let r and s be between or equal to 1 to m. Let Irs be the matrix whose rs-composition is 1 and such that all other components are equal to 0.

(a) What is IrsA.

I'm totally confused, can someone explain where to begin.

I don't know what this bit means.
Quote:

...Let Irs be the matrix whose rs-composition is 1...
Have you been given an explanation of this? It sounds as if $\displaystyle I_{rs}$ is a binary matrix (all its elements are either $\displaystyle 0$ or $\displaystyle 1$) but more than that, I can't say.

All I can say for sure so far, is that, if $\displaystyle I_{rs}A$ is meaningful, then the number of columns in $\displaystyle I_{rs}$ is the same as the number of rows in $\displaystyle A$. So $\displaystyle I_{rs}$ is a $\displaystyle l\times m$ matrix, for some value of $\displaystyle l$, and the order of the resulting product will be $\displaystyle l \times n$.

Can anyone else shed any light on this?

Let $\displaystyle A = (a_{ij})$ be an m x n matrix. Let r and s be between or equal to 1 to m. Let $\displaystyle I_{rs}$ be the m×m matrix whose (r,s)-component is 1 and such that all other components are equal to 0.
(a) What is $\displaystyle I_{rs}A$?
First, as Grandad says, the number of columns of $\displaystyle I_{rs}$ must be the same as the number of rows of A (otherwise the product $\displaystyle I_{rs}A$ is not defined). Second, I'm guessing that $\displaystyle I_{rs}$ is meant to be a square matrix, so that the product $\displaystyle I_{rs}A$ will have the same number of rows and columns as A.
If the only nonzero component of $\displaystyle I_{rs}$ is a 1 in row r and column s, then you should be able to see that the only nonzero components of $\displaystyle I_{rs}A$ will be in row r. What's more, the element in the (r,j)-position in $\displaystyle I_{rs}A$ will be the (s,j)-component of A. So row r of $\displaystyle I_{rs}A$ will be the same as row s of A, and all the other rows of $\displaystyle I_{rs}A$ will consist of zeros.