1. ## 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.

2. Hello millerst
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.
...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.