Question

Consider two matrices A and B whose product AB is defined. Describe the $i_{th}$row of the product AB in terms of the rows A and the matrix B.

Short answer

The $i_{th}$row of AB is the product of the $i_{th}$row of A and the matrix B .

Step-by-step solution

Concept Introduction

Let us assume that B be $n\times p$matrix and A be a matrix $p\times m$. The $i jth$entry of BA is the dot product of the $ijth$row of B the with the $i jth$column of A.

$BA=\left[\begin{array}{cccc}{b}_{11}& b_{12}& \cdots & b_{1p}\\ b_{21}& b_{22}& \cdots & b_{2p}\\ ⋮& ⋮& & ⋮\\ b_{ij}& b_{i2}& \cdots & b_{ip}\\ ⋮& ⋮& \cdots & ⋮\\ b_{n1}& b_{n2}& \cdots & b_{np}\end{array}\right]\left[\begin{array}{cccccc}{a}_{11}& a_{12}& \cdots & a_{ij}& \cdots & a_{1m}\\ a_{21}& a_{22}& \cdots & a_{2j}& \cdots & a_{2m}\\ ⋮& ⋮& & ⋮& & ⋮\\ a_{p1}& a_{p2}& \cdots & a_{pj}& \cdots & a_{pm}\end{array}\right]$

Is the $n\times m$matrix whose $i jth$entry is

$b_{i1}{a}_{1j}+b_{i2}{a}_{2j}+.....+b_{ip}{a}_{pj}=\sum _{k=1}^P{b}_{ik}{a}_{kj}$

Product in terms of the rows and the matrix

Let A be an $n\times p$matrix and B be a $p\times m$matrix. Then product AB is defined. Let $\overrightarrow{r_i}$be the $i-throw$of the matrix A , for $i=1,......n.$Then we can write A in terms of its rows:

$A=\left[\begin{array}{c}\overrightarrow{r_1}\\ .\\ .\\ .\\ \overrightarrow{r_1}\\ .\\ .\\ .\\ \overrightarrow{r_n}\end{array}\right]$

Notice that each $\overrightarrow{r_i}$is a $1\times p$matrix. That means we can partition A into n many $1\times p$block matrices. Now using Theorem 2.3.9 on page 32, we have

$A=\left[\begin{array}{c}\overrightarrow{r_1}\\ .\\ .\\ .\\ \overrightarrow{r_1}\\ .\\ .\\ .\\ \overrightarrow{r_n}\end{array}\right],B=\left[\begin{array}{c}\overrightarrow{r_1}B\\ .\\ .\\ .\\ \overrightarrow{r_{1}B}\\ .\\ .\\ .\\ \overrightarrow{r_n}B\end{array}\right](aproductofpartitionedmatrices)$

Therefore, the $ith$row of AB is the product of the $ith$row of A and the matrix B.