Question

Suppose the first two columns, \({{\bf{b}}_1}\) and \({{\bf{b}}_2}\), of Bare equal. What can you say about the columns of AB(if ABis defined)? Why?

Short answer

If thefirst two columns,\({{\bf{b}}_1}\)and\({{\bf{b}}_2}\), of Bare equal,the first two columns of AB are also equal.

Step-by-step solution

Definition of matrix multiplication

Consider two matrices P and Q of order\(m \times n\)and\(n \times p\), respectively. The order of the product PQ matrix is\(m \times p\).

Let\({{\bf{q}}_1},{{\bf{q}}_2},...,{{\bf{q}}_n}\)be the columns of the matrix Q. Then, the product PQ is obtained as shown below:

\(\begin{aligned}{c}PQ = P\left( {\begin{aligned}{*{20}{c}}{{{\bf{q}}_1}}&{{{\bf{q}}_2}}& \cdots &{{{\bf{q}}_n}}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{P{{\bf{q}}_1}}&{P{{\bf{q}}_2}}& \cdots &{P{{\bf{q}}_n}}\end{aligned}} \right)\end{aligned}\)

The matrix product AB

It is given that thefirst two columns,\({{\bf{b}}_1}\)and\({{\bf{b}}_2}\), of Bare equal. So, the matrix Bcan be represented as:

\(B = \left( {\begin{aligned}{*{20}{c}}{{{\bf{b}}_1}}&{{{\bf{b}}_1}}&{{{\bf{b}}_3}}& \cdots &{{{\bf{b}}_p}}\end{aligned}} \right)\)

Obtain the product AB as shown below:

\(\begin{aligned}{c}AB = A\left( {\begin{aligned}{*{20}{c}}{{{\bf{b}}_1}}&{{{\bf{b}}_1}}&{{{\bf{b}}_3}}& \cdots &{{{\bf{b}}_p}}\end{aligned}} \right)\\ = \left( {A\begin{aligned}{*{20}{c}}{{{\bf{b}}_1}}&{A{{\bf{b}}_1}}&{A{{\bf{b}}_3}}& \cdots &{A{{\bf{b}}_p}}\end{aligned}} \right)\end{aligned}\)

It is observed that the first two columns of AB, that is,\(A{{\bf{b}}_1}\)and \(A{{\bf{b}}_2}\) are equal.

Thus, if thefirst two columns,\({{\bf{b}}_1}\)and\({{\bf{b}}_2}\), of Bare equal,the first two columns of AB are also equal.