Suppose the second column of Bis all zeros. What can you

say about the second column of AB?

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:

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

It is given thatthe second column of Bis all zeros. So,\({{\bf{b}}_2} = {\bf{0}}\), and the matrix Bcan be represented as follows:

\(B = \left( {\begin{aligned}{*{20}{c}}{{{\bf{b}}_1}}&{\bf{0}}&{{{\bf{b}}_3}}& \cdots &{{{\bf{b}}_p}}\end{aligned}} \right)\)

Obtain the product AB as shown below:

\(\begin{aligned}{c}AB = A\left( {\begin{aligned}{*{20}{c}}{{{\bf{b}}_1}}&{\bf{0}}&{{{\bf{b}}_3}}& \cdots &{{{\bf{b}}_p}}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{A{{\bf{b}}_1}}&{A\left( {\bf{0}} \right)}&{A{{\bf{b}}_3}}& \cdots &{A{{\bf{b}}_p}}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{A{{\bf{b}}_1}}&{\bf{0}}&{A{{\bf{b}}_3}}& \cdots &{A{{\bf{b}}_p}}\end{aligned}} \right)\end{aligned}\)

So, if\({{\bf{b}}_2} = {\bf{0}}\), then\(A{{\bf{b}}_2} = {\bf{0}}\).

Thus, ifthe second column of Bis all zeros, thesecond column of the product AB is also zero.