Let \(A = \left( {\begin{aligned}{*{20}{c}}3&{ - 6}\\{ - 1}&2\end{aligned}} \right)\). Construct a \({\bf{2}} \times {\bf{2}}\) matrix Bsuch that ABis the zero matrix. Use two different nonzero columns for B.

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\(B = \left( {\begin{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right)\).

It is given thatABis the zero matrix. So,\(AB = 0\).

Obtain the product of\(A = \left( {\begin{aligned}{*{20}{c}}3&{ - 6}\\{ - 1}&2\end{aligned}} \right)\)and\(B = \left( {\begin{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right)\)as shown below:

\(\begin{aligned}{c}AB = \left( {\begin{aligned}{*{20}{c}}3&{ - 6}\\{ - 1}&2\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{3a - 6c}&{3b - 6d}\\{ - a + 2c}&{ - b + 2d}\end{aligned}} \right)\end{aligned}\)

Put\(AB = 0\)as shown below:

\(\left( {\begin{aligned}{*{20}{c}}{3a - 6c}&{3b - 6d}\\{ - a + 2c}&{ - b + 2d}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}0&0\\0&0\end{aligned}} \right)\)

Compare the equations in the matrix equation\(\left( {\begin{aligned}{*{20}{c}}{3a - 6c}&{3b - 6d}\\{ - a + 2c}&{ - b + 2d}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}0&0\\0&0\end{aligned}} \right)\)as shown below:

\(\begin{aligned}{c}3a - 6c = 0\\3b - 6d = 0\\ - a + 2c = 0\\ - b + 2d = 0\end{aligned}\)

Let the value of\(a = 2\)and\(b = 6\). Then, the value of\(c\)and\(d\)are shown below:

\(\begin{aligned}{c} - \left( 2 \right) + 2c = 0\\2c = 2\\c = 1\end{aligned}\)

And the value of d is shown below:

\(\begin{aligned}{c} - \left( 6 \right) + 2d = 0\\2d = 6\\d = 3\end{aligned}\)

Now, substitute all the values \(a = 2\),\(b = 6\),\(c = 1\), and \(d = 3\) in the matrix. \(B = \left( {\begin{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right)\).

\(B = \left( {\begin{aligned}{*{20}{c}}2&6\\1&3\end{aligned}} \right)\)

Also, the columns of\(B\)are nonzero.

Thus, matrix \(B = \left( {\begin{aligned}{*{20}{c}}2&6\\1&3\end{aligned}} \right)\).