If \(A = \left( {\begin{aligned}{*{20}{c}}1&{ - 2}\\{ - 2}&5\end{aligned}} \right)\) and \(AB = \left( {\begin{aligned}{*{20}{c}}{ - 1}&2&{ - 1}\\6&{ - 9}&3\end{aligned}} \right)\), determine the first and second column of 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:

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

Let\({{\bf{b}}_1},{{\bf{b}}_2},{{\bf{b}}_3}\)be the columns of matrix B. Then, the product AB is obtained as shown below:

\(\begin{aligned}{c}AB = A\left( {\begin{aligned}{*{20}{c}}{{{\bf{b}}_1}}&{{{\bf{b}}_2}}&{{{\bf{b}}_3}}\end{aligned}} \right)\\ = \left( {A\begin{aligned}{*{20}{c}}{{{\bf{b}}_1}}&{A{{\bf{b}}_2}}&{A{{\bf{b}}_3}}\end{aligned}} \right)\end{aligned}\)

And the given product AB is\(AB = \left( {\begin{aligned}{*{20}{c}}{ - 1}&2&{ - 1}\\6&{ - 9}&3\end{aligned}} \right)\).

Compare \(AB = \left( {A\begin{aligned}{*{20}{c}}{{{\bf{b}}_1}}&{A{{\bf{b}}_2}}&{A{{\bf{b}}_3}}\end{aligned}} \right)\) and \(AB = \left( {\begin{aligned}{*{20}{c}}{ - 1}&2&{ - 1}\\6&{ - 9}&3\end{aligned}} \right)\). So, \(A{{\bf{b}}_1} = \left( {\begin{aligned}{*{20}{c}}{ - 1}\\6\end{aligned}} \right)\), and \(A{{\bf{b}}_2} = \left( {\begin{aligned}{*{20}{c}}2\\{ - 9}\end{aligned}} \right)\).

Let,\({{\bf{b}}_1} = \left( {\begin{aligned}{*{20}{c}}{{x_1}}\\{{x_2}}\end{aligned}} \right)\). Then, it can be written as shown below:

\(\begin{aligned}{c}A{{\bf{b}}_1} = \left( {\begin{aligned}{*{20}{c}}{ - 1}\\6\end{aligned}} \right)\\\left( {\begin{aligned}{*{20}{c}}1&{ - 2}\\{ - 2}&5\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{{x_1}}\\{{x_2}}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}{ - 1}\\6\end{aligned}} \right)\end{aligned}\)

In augmented form,

\(\left( {\begin{aligned}{*{20}{c}}1&{ - 2}&{ - 1}\\{ - 2}&5&6\end{aligned}} \right)\)

Use\({x_1}\)term in the first equation to eliminate\( - 2{x_1}\)term from the second equation. Add 2 times row 1 to row 2.

\(\left( {\begin{aligned}{*{20}{c}}1&{ - 2}&{ - 1}\\{ - 2}&5&6\end{aligned}} \right) \sim \left( {\begin{aligned}{*{20}{c}}1&{ - 2}&{ - 1}\\0&1&4\end{aligned}} \right)\)

Use\({x_2}\)term in the second equation to eliminate\( - 2{x_2}\)term from the first equation. Add 2 times row 2 to row 1.

\(\left( {\begin{aligned}{*{20}{c}}1&{ - 2}&{ - 1}\\0&1&4\end{aligned}} \right) \sim \left( {\begin{aligned}{*{20}{c}}1&0&7\\0&1&4\end{aligned}} \right)\)

Thus, the row reduced echelon form is\(\left( {\begin{aligned}{*{20}{c}}1&0&7\\0&1&4\end{aligned}} \right)\).

So, \({x_1} = 7\), and \({x_2} = 4\).

Therefore, the first column of matrix B is \(\left( {\begin{aligned}{*{20}{c}}7\\4\end{aligned}} \right)\).

Let,\({{\bf{b}}_2} = \left( {\begin{aligned}{*{20}{c}}{{x_1}}\\{{x_2}}\end{aligned}} \right)\). Then, it can be written as shown below:

\(\begin{aligned}{c}A{{\bf{b}}_2} = \left( {\begin{aligned}{*{20}{c}}2\\{ - 9}\end{aligned}} \right)\\\left( {\begin{aligned}{*{20}{c}}1&{ - 2}\\{ - 2}&5\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{{x_1}}\\{{x_2}}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}2\\{ - 9}\end{aligned}} \right)\end{aligned}\)

In augmented form,

\(\left( {\begin{aligned}{*{20}{c}}1&{ - 2}&2\\{ - 2}&5&{ - 9}\end{aligned}} \right)\)

Use\({x_1}\)term in the first equation to eliminate\( - 2{x_1}\)term from the second equation. Add 2 times row 1 to row 2.

\(\left( {\begin{aligned}{*{20}{c}}1&{ - 2}&2\\{ - 2}&5&{ - 9}\end{aligned}} \right) \sim \left( {\begin{aligned}{*{20}{c}}1&{ - 2}&2\\0&1&{ - 5}\end{aligned}} \right)\)

Use\({x_2}\)term in the second equation to eliminate\( - 2{x_2}\)term from the first equation. Add 2 times row 2 to row 1.

\(\left( {\begin{aligned}{*{20}{c}}1&{ - 2}&2\\0&1&{ - 5}\end{aligned}} \right) \sim \left( {\begin{aligned}{*{20}{c}}1&0&{ - 8}\\0&1&{ - 5}\end{aligned}} \right)\)

Thus, the row reduced echelon form is\(\left( {\begin{aligned}{*{20}{c}}1&0&{ - 8}\\0&1&{ - 5}\end{aligned}} \right)\).

So, \({x_1} = - 8\), and \({x_2} = - 5\).

Therefore, the first column of matrix B is \(\left( {\begin{aligned}{*{20}{c}}{ - 8}\\{ - 5}\end{aligned}} \right)\).

Thus, the first and second columns of matrix B are \(\left( {\begin{aligned}{*{20}{c}}7\\4\end{aligned}} \right)\) and \(\left( {\begin{aligned}{*{20}{c}}{ - 8}\\{ - 5}\end{aligned}} \right)\), respectively.