14. For A as in exercise 12 i.e., \[A = \left[ {\begin{array}{*{20}{c}}{\bf{1}}&{\bf{2}}&{\bf{3}}\\{\bf{4}}&{\bf{5}}&{\bf{7}}\\{ - {\bf{5}}}&{ - {\bf{1}}}&{\bf{0}}\\{\bf{2}}&{\bf{7}}&{{\bf{11}}}\end{array}} \right]\], find a nonzero vector in Nul A and a nonzero vector in Col A.

First, solve the equation \[Ax = 0\].

Its augmented matrix is:

\[\left[ {\begin{array}{*{20}{c}}A&0\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&2&3&0\\4&5&7&0\\{ - 5}&{ - 1}&0&0\\2&7&{11}&0\end{array}} \right]\]

At row 2, multiply row 1 by 4 and subtract it from row 2, i.e., \[{R_2} \to {R_2} - 4{R_1}\], at row 3. Multiply row 1 by 5 and add it to row 3, i.e., \[{R_3} \to {R_3} + 5{R_1}\], and at row 4, multiply row 1 by 2 and subtract it from row 4, i.e., \[{R_4} \to {R_4} - 2{R_1}\]. Therefore,

\[ \sim \left[ {\begin{array}{*{20}{c}}1&2&3&0\\0&{ - 3}&{ - 5}&0\\0&9&{15}&0\\0&3&5&0\end{array}} \right]\]

At row 3, multiply row 2 by 3 and add it to row 3, i.e., \[{R_3} \to {R_3} + 3{R_2}\], and at row 4, add row 2 to row 4, i.e., \[{R_4} \to {R_4} + {R_2}\]. Therefore,

\[ \sim \left[ {\begin{array}{*{20}{c}}1&2&3&0\\0&{ - 3}&{ - 5}&0\\0&0&0&0\\0&0&0&0\end{array}} \right]\]

At row 2, divide row 2 by -3.

\[ \sim \left[ {\begin{array}{*{20}{c}}1&2&3&0\\0&1&{{5 \mathord{\left/

{\vphantom {5 3}} \right.

\kern-\nulldelimiterspace} 3}}&0\\0&0&0&0\\0&0&0&0\end{array}} \right]\]

This implies that

\[\begin{array}{c}{x_1} + 2{x_2} + 3{x_3} = 0\\{x_2} + {5 \mathord{\left/

{\vphantom {5 3}} \right.

\kern-\nulldelimiterspace} 3}{x_3} = 0\\0 = 0\end{array}\]

Thus, the system is consistent.

Therefore, \[{x_2} = - {5 \mathord{\left/

{\vphantom {5 3}} \right.

\kern-\nulldelimiterspace} 3}{x_3}\].This implies that

\[\begin{array}{c}{x_1} + 2\left( { - {5 \mathord{\left/

{\vphantom {5 3}} \right.

\kern-\nulldelimiterspace} 3}{x_3}} \right) + 3{x_3} = 0\\{x_1} + {{\left( { - 10 + 9} \right)} \mathord{\left/

{\vphantom {{\left( { - 10 + 9} \right)} 3}} \right.

\kern-\nulldelimiterspace} 3}{x_3} = 0\\{x_1} = {1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}{x_3}\end{array}\]

The general solution is\[{x_1} = {1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}{x_3}\], and \[{x_2} = - {5 \mathord{\left/

{\vphantom {5 3}} \right.

\kern-\nulldelimiterspace} 3}{x_3}\]with \[{x_3}\] free. So, the nonzero vector in Nul A is given by \[{x_3}\].

Choose \[{x_3} = 3\], \[{x_1} = 1\], and \[{x_2} = - 5\] to obtain a nonzero vector \[\left( {1, - 5,3} \right)\] in Nul A.

By definition of column space, Col A is the span of \[\left\{ {\left( {1,4, - 5,2} \right),\left( {2,5, - 1,7} \right),\left( {3,7,0,11} \right)} \right\}\].

So, every column of A belongs to Col A. Choose any nonzero column of A. So, \[\left( {1,4, - 5,2} \right)\] is a nonzero vector in Col A.