91Ó°ÊÓ

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

Its augmented matrix is:

\[\left[ {\begin{array}{*{20}{c}}A&0\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}3&2&1&{ - 5}&0\\{ - 9}&{ - 4}&1&7&0\\9&2&{ - 5}&1&0\end{array}} \right]\]

At row 2, multiply row 1 by 3 and add it to row 2, i.e., , and at row 3, multiply row 1 by 3 and subtract it from row 3, i.e., \[{R_3} \to {R_3} - 3{R_1}\]. Then,

\[ \sim \left[ {\begin{array}{*{20}{c}}3&2&1&{ - 5}&0\\0&2&4&{ - 8}&0\\0&{ - 4}&{ - 8}&{16}&0\end{array}} \right]\]

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

\[ \sim \left[ {\begin{array}{*{20}{c}}3&2&1&{ - 5}&0\\0&2&4&{ - 8}&0\\0&0&0&0&0\end{array}} \right]\]

At row 1, subtract row 2 from row 1, i.e., \[{R_1} \to {R_1} - {R_2}\].

\[ \sim \left[ {\begin{array}{*{20}{c}}3&0&{ - 3}&3&0\\0&2&4&{ - 8}&0\\0&0&0&0&0\end{array}} \right]\]

At row 1, divide row 1 by 3, and at row 2, divide row 2 by 2.

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

This implies that

\[\begin{array}{c}{x_1} - {x_3} + {x_4} = 0\\{x_2} + 2{x_3} - 4{x_4} = 0\\0 = 0\end{array}\]

Thus, the system is consistent.

The general solution is \[{x_1} = {x_3} - {x_4}\], and with \[{x_3},\] and are free. So, the nonzero vector in Nul A is given by \[{x_3},\] and .

Choose and then , and \[{x_2} = 6\] to obtain a nonzero vector \[\left( {1,2, - 1,6} \right)\] in Nul A.

By definition of column space, Col A is the span of \[\left\{ {\left( {3, - 9,9} \right),\left( {2, - 4,2} \right),\left( {1,1, - 5} \right),\left( { - 5,7,1} \right)} \right\}\].

So, every column of A belongs to Col A. Choose any nonzero column of A. So, is a nonzero vector in Col A.