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The set of alllinear combinations of the columns of matrix A is Col A.It is called thecolumn space of A. Pivot columns are thebasis for Col A.

The set of allhomogeneous equationsolutions,\(A{\bf{x}} = 0\), is NulA.0It is called thenull spaceof A.

To identify the pivot and the pivot position, observe the leftmost column (nonzero column) of the matrix, that is, the pivot column. At the top of this column, 1 is the pivot.

B = \(\left[ {\begin{array}{*{20}{c}} {\boxed1}&0&6&5 \\ 0&{\boxed2}&5&3 \\ 0&0&0&0 \end{array}} \right]\)

Here, the first and second columns have pivot elements.

The corresponding columns of matrix A are shown below:

\(\left[ {\begin{array}{*{20}{c}}{ - 2}\\2\\{ - 3}\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}4\\{ - 6}\\8\end{array}} \right]\)

The column space is shown below:

\({\rm{Col }}A = \left\{ {\left[ {\begin{array}{*{20}{c}}{ - 2}\\2\\{ - 3}\end{array}} \right],\left[ {\begin{array}{*{20}{c}}4\\{ - 6}\\8\end{array}} \right]} \right\}\)

Thus, the basis for Col Ais \(\left\{ {\left[ {\begin{array}{*{20}{c}}{ - 2}\\2\\{ - 3}\end{array}} \right],\left[ {\begin{array}{*{20}{c}}4\\{ - 6}\\8\end{array}} \right]} \right\}\).

It is given that there are fourcolumns in the given matrix.That means there should be four entries in vector x.

Thus, the equation \(A{\bf{x}} = 0\) can be written as shown below:

\(\begin{array}{c}Ax = 0\\\left[ {\begin{array}{*{20}{c}}1&0&6&5\\0&2&5&3\\0&0&0&0\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right]\end{array}\)

So, the system of equations is

\(\begin{array}{c}{x_1} + 6{x_3} + 5{x_4} = 0,\\2{x_2} + 5{x_3} + 3{x_4} = 0.\end{array}\)

From the above equations,\({x_1}\)and \({x_2}\) correspondto the pivot positions. So, \({x_1}\) and\({x_2}\) are the basic variables, and\({x_3}\)and\({x_4}\) are the free variables.

Let \({x_3} = a\), \({x_4} = b\).

Substitute the values\({x_3} = a\)and \({x_4} = b\) in the equation \({x_1} + 6{x_3} + 5{x_4} = 0\) to obtain the general solution.

\(\begin{array}{c}{x_1} + 6\left( a \right) + 5\left( b \right) = 0\\{x_1} = - 6a - 5b\end{array}\)

Substitute the value \({x_3} = a\) and \({x_4} = b\) in the equation \(2{x_2} + 5{x_3} + 3{x_4} = 0\) to obtain the general solution.

\(\begin{array}{c}2{x_2} + 5\left( a \right) + 3\left( b \right) = 0\\{x_2} = - \frac{5}{2}a - \frac{3}{2}b\end{array}\)

Obtain the vector in the parametric form using \({x_1} = - 6a - 5b\), \({x_2} = - \frac{5}{2}a - \frac{3}{2}b\), \({x_3} = a\), and \({x_4} = b\).

\(\begin{array}{c}\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 6a - 5b}\\{ - \frac{5}{2}a - \frac{3}{2}b}\\a\\b\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{ - 6a}\\{ - \frac{5}{2}}\\a\\0\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{ - 5b}\\{ - \frac{3}{2}b}\\0\\b\end{array}} \right]\\ = a\left[ {\begin{array}{*{20}{c}}{ - 6}\\{ - \frac{5}{2}}\\1\\0\end{array}} \right] + b\left[ {\begin{array}{*{20}{c}}{ - 5}\\{ - \frac{3}{2}}\\0\\1\end{array}} \right]\\ = {x_3}\left[ {\begin{array}{*{20}{c}}{ - 6}\\{ - \frac{5}{2}}\\1\\0\end{array}} \right] + {x_4}\left[ {\begin{array}{*{20}{c}}{ - 5}\\{ - \frac{3}{2}}\\0\\1\end{array}} \right]\end{array}\)

The basis of Nul A is shown below:

\({\rm{Nul }}A = \left\{ {\left[ {\begin{array}{*{20}{c}}{ - 6}\\{ - \frac{5}{2}}\\1\\0\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - 5}\\{ - \frac{3}{2}}\\0\\1\end{array}} \right]} \right\}\)

Thus, the basis for NulAis \(\left\{ {\left[ {\begin{array}{*{20}{c}}{ - 6}\\{ - \frac{5}{2}}\\1\\0\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - 5}\\{ - \frac{3}{2}}\\0\\1\end{array}} \right]} \right\}\).