Exercises 9–12 display a matrix Aand an echelon form of A. Find bases for Col Aand Nul A, and then state the dimensions of these subspaces.

\(A = \left[ {\begin{array}{*{20}{c}}1&2&{ - 4}&3&3\\5&{10}&{ - 9}&{ - 7}&8\\4&8&{ - 9}&{ - 2}&7\\{ - 2}&{ - 4}&5&0&{ - 6}\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&2&{ - 4}&3&3\\0&0&1&{ - 2}&0\\0&0&0&0&{ - 5}\\0&0&0&0&0\end{array}} \right]\)a

The set of all linear combinations of the columns of matrix A is Col A, or it is called the column space of A. Pivot columns are the bases for Col A.

The set of all homogeneous equation solutions\(A{\bf{x}} = 0\)is Nul A, or it is called the null space of A.

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

It is observed that the first, third, and fifth columns have pivot elements.

The corresponding columns of matrix A are shown below:

\(\left[ {\begin{array}{*{20}{c}}1\\5\\4\\{ - 2}\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}{ - 4}\\{ - 9}\\{ - 9}\\5\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}3\\8\\7\\{ - 6}\end{array}} \right]\)

The column space is given as shown below:

\({\rm{Col }}A = \left\{ {\left[ {\begin{array}{*{20}{c}}1\\5\\4\\{ - 2}\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - 4}\\{ - 9}\\{ - 9}\\5\end{array}} \right],\left[ {\begin{array}{*{20}{c}}3\\8\\7\\{ - 6}\end{array}} \right]} \right\}\)

Thus, the bases for Col A are \(\left\{ {\left[ {\begin{array}{*{20}{c}}1\\5\\4\\{ - 2}\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - 4}\\{ - 9}\\{ - 9}\\5\end{array}} \right],\left[ {\begin{array}{*{20}{c}}3\\8\\7\\{ - 6}\end{array}} \right]} \right\}\).

It is given that there are 5 columns in the given matrix, which means there should be 5 entries in vector x.

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

\(\begin{array}{c}Ax = 0\\\left[ {\begin{array}{*{20}{c}}1&2&{ - 4}&3&3\\0&0&1&{ - 2}&0\\0&0&0&0&{ - 5}\\0&0&0&0&0\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\\{{x_5}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\\0\\0\end{array}} \right]\end{array}\)

The augmented matrix is shown below:

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

Multiply row 3 by \( - \frac{1}{5}\).

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

Add \( - 3\) times row 3 to row 1.

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

Add 4 times row 2 to row 1.

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

So, the system of equations is as shown below:

\(\begin{array}{c}{x_1} + 2{x_2} - 5{x_4} = 0\\{x_3} - 2{x_4} = 0\\{x_5} = 0\end{array}\)

From the above equations, \({x_1}\), \({x_3}\), and \({x_5}\) correspond to the pivot positions. So, \({x_1}\), \({x_3}\), and \({x_5}\) are the basic variables, and \({x_2}\)and \({x_4}\) are the free variables.

Let \({x_2} = a\), \({x_4} = b\).

Substitute the values \({x_2} = a\) and \({x_4} = b\) in the equation \({x_1} + 2{x_2} - 5{x_4} = 0\) to obtain the general solution.

\(\begin{array}{c}{x_1} + 2\left( a \right) - 5\left( b \right) = 0\\{x_1} = - 2a + 5b\end{array}\)

Substitute the value \({x_4} = b\) in the equation \({x_3} - 2{x_4} = 0\) to obtain the general solution.

\(\begin{array}{c}{x_3} - 2\left( b \right) = 0\\{x_3} = 2b\end{array}\)

Obtain the vector in the parametric form by using \({x_1} = - 2a + 5b\), \({x_2} = a\), \({x_3} = 2b\), \({x_4} = b\), and \({x_5} = 0\).

\[\begin{array}{c}\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\\{{x_5}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 2a + 5b}\\a\\{2b}\\b\\0\end{array}} \right]\\ = a\left[ {\begin{array}{*{20}{c}}{ - 2}\\1\\0\\0\\0\end{array}} \right] + b\left[ {\begin{array}{*{20}{c}}5\\0\\2\\1\\0\end{array}} \right]\\ = {x_3}\left[ {\begin{array}{*{20}{c}}{ - 2}\\1\\0\\0\\0\end{array}} \right] + {x_5}\left[ {\begin{array}{*{20}{c}}5\\0\\2\\1\\0\end{array}} \right]\end{array}\]

Nul A is shown below:

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

Thus, the bases for Nul A are \(\left\{ {\left[ {\begin{array}{*{20}{c}}{ - 2}\\1\\0\\0\\0\end{array}} \right],\left[ {\begin{array}{*{20}{c}}5\\0\\2\\1\\0\end{array}} \right]} \right\}\).

It is observed that matrix A has 3 pivot columns; so the dimension of Col A is 3. Thus, Col A= 3.

Also, it is observed that the homogeneous equation \(A{\bf{x}} = 0\) has two free variables; so the dimension of Nul A is 2. Thus, Nul A= 2.