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For the matrix A below, find a nonzero vector in nul A and a nonzero vector in col A.

\[ A = \begin{bmatrix} 1 & -2 & 5 & 6 \\ 5 & 1 & -10 & 15 \\ 1 & -2 & 8 & 4 \end{bmatrix} \]

This question aims to find the null space which represents the set of all solutions to the homogeneous equation and column space which represents the range of a given vector.

The concepts that we need to solve this question are null space, column space, homogeneous equation of vectors, and linear transformations. The null space of a vector is written as $Nul A$ is a set of all possible solutions to the homogeneous equation $Ax=0$. The column space of a vector is written as $Col A$ is the set of all possible linear combinations or range of the given matrix.

Expert Anwer

The homogeneous equation is given as:

\[ AX = 0 \]

The matrix $A$ is given in the question and $X$ is a column vector with $4$ unknown variables. We can assume matrix $X$ to be:

\[ X = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} \]

Using row operations on matrix $A$ to reduce the matrix to echelon form.

\[ R_2 \rightarrow R_2 -\ 5R_1, \hspace{0.3in} R_3 \rightarrow R_3 -\ R_1 \]

\[ A = \begin{bmatrix} 1 & -2 & 5 & 6 \\ 0 & 1 & -35 & -15 \\ 0 & 0 & 3 & -2 \end{bmatrix} \]

\[ R_2 \rightarrow R_2/11, \hspace{0.3in} R_1 \rightarrow R_1 + 2R_2 \]

\[ A = \begin{bmatrix} 1 & 0 & -15/11 & 36/11 \\ 0 & 1 & -35/11 & -15/11 \\ 0 & 0 & 3 & -2 \end{bmatrix} \]

\[ R_3 \rightarrow R_3/3,  \hspace{0.3in} R_1 \rightarrow R_1 + 15R_2/11 \]

\[ A = \begin{bmatrix} 1 & 0 & 0 & 26/11 \\ 0 & 1 & -35/11 & -15/11 \\ 0 & 0 & 1 & -2/3 \end{bmatrix} \]

\[ R_1 \rightarrow R_1 – 35R_3/11 \]

\[ A = \begin{bmatrix} 1 & 0 & 0 & 26/11 \\ 0 & 1 & 0 & -115/33 \\ 0 & 0 & 1 & -2/3 \end{bmatrix} \]

The matrix $A$ contains $2$ pivot columns and $2$ free columns. Substituting the values in homogeneous equation, we get:

\[ A = \begin{bmatrix} 1 & 0 & 0 & 26/11 \\ 0 & 1 & 0 & -115/33 \\ 0 & 0 & 1 & -2/3 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} \]

Solving for unknown variables, we get:

\[ x_1 + \dfrac{26}{11}x_4 = 0 \longrightarrow x_1 = -\dfrac{26}{11} \]

\[ x_2 -\ \dfrac{115}{33}x_4 = 0 \longrightarrow x_2 = \dfrac{115}{33} \]

\[ x_3 -\ \dfrac{2}{3}x_4 = 0 \longrightarrow x_3 = \dfrac{2}{3} \]

The parametric solution is given as:

\[ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} -\dfrac{26}{11}x_4 \\ \dfrac{115}{33}x_4 \\ \dfrac{2}{3}x_4 \\ x_4 \end{bmatrix} \]

\[ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} -\dfrac{26}{11} \\ \dfrac{115}{33} \\ \dfrac{2}{3} \\ 1 \end{bmatrix} x_4 \]

Numerical Result

The nonzero vector in $Nul A$ is:

\[ \begin{Bmatrix} \begin{bmatrix} -\dfrac{26}{11} \\ \dfrac{115}{33} \\ \dfrac{2}{3} \\ 1 \end{bmatrix} \end{Bmatrix} \]

The pivot columns in the echelon form of matrix $A$ points to $Col A$, which are given as:

\[ \begin{Bmatrix} \begin{bmatrix} 1 \\ 5 \\ 1 \end{bmatrix} , \begin{bmatrix} -2 \\ 1 \\ -2 \end{bmatrix}, \begin{bmatrix} 5 \\ -10 \\ 8 \end{bmatrix}  \end{Bmatrix} \]

Example

Find the column space of the given matrix below:

\[ \begin{bmatrix} -3 & 2 \\ -5 & -9 \end{bmatrix} \]

The echelon form of the given matrix found to be:

\[ \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \]

The $Col$ space of the given matrix is given as:

\[ \begin{Bmatrix} \begin{bmatrix} -3 \\ -5 \end{bmatrix} , \begin{bmatrix} 2 \\ -9 \end{bmatrix}  \end{Bmatrix} \]

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