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Assume that A is row equivalent to B. Find bases for Nul A and Col A.

\[ A = \begin{bmatrix} 4 & -3 & -17 & 27 \\ 2 & 3 & 5 & -9 \\ -8 & -9 & -11 & 21 \end{bmatrix} \]

\[ B = \begin{bmatrix} 1 & 0 & -2 & 3 \\ 0 & 1 & 3 & -5 \\ 0 & -15 & -45 & 75 \end{bmatrix} \]

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

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

Expert Anwer

To calculate the $Col A$ and $Nul A$ of the given vector $A$, we need the vector’s row-reduced echelon form. Vector $B$ is the row equivalent matrix of $A$, which is given as:

\[ B = \begin{bmatrix} 1 & 0 & -2 & 3 \\ 0 & 1 & 3 & -5 \\ 0 & -15 & -45 & 75 \end{bmatrix} \]

Applying row operation as:

\[ R_3 = R_3 + 15R_2 \]

\[ B = \begin{bmatrix} 1 & 0 & -2 & 3 \\ 0 & 1 & 3 & -5 \\ 0 & 0 & 0 & 0 \end{bmatrix} \]

Now the $B$ matrix is the row-reduced echelon form of $A$. We can write it in equation form as:

\[ x_1 -\ 2x_3  + 3x_4 = 0 \hspace{0.3in} \longrightarrow \hspace{0.3in} x_1 = 2x_3 -\ 3x_4 \]

\[ x_2 + 3x_3  -\ 5x_4 = 0 \hspace{0.3in} \longrightarrow \hspace{0.3in} x_2 = -3x_3 + 5x_4 \]

Here, $x_3$ and $x_4$ are the free variables.

\[ x_3 \begin{bmatrix} 2 \\ -3 \\ 1 \\ 0 \end{bmatrix} + x_4  \begin{bmatrix} -3 \\ 5 \\ 0 \\ 1 \end{bmatrix} \]

The basis for $Nul A$ are given as:

\[ \begin{bmatrix} 2 \\ -3 \\ 1 \\ 0 \end{bmatrix} , \begin{bmatrix} -3 \\ 5 \\ 0 \\ 1 \end{bmatrix} \]

There are two pivot columns in the row-reduced echelon form of matrix $A$. Hence, the basis for $Col A$ are those two columns of the original matrix which are given as:

\[ \begin{bmatrix} 2 \\ -8 \\ 4 \end{bmatrix} , \begin{bmatrix} 3 \\ 9 \\ -3 \end{bmatrix} \]

Numerical Results

The basis for $Nul A$ are given as:

\[ \begin{bmatrix} 2 \\ -3 \\ 1 \\ 0 \end{bmatrix} , \begin{bmatrix} -3 \\ 5 \\ 0 \\ 1 \end{bmatrix} \]

The basis for $Col A$ are given as:

\[ \begin{bmatrix} 4 \\ 2 \\ -8 \end{bmatrix} , \begin{bmatrix} -3 \\ 3 \\ -9 \end{bmatrix} \]

Example

Matrix $B$ is given as the row-reduced echelon form of the matrix $A$. Find $Nul A$ of matrix $A$.

\[ A = \begin{bmatrix} 4 & -3 & -17 \\ 2 & 3 & 5 \end{bmatrix} \]

\[ B = \begin{bmatrix} 1 & 0 & -2 \\ 0 & 1 & 3 \end{bmatrix} \]

The parametric solution is given as:

\[ x_1 -\ 2x_3 = 0 \longrightarrow x_1 = 2x_3 \]

\[ x_2 + 3x_3 = 0 \longrightarrow x_2 = -3x_3 \]

\[ \begin{bmatrix} 2 \\ -3 \\ 1 \end{bmatrix} \]

The above column matrix is the $Nul A$ of the given matrix $A$.

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