Find the bases for $$ \operatorname{Col} \operatorname{A} $$ and $$ \operatorname{Nul} \mathrm{A} $$, and then state the dimension of these subspaces for the matrix $$ A $$ and an echelon form of $$ A $$ below. $$ A=\left[\begin{array}{rrrrr} 1 & 4 & -2 & -1 & 3 \ 5 & 20 & 1 & -27 & 8 \ 4 & 16 & -1 & -18 & 7 \ -2 & -8 & 1 & 8 & -6 \end{array}\right] \sim\left[\begin{array}{rrrrr} 1 & 4 & -2 & -1 & 3 \ 0 & 0 & 1 & -2 & 0 \ 0 & 0 & 0 & 0 & -5 \ 0 & 0 & 0 & 0 & 0 \end{array}\right] $$ A basis for Col A is given by $$ \square $$ B. (Use a comma to separate vectors as needed.) The dimension of Col A is $$ \square $$ $$ \square $$. A basis for Nul A is given by $$ \square $$ B. (Use a comma to separate vectors as needed.) The dimension of Nul A is $$ \square $$ .

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The echelon form shows pivots in columns 1, 3, and 5. This means:

1. The basis for $$\operatorname{Col} A$$ is given by the corresponding columns of the original matrix:
   $$
   \begin{pmatrix} 1 \ 5 \ 4 \ -2 \end{pmatrix}, \quad
   \begin{pmatrix} -2 \ 1 \ -1 \ 1 \end{pmatrix}, \quad
   \begin{pmatrix} 3 \ 8 \ 7 \ -6 \end{pmatrix}
   $$
   The dimension of $$\operatorname{Col} A$$ is $$3$$.

2. To find the basis for $$\operatorname{Nul} A$$, we solve $$A\mathbf{x}=\mathbf{0}$$.

Since the pivots are in columns 1, 3, and 5, let the free variables be:
   $$
   x_2 = s \quad \text{and} \quad x_4 = t.
   $$
   From the echelon form we have the following equations:

- **Row 1:** 
     $$
     x_1 + 4x_2 - 2x_3 - x_4 + 3x_5 = 0 \quad \Longrightarrow \quad x_1 = -4x_2 + 2x_3 + x_4 - 3x_5.
     $$
   - **Row 2:** 
     $$
     x_3 - 2x_4 = 0 \quad \Longrightarrow \quad x_3 = 2x_4 = 2t.
     $$
   - **Row 3:** 
     $$
     -5x_5 = 0 \quad \Longrightarrow \quad x_5 = 0.
     $$

Substitute $$x_3 = 2t$$ and $$x_5 = 0$$ into the equation for $$x_1$$:
   $$
   x_1 = -4s + 2(2t) + t = -4s + 4t + t = -4s + 5t.
   $$

So the general solution is:
   $$
   \begin{pmatrix}
   x_1 \ x_2 \ x_3 \ x_4 \ x_5
   \end{pmatrix}
   =
   \begin{pmatrix}
   -4s+5t \ s \ 2t \ t \ 0
   \end{pmatrix}
   = s\begin{pmatrix} -4 \ 1 \ 0 \ 0 \ 0 \end{pmatrix} + t\begin{pmatrix} 5 \ 0 \ 2 \ 1 \ 0 \end{pmatrix}.
   $$
   Thus, a basis for $$\operatorname{Nul} A$$ is:
   $$
   \begin{pmatrix} -4 \ 1 \ 0 \ 0 \ 0 \end{pmatrix}, \quad \begin{pmatrix} 5 \ 0 \ 2 \ 1 \ 0 \end{pmatrix}.
   $$
   The dimension of $$\operatorname{Nul} A$$ is $$2$$.

Summary:

A basis for Col A is given by  
$$\begin{pmatrix} 1 \ 5 \ 4 \ -2 \end{pmatrix}, \begin{pmatrix} -2 \ 1 \ -1 \ 1 \end{pmatrix}, \begin{pmatrix} 3 \ 8 \ 7 \ -6 \end{pmatrix}$$.  
The dimension of Col A is $$3$$.

A basis for Nul A is given by  
$$\begin{pmatrix} -4 \ 1 \ 0 \ 0 \ 0 \end{pmatrix}, \begin{pmatrix} 5 \ 0 \ 2 \ 1 \ 0 \end{pmatrix}$$.  
The dimension of Nul A is $$2$$.
A basis for Col A is given by $$\begin{pmatrix} 1 \ 5 \ 4 \ -2 \end{pmatrix}, \begin{pmatrix} -2 \ 1 \ -1 \ 1 \end{pmatrix}, \begin{pmatrix} 3 \ 8 \ 7 \ -6 \end{pmatrix}$$. The dimension of Col A is $$3$$. A basis for Nul A is given by $$\begin{pmatrix} -4 \ 1 \ 0 \ 0 \ 0 \end{pmatrix}, \begin{pmatrix} 5 \ 0 \ 2 \ 1 \ 0 \end{pmatrix}$$. The dimension of Nul A is $$2$$.

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