Given the matrix $A$, we will address each part step by step:

$A = \begin{bmatrix} 2 & 2 & 2 & 8 \\ 2 & 2 & 3 & 10 \end{bmatrix}$

(a) Find the Reduced Row Echelon Form (RREF) of $A$:

To find the RREF, we apply row operations to transform the matrix into its simplest form, where:

1. The leading entry in each row is 1.
2. Each leading 1 is the only nonzero entry in its column.
3. The leading 1 of the lower rows is further to the right than that of the row above.

Applying elementary row operations:

$\text{Step 1: Subtract Row 1 from Row 2}$ $\begin{bmatrix} 2 & 2 & 2 & 8 \\ 0 & 0 & 1 & 2 \end{bmatrix}$

$\text{Step 2: Scale Row 1 by dividing it by 2}$ $\begin{bmatrix} 1 & 1 & 1 & 4 \\ 0 & 0 & 1 & 2 \end{bmatrix}$

$\text{Step 3: Subtract Row 2 from Row 1}$ $\begin{bmatrix} 1 & 1 & 0 & 2 \\ 0 & 0 & 1 & 2 \end{bmatrix}$

So, the RREF of $A$ is:

$\boxed{ \begin{bmatrix} 1 & 1 & 0 & 2 \\ 0 & 0 & 1 & 2 \end{bmatrix} }$

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(b) Find the Rank of $A$:

The rank of a matrix is the number of leading 1's in its RREF.

From the RREF $\begin{bmatrix} 1 & 1 & 0 & 2 \\ 0 & 0 & 1 & 2 \end{bmatrix}$, we observe that there are 2 leading 1's, so the rank of $A$ is:

$\boxed{ 2 }$

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(c) Find a Basis of the Column Space $C(A)$ expressed in terms of the columns of $A$:

The column space $C(A)$ is spanned by the linearly independent columns of the original matrix $A$. From the RREF, the pivot columns (the columns corresponding to leading 1's) are the first and third columns.

Thus, the basis for the column space consists of the first and third columns of $A$:

$\text{Basis for } C(A) = \left\{ \begin{bmatrix} 2 \\ 2 \end{bmatrix}, \begin{bmatrix} 2 \\ 3 \end{bmatrix} \right\}$

So, the basis of the column space is:

$\boxed{ \left\{ \begin{bmatrix} 2 \\ 2 \end{bmatrix}, \begin{bmatrix} 2 \\ 3 \end{bmatrix} \right\} }$

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(d) Find a Basis of the Null Space $N(A)$:

To find the null space, we solve $A \mathbf{x} = 0$. Using the RREF of $A$, we get the following system of equations:

x_1 + x_2 + 2x_4 &= 0 \\ x_3 + 2x_4 &= 0 \end{aligned}$$ Express $$x_1$$ and $$x_3$$ in terms of free variables $$x_2$$ and $$x_4$$: $$x_1 = -x_2 - 2x_4, \quad x_3 = -2x_4$$ Thus, the solution vector $$\mathbf{x}$$ is: $$\mathbf{x} = \begin{bmatrix} -x_2 - 2x_4 \\ x_2 \\ -2x_4 \\ x_4 \end{bmatrix} = x_2 \begin{bmatrix} -1 \\ 1 \\ 0 \\ 0 \end{bmatrix} + x_4 \begin{bmatrix} -2 \\ 0 \\ -2 \\ 1 \end{bmatrix}$$ The **basis of the null space** is: $$\boxed{ \left\{ \begin{bmatrix} -1 \\ 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} -2 \\ 0 \\ -2 \\ 1 \end{bmatrix} \right\} }$$ --- Would you like further details or have any questions about these calculations? ### Related Questions: 1. How would you compute the RREF for a larger matrix? 2. What is the geometric interpretation of the column space? 3. How do you find the rank of a matrix using determinant properties? 4. Can you explain the relationship between the rank and nullity of a matrix? 5. What would the RREF look like for a matrix with dependent rows? ### Tip: The rank of a matrix is always less than or equal to the smaller dimension of the matrix (min(rows, columns)).