To determine whether the vectors $(1111)$, $(1232)$, $(2564)$, and $(2685)$ form a basis of $\mathbb{R}^4$, we need to check if they are linearly independent and if they span $\mathbb{R}^4$.

Step 1: Check for Linear Independence

For these vectors to form a basis, they must be linearly independent. This means that the only solution to the equation:

$c_1 \begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix} + c_2 \begin{pmatrix} 1 \\ 2 \\ 3 \\ 2 \end{pmatrix} + c_3 \begin{pmatrix} 2 \\ 5 \\ 6 \\ 4 \end{pmatrix} + c_4 \begin{pmatrix} 2 \\ 6 \\ 8 \\ 5 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}$

must be $c_1 = c_2 = c_3 = c_4 = 0$.

This can be written as the following system of linear equations:

1. & \quad c_1 + c_2 + 2c_3 + 2c_4 = 0 \\ 2. & \quad c_1 + 2c_2 + 5c_3 + 6c_4 = 0 \\ 3. & \quad c_1 + 3c_2 + 6c_3 + 8c_4 = 0 \\ 4. & \quad c_1 + 2c_2 + 4c_3 + 5c_4 = 0 \end{aligned}$$ We can write this system as a matrix and row-reduce it to determine if the system has a non-trivial solution. ### Step 2: Set up the Augmented Matrix $$\begin{pmatrix} 1 & 1 & 2 & 2 & 0 \\ 1 & 2 & 5 & 6 & 0 \\ 1 & 3 & 6 & 8 & 0 \\ 1 & 2 & 4 & 5 & 0 \end{pmatrix}$$ ### Step 3: Row-Reduce the Matrix We will perform Gaussian elimination on this matrix to check for linear independence. 1. Subtract row 1 from rows 2, 3, and 4 to eliminate the first column: $$\begin{pmatrix} 1 & 1 & 2 & 2 & 0 \\ 0 & 1 & 3 & 4 & 0 \\ 0 & 2 & 4 & 6 & 0 \\ 0 & 1 & 2 & 3 & 0 \end{pmatrix}$$ 2. Subtract row 2 from row 3 and row 2 from row 4: $$\begin{pmatrix} 1 & 1 & 2 & 2 & 0 \\ 0 & 1 & 3 & 4 & 0 \\ 0 & 0 & -2 & -2 & 0 \\ 0 & 0 & -1 & -1 & 0 \end{pmatrix}$$ 3. Multiply row 3 by $$-\frac{1}{2}$$ and row 4 by $$-1$$: $$\begin{pmatrix} 1 & 1 & 2 & 2 & 0 \\ 0 & 1 & 3 & 4 & 0 \\ 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 & 0 \end{pmatrix}$$ 4. Subtract row 3 from row 4 to eliminate the third column: $$\begin{pmatrix} 1 & 1 & 2 & 2 & 0 \\ 0 & 1 & 3 & 4 & 0 \\ 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}$$ The matrix has a row of all zeros, which means that the system has a non-trivial solution, implying that the vectors are **linearly dependent**. ### Step 4: Determine the Dimension of the Subspace Since the matrix has 3 pivot positions, the rank of the matrix is 3. This means that the vectors span a subspace of $$\mathbb{R}^4$$ with dimension 3. ### Conclusion: The vectors $$(1111)$$, $$(1232)$$, $$(2564)$$, and $$(2685)$$ **do not form a basis** of $$\mathbb{R}^4$$ because they are linearly dependent. The dimension of the subspace they span is **3**. --- Would you like any additional details, or do you have further questions? Here are 5 related questions: 1. How do you determine the rank of a matrix using row reduction? 2. What is the difference between a basis and a spanning set? 3. How can you check if a set of vectors is linearly independent? 4. What is the process for finding the null space of a matrix? 5. Can a set of vectors in $$\mathbb{R}^4$$ with fewer than 4 vectors form a basis? **Tip**: When working with systems of linear equations, always check if there is a row of zeros in the final matrix; this indicates linear dependence!