From $ b = -2 $, substitute into 2: $ c = 6 + 2a $. Let $ a $ be a free variable. Choosing $ a = 0 $, then $ c = 6 $. Thus, $ \mathbf{v} = \begin{pmatrix} 0 \\ -2 \\ 6 \end{pmatrix} $. Final answer: $ \boxed{\begin{pmatrix} 0 \\ -2 \\ 6 \end{pmatrix}} $.**Question 1:

["Transforming Linear Equations: A Step-by-Step Substitution Example with Vector Values", "Understanding how to substitutionally manipulate linear equations is fundamental in algebra and linear algebra. This article explores a concrete example of substituting a scalar value into a linear equation, evaluating a derived expression, and constructing a vector—a process essential in solving systems, optimization, and geometric modeling.", "---", "### Problem Statement", "Start with the scalar equation:\n$$\nb = -2\n$$\nSubstitute $ b $ into the expression:\n$$\nc = 6 + 2a\n$$\nLet $ a $ be a free variable. Then, setting $ a = 0 $, compute the corresponding $ c $:\n$$\nc = 6 + 2(0) = 6\n$$\nThus, the vector solution is:\n$$\n\mathbf{v} = \begin{pmatrix} 0 \ -2 \ 6 \end{pmatrix}\n$$\nThe final boxed answer is:\n$$\n\boxed{\begin{pmatrix} 0 \ -2 \ 6 \end{pmatrix}}\n$$", "---", "### Step-by-Step Explanation", "1. Fix $ b = -2 $ as a given condition\n Though $ b $ is defined independently, substituting into $ c = 6 + 2a $ isolates $ c $ in terms of $ a $. Choosing $ a = 0 $ serves as a representative input to compute a concrete value.", "2. Substitute $ a = 0 $ into $ c = 6 + 2a $\n This step reduces the equation to a numerical value:\n $$\n c = 6 + 2(0) = 6\n $$", "3. Form the column vector using $ a = 0 $, $ b = -2 $, $ c = 6 $\n With $ a $ treated as a scalar parameter, the vector $ \mathbf{v} $ captures the output of the linear transformation. The components follow naturally:\n $$\n \mathbf{v} = \begin{pmatrix} a \ b \ c \end{pmatrix} = \begin{pmatrix} 0 \ -2 \ 6 \end{pmatrix}\n $$", "This approach demonstrates how free variables generate data points in vector spaces—key for applications in machine learning, physics, and engineering.", "---", "### Why This Matters", "Substituting variables into equations is the building block of solving linear systems, parameterized models, and geometric transformations. By fixing known values and analyzing free variables, we bridge abstract expressions with tangible numerical results.", "---", "### Summary", "- Substitution anchors dependent variables to given constants.\n- Treating variables as parameters allows vectorization in higher-dimensional spaces.\n- Simple examples like $ \mathbf{v} = \begin{pmatrix} 0 \ -2 \ 6 \end{pmatrix} $ lay groundwork for advanced linear algebra.", "This method exemplifies clarity, precision, and practicality—hallmarks of effective mathematical communication."]