Step 1: Concept Overview:
A set of \(n\) vectors in an \(n\)-dimensional space (e.g., 3 vectors in \(\mathbb{R}^3\)) is linearly dependent if the determinant of the matrix formed by these vectors is zero.
Step 2: Method:
Form a 3x3 matrix \(A\) using the given vectors as columns and solve \( \det(A) = 0 \) to find \(p\).
\[ A = \begin{pmatrix} 1 & 1 & p
-1 & 2 & 0
3 & -3 & 1 \end{pmatrix} \]
Step 3: Solution:
Calculate the determinant of \(A\) using cofactor expansion along the third column:
\[ \det(A) = p \cdot \begin{vmatrix} -1 & 2
3 & -3 \end{vmatrix} - 0 \cdot \begin{vmatrix} 1 & 1
3 & -3 \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & 1
-1 & 2 \end{vmatrix} \]
\[ = p((-1)(-3) - (2)(3)) - 0 + 1((1)(2) - (1)(-1)) \]
\[ = p(3 - 6) + (2 + 1) \]
\[ = p(-3) + 3 = -3p + 3 \]
For linear dependence, the determinant is zero:
\[ -3p + 3 = 0 \]
\[ 3 = 3p \]
\[ p = 1 \]
Step 4: Answer:
The value of \(p\) for which the vectors are linearly dependent is 1.