Question

If the vectors \( \begin{pmatrix} 1 \\ -1 \\ 3 \end{pmatrix}, \begin{pmatrix} 1 \\ 2 \\ -3 \end{pmatrix}, \begin{pmatrix} p \\ 0 \\ 1 \end{pmatrix} \) are linearly dependent, then the value of p is:

• A. 2
• B. 4
• C. 1
• D. 6

Hint:

When calculating determinants for this purpose, always look for a row or column with the most zeros to expand along. This minimizes the number of cofactors you need to compute.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
A set of \(n\) vectors in an \(n\)-dimensional space (here, 3 vectors in \(\mathbb{R}^3\)) is linearly dependent if and only if the determinant of the matrix formed by these vectors as columns (or rows) is zero.

Step 2: Key Formula or Approach:
We will form a 3x3 matrix \(A\) using the given vectors as columns and set its determinant to zero to solve for \(p\). \[ A = \begin{pmatrix} 1 & 1 & p
-1 & 2 & 0
3 & -3 & 1 \end{pmatrix} \] We need to solve \( \det(A) = 0 \).

Step 3: Detailed Explanation:
We calculate the determinant of \(A\), for instance, by cofactor expansion along the third column, as it contains a zero which simplifies the calculation. \[ \det(A) = p \cdot \begin{vmatrix} -1 & 2
3 & -3 \end{vmatrix} - 0 \cdot \begin{vmatrix} 1 & 1
3 & -3 \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & 1
-1 & 2 \end{vmatrix} \] \[ = p((-1)(-3) - (2)(3)) - 0 + 1((1)(2) - (1)(-1)) \] \[ = p(3 - 6) + (2 + 1) \] \[ = p(-3) + 3 = -3p + 3 \] For the vectors to be linearly dependent, the determinant must be zero: \[ -3p + 3 = 0 \] \[ 3 = 3p \] \[ p = 1 \]
Step 4: Final Answer:
The value of \(p\) for which the vectors are linearly dependent is 1.