Question

The values of \(\lambda\) such that \((x,y,z) \ne (0,0,0)\) and \[ (\hat{i} + \hat{j} + 3\hat{k})x + (3\hat{i} - 3\hat{j} + \hat{k})y + (-4\hat{i} + 5\hat{j})z = \lambda(x\hat{i} + y\hat{j} + z\hat{k}) \]

• A. \(0,1\)
• B. \(-1,1\)
• C. \(-1,0\)
• D. \(-2,0\)

Hint:

For vector equations of form \(A\vec{X}=\lambda \vec{X}\), directly form matrix and solve determinant.

Answer 1

The Correct Option is C

Concept: This represents an eigenvalue problem: \[ A\vec{X} = \lambda \vec{X} \Rightarrow (A - \lambda I)\vec{X} = 0 \] Non-trivial solution exists when: \[ |A - \lambda I| = 0 \]

Step 1: Form matrix.
\[ A = \begin{bmatrix} 1 & 3 & -4 1 & -3 & 5 3 & 1 & 0 \end{bmatrix} \]

Step 2: Find determinant.
\[ \left| \begin{matrix} 1-\lambda & 3 & -4 1 & -3-\lambda & 5 3 & 1 & -\lambda \end{matrix} \right| = 0 \]

Step 3: Solve characteristic equation.
Expanding gives: \[ \lambda(\lambda+1)(\lambda-0)=0 \] \[ \Rightarrow \lambda = -1,\ 0 \] Conclusion: \[ {(C)\ -1,\ 0} \]