Question

The number of values of $k$ for which the following system of linear equations: \[ \begin{pmatrix} 4 & k & 2\\ k & 4 & 1\\ 2 & 2 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \] possess a non-zero solution, is _______

• A. 1
• B. 2
• C. 3
• D. Infinite

Hint:

For any homogeneous system $A X = 0$:
Non-trivial/non-zero solution $\iff \det(A) = 0$.
Setting the determinant of the $3 \times 3$ matrix to zero results in a quadratic equation in $k$, which yields exactly 2 roots.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given a homogeneous system of linear equations \[ AX=0. \] We need to determine the number of values of \(k\) for which the system possesses a non-zero (non-trivial) solution. 

Step 2: Key Formula or Approach:
A homogeneous system of linear equations has a non-zero solution if and only if the determinant of the coefficient matrix is zero. \[ \det(A)=0. \] 
Step 3: Detailed Explanation: 

• The coefficient matrix is \[ A= \begin{pmatrix} 4 & k & 2\\ k & 4 & 1\\ 2 & 2 & 1 \end{pmatrix}. \] 
• Compute the determinant of \(A\): \[ \det(A) = 4 \begin{vmatrix} 4 & 1\\ 2 & 1 \end{vmatrix} - k \begin{vmatrix} k & 1\\ 2 & 1 \end{vmatrix} + 2 \begin{vmatrix} k & 4\\ 2 & 2 \end{vmatrix}. \] 
• Evaluate the \(2\times2\) determinants: \[ \det(A) = 4\left(4\cdot1-1\cdot2\right) - k\left(k\cdot1-1\cdot2\right) + 2\left(k\cdot2-4\cdot2\right). \] \[ = 4(2)-k(k-2)+2(2k-8). \] 
• Expand and simplify: \[ 8-k^2+2k+4k-16=0. \] \[ -k^2+6k-8=0. \] Multiply the entire equation by \(-1\): \[ k^2-6k+8=0. \] 
• Factorize the quadratic equation: \[ (k-2)(k-4)=0. \] Hence, \[ k=2 \quad\text{or}\quad k=4. \] Thus, there are two distinct values of \(k\). 
Step 4: Final Answer:
\[ \boxed{\text{The number of values of }k\text{ is }2.} \]