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Matrices & Determinants: Equation Pmatrix Pmatrix Pmatrix Pmatrix Pmatrix Pmatrix Sol

JEE Maths question with a full step-by-step solution.

Question
The equation (12213434k)(xyz)=(000)\begin{pmatrix} 1 & 2 & 2 \\ 1 & 3 & 4 \\ 3 & 4 & k \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} has a solution (x,y,z)(x, y, z) besides (0,0,0)(0, 0, 0). The value of kk equals:
A00
B11
C22correct
D33
Solution
Step 1: Condition for a non-trivial solution A homogeneous system has a non-trivial solution if and only if the coefficient determinant is zero:
∣12213434k∣=0\begin{vmatrix} 1 & 2 & 2 \\ 1 & 3 & 4 \\ 3 & 4 & k \end{vmatrix} = 0
 Step 2: Expand along the first row
1(3k−16)−2(k−12)+2(4−9)=01(3k - 16) - 2(k - 12) + 2(4 - 9) = 0
3k−16−2k+24−10=0  ⟹  k−2=0  ⟹  k=23k - 16 - 2k + 24 - 10 = 0 \implies k - 2 = 0 \implies k = 2
 Answer: (3)
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