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The set of values of K for which the system of equations [2314501K3][xyz]=[567]\begin{bmatrix} 2 & 3 & 1 \\ 4 & 5 & 0 \\ 1 & K & 3 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 5 \\ 6 \\ 7 \end{bmatrix} gives a unique solution is :

Solution

✅ Correct Option: 4

Determinant of coefficient matrix: 2(15)−3(12)+1(4K−5)=30−36+4K−5=4K−112(15)-3(12)+1(4K-5)=30-36+4K-5=4K-11.

For unique solution, determinant must be nonzero: K≠11/4K\neq 11/4.

So K∈R−{11/4}K\in\mathbb{R}-\{11/4\}.

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