Question:
If the system of equations \(x + 2y – 3z = 1\), \((k+3)z = 3\), \((2k+1)x + z = 0\) is inconsistent, then the value of \(k\) is:
यदि समीकरण निकाय \(x + 2y – 3z = 1\), \((k+3)z = 3\), \((2k+1)x + z = 0\) असंगत है, तो \(k\) का मान है:
Options:
- -3
- 1/2
- 0
- 2
- -3
- 1/2
- 0
- 2
Correct Answer: a)
Explanation:
In matrix form, the system of equations has coefficient determinant:
\[\Delta = \begin{vmatrix} 1 & 2 & -3 \\ 0 & 0 & k+3 \\ 2k+1 & 0 & 1 \end{vmatrix}\]
Expanding along column 2:
\[\Delta = -2 \begin{vmatrix} 0 & k+3 \\ 2k+1 & 1 \end{vmatrix} = -2 [0 – (k+3)(2k+1)] = 2(k+3)(2k+1)\]
For the system to be inconsistent, \(\Delta = 0 \implies k = -3\) or \(k = -1/2\).
For \(k = -3\), equation 2 becomes \(0 \cdot z = 3\), which is impossible (inconsistent).
Hence, the value of \(k\) is \(-3\).
Hence, the correct option is (a).
गुणांक सारणिक:
\[\Delta = 2(k+3)(2k+1) = 0 \implies k = -3 \text{ या } k = -1/2\]
\(k = -3\) रखने पर द्वितीय समीकरण \(0 \cdot z = 3\) बन जाता है, जो सर्वथा असंभव (असंगत) है।
अतः \(k = -3\)।
अतः सही विकल्प (a) है।