JEE Mains · Maths · STD 12 - 3 and 4 . metrices and determinant
The system of equations \(kx + y + z =1\) \(x + ky + z = k\) and \(x + y + zk = k ^{2}\) has no solution if \(k\) is equal to
- A \(0\)
- B \(1\)
- C \(-1\)
- D \(-2\)
Answer & Solution
Correct Answer
(D) \(-2\)
Step-by-step Solution
Detailed explanation
\(k x+y+z=1\) \(x+k y+z=k\) \(x+y+z k=k^{2}\) \(\Delta=\left|\begin{array}{ccc} K & 1 & 1 \\ 1 & K & 1 \\ 1 & 1 & K \end{array}\right|= K \left( K ^{2}-1\right)-1( K -1)+1(1- K )\) \(= K ^{3}- K - K +1+1- K\) \(= K ^{3}-3 K +2\) \(=( K -1)^{2}( K +2)\) For \(K =1\)…
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