JEE Mains · Maths · STD 12 - 3 and 4 . metrices and determinant
Let \(a, b, c, d\) be in arithmetic progression with common difference \(\lambda\). If \(\left|\begin{array}{lll} x+a-c & x+b & x+a \\ x-1 & x+c & x+b \\ x-b+d & x+d & x+c \end{array}\right|=2\) then value of \(\lambda^{2}\) is equal to \(.....\)
- A \(4\)
- B \(1\)
- C \(9\)
- D \(16\)
Answer & Solution
Correct Answer
(B) \(1\)
Step-by-step Solution
Detailed explanation
\(\left|\begin{array}{lll}x+a-c & x+b & x+a \\ x-1 & x+c & x+b \\ x-b+d & x+d & x+c\end{array}\right|=2\) \(C_{2} \rightarrow C_{2}-C_{3}\) \(\left|\begin{array}{ccc}x-2 \lambda & \lambda & x+a \\ x-1 & \lambda & x+b \\ x+2 \lambda & \lambda & x+C\end{array}\right|=2\)…
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