TS EAMCET · Maths · Determinants
\(\left|\begin{array}{ccc}x+2 & x+3 & x+5 \\ x+4 & x+6 & x+9 \\ x+8 & x+11 & x+15\end{array}\right|\) is equal to
- A \(3 x^2+4 x+5\)
- B \(x^3+8 x+2\)
- C \(0\)
- D \(-2\)
Answer & Solution
Correct Answer
(D) \(-2\)
Step-by-step Solution
Detailed explanation
Let \(\Delta=\left|\begin{array}{lll}x+2 & x+3 & x+5 \\ x+4 & x+6 & x+9 \\ x+8 & x+11 & x+15\end{array}\right|\) Apply operations \(R_2 \rightarrow R_2-R_1, R_3 \rightarrow R_3-R_1\), we get…
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