JEE Mains · Maths · STD 12 - 3 and 4 . metrices and determinant
If \(\left| {\begin{array}{*{20}{c}}
{a - b - c}&{2a}&{2a}\\
{2b}&{b - c - a}&{2b}\\
{2c}&{2c}&{c - a - b}
\end{array}} \right|\) \( = \left( {a + b + c} \right)\,{\left( {x + a + b + c} \right)^2}\) , \(x \ne 0\) and \(a + b + c \ne 0\), then \(x\) is equal to
- A \(abc\)
- B \( - 2 \left( {a + b + c} \right)\)
- C \( 2 \left( {a + b + c} \right)\)
- D \( - \left( {a + b + c} \right)\)
Answer & Solution
Correct Answer
(B) \( - 2 \left( {a + b + c} \right)\)
Step-by-step Solution
Detailed explanation
\({R_1} \to {R_1} + {R_2} + {R_3}\) \(\left( {a + b + c} \right)\left| {\begin{array}{*{20}{c}} 1&1&1\\ {2b}&{b - a - c}&{2b}\\ {2c}&{2c}&{c - a - b} \end{array}} \right|\) \({c_3} \to {c_3} - {c_1},{c_2} \to {c_2} - {c_1}\)…
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