3 and 4 .Determinants and Matrices
hard

જો $\left| {\begin{array}{*{20}{c}}
{a - b - c}&{2a}&{2a}\\
{2b}&{b - c - a}&{2b}\\
{2c}&{2c}&{c - a - b}
\end{array}} \right|$ $ = \left( {a + b + c} \right)\,{\left( {x + a + b + c} \right)^2}$ , $x   \ne  0$ અને $a + b + c \ne 0$, તો $x$ મેળવો.

A

$abc$

B

$ - 2 \left( {a + b + c} \right)$ 

C

$ 2 \left( {a + b + c} \right)$

D

$ - \left( {a + b + c} \right)$

(JEE MAIN-2019)

Solution

${R_1} \to {R_1} + {R_2} + {R_3}$

$\left( {a + b + c} \right)\left| {\begin{array}{*{20}{c}}
1&1&1\\
{2b}&{b – a – c}&{2b}\\
{2c}&{2c}&{c – a – b}
\end{array}} \right|$

${c_3} \to {c_3} – {c_1},{c_2} \to {c_2} – {c_1}$

$ = \left( {a + b + c} \right)\left| {\begin{array}{*{20}{c}}
1&0&0\\
{2b}&{ – \left( {a + b + c} \right)}&0\\
{2c}&0&{ – \left( {a + b + c} \right)}
\end{array}} \right|$

$ = {\left( {a + b + c} \right)^3}$

$ = \left( {a + b + c} \right){\left( {x + a + b + c} \right)^2}$

Standard 12
Mathematics

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