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STD 12 - 3 and 4 . determinant and metrices — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 3 and 4 . determinant and metrices4 Marks
MCQ
The value of $x$  obtained from the equation $\left| {\,\begin{array}{*{20}{c}}{x + \alpha }&\beta &\gamma \\\gamma &{x + \beta }&\alpha \\\alpha &\beta &{x + \gamma }\end{array}\,} \right| = 0$ will be
  • ✓
    $0 $ and $ - (\alpha + \beta + \gamma )$
  • B
    $0$  and $(\alpha + \beta + \gamma )$
  • C
    $1$ and $(\alpha - \beta - \gamma )$
  • D
    $0 $ and $({\alpha ^2} + {\beta ^2} + {\gamma ^2})$
✓

Answer

Correct option: A.
$0 $ and $ - (\alpha + \beta + \gamma )$
Equation given, $\left| {\,\begin{array}{*{20}{c}}{x + \alpha + \beta + \gamma }&\beta &\gamma \\{x + \alpha + \beta + \gamma }&{x + \beta }&\alpha \\{x + \alpha + \beta + \gamma }&\beta &{x + \gamma }\end{array}\,} \right| = 0,$
$[{C_1} \to {C_1} + ({C_2} + {C_3})]$
or $(x + \alpha + \beta + \gamma )\,\left| {\,\begin{array}{*{20}{c}}1&\beta &\gamma \\1&{x + \beta }&\alpha \\1&\beta &{x + \gamma }\end{array}\,\,} \right|\, = 0$
or $(x + \alpha + \beta + \gamma )\,\left| {\,\begin{array}{*{20}{c}}1&\beta &\gamma \\0&x&{\alpha - \gamma }\\0&0&x\end{array}\,} \right|\, = 0,$
$\left[ {R_2} \to {R_2} - {R_1} {R_3} \to {R_3} - {R_1} \right]$
or $(x + \alpha + \beta + \gamma )[{x^2} - 0] = 0$
or ${x^2}(x + \alpha + \beta + \gamma ) = 0$
$\therefore $ $x = 0$ or $x = - (\alpha + \beta + \gamma )$.

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