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समीकरण $\left| {\,\begin{array}{*{20}{c}}{x + \alpha }&\beta &\gamma \\\gamma &{x + \beta }&\alpha \\\alpha &\beta &{x + \gamma }\end{array}\,} \right| = 0$ से प्राप्त $x$ के मान होंगे
$0 $ और $ - (\alpha + \beta + \gamma )$
$0 $ और $(\alpha + \beta + \gamma )$
$1$ और $(\alpha - \beta - \gamma )$
$0 $ और $({\alpha ^2} + {\beta ^2} + {\gamma ^2})$
Solution
(a) दिया गया समीकरण है $\left| {\,\begin{array}{*{20}{c}}{x + \alpha + \beta + \gamma }&\beta &\gamma \\{x + \alpha + \beta + \gamma }&{x + \beta }&\alpha \\{x + \alpha + \beta + \gamma }&\beta &{x + \gamma }\end{array}\,} \right| = 0$,
$[{C_1} \to {C_1} + ({C_2} + {C_3})]$
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 $(x + \alpha + \beta + \gamma )\,\left| {\,\begin{array}{*{20}{c}}1&\beta &\gamma \\0&x&{\alpha – \gamma }\\0&0&x\end{array}\,} \right|\, = \,0$,
$\left[ \begin{array}{l}{R_2} \to {R_2} – {R_1}\\{R_3} \to {R_3} – {R_1}\end{array} \right]$
or $(x + \alpha + \beta + \gamma )[{x^2} – 0] = 0$
or ${x^2}(x + \alpha + \beta + \gamma ) = 0$
$\therefore $ $x = 0$ or $x = – (\alpha + \beta + \gamma )$.
