3 and 4 .Determinants and Matrices
easy

સમીકરણ $\left| {\,\begin{array}{*{20}{c}}{x + a}&b&c\\b&{x + c}&a\\c&a&{x + b}\end{array}\,} \right| = 0$ નું કોઈ એક બીજ મેળવો.

A

$ - (a + b)$

B

$ - (b + c)$

C

$ - a$

D

$ - (a + b + c)$

Solution

(d) $\left| {\,\begin{array}{*{20}{c}}{x + a}&b&c\\b&{x + c}&a\\c&a&{x + b}\end{array}\,} \right| = 0$

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

                                                                                       ${C_1} → {C_1}+{C_2}+{C_3}$

$ \Rightarrow $ $x = – (a + b + c)$ is one of the root of the equation.

Standard 12
Mathematics

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