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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
