$\left| {\,\begin{array}{*{20}{c}}{a + b}&{b + c}&{c + a}\\{b + c}&{c + a}&{a + b}\\{c + a}&{a + b}&{b + c}\end{array}\,} \right| = K\,\,\left| {\,\begin{array}{*{20}{c}}a&b&c\\b&c&a\\c&a&b\end{array}\,} \right|\,,$ तो $K = $
$\left| {\,\begin{array}{*{20}{c}}{a + b}&{b + c}&{c + a}\\{b + c}&{c + a}&{a + b}\\{c + a}&{a + b}&{b + c}\end{array}\,} \right| = K\,\,\left| {\,\begin{array}{*{20}{c}}a&b&c\\b&c&a\\c&a&b\end{array}\,} \right|\,,$ तो $K = $
- A
$1$
- B
$2$
- C
$3$
- D
$4$
Similar Questions
$2\,\,\left| {\,\begin{array}{*{20}{c}}1&1&1\\a&b&c\\{{a^2} - bc}&{{b^2} - ac}&{{c^2} - ab}\end{array}\,} \right| = $
$2\,\,\left| {\,\begin{array}{*{20}{c}}1&1&1\\a&b&c\\{{a^2} - bc}&{{b^2} - ac}&{{c^2} - ab}\end{array}\,} \right| = $
सारणिक $\left| {\,\begin{array}{*{20}{c}}0&{{b^3} - {a^3}}&{{c^3} - {a^3}}\\{{a^3} - {b^3}}&0&{{c^3} - {b^3}}\\{{a^3} - {c^3}}&{{b^3} - {c^3}}&0\end{array}\,} \right|$ का मान है
सारणिक $\left| {\,\begin{array}{*{20}{c}}0&{{b^3} - {a^3}}&{{c^3} - {a^3}}\\{{a^3} - {b^3}}&0&{{c^3} - {b^3}}\\{{a^3} - {c^3}}&{{b^3} - {c^3}}&0\end{array}\,} \right|$ का मान है
यदि $a, b, c$ समांतर श्रेढ़ी में हों तो सारणिक
$\left|\begin{array}{lll}x+2 & x+3 & x+2 a \\ x+3 & x+4 & x+2 b \\ x+4 & x+5 & x+2 c\end{array}\right|$ का मान होगा|:
यदि $a, b, c$ समांतर श्रेढ़ी में हों तो सारणिक
$\left|\begin{array}{lll}x+2 & x+3 & x+2 a \\ x+3 & x+4 & x+2 b \\ x+4 & x+5 & x+2 c\end{array}\right|$ का मान होगा|:
यदि $\left| {\,\begin{array}{*{20}{c}}{y + z}&{x - z}&{x - y}\\{y - z}&{z - x}&{y - x}\\{z - y}&{z - x}&{x + y}\end{array}\,} \right| = k\,xyz$,तो $ k$ का मान है
यदि $\left| {\,\begin{array}{*{20}{c}}{y + z}&{x - z}&{x - y}\\{y - z}&{z - x}&{y - x}\\{z - y}&{z - x}&{x + y}\end{array}\,} \right| = k\,xyz$,तो $ k$ का मान है
$\left| {\,\begin{array}{*{20}{c}}{x + 1}&{x + 2}&{x + 4}\\{x + 3}&{x + 5}&{x + 8}\\{x + 7}&{x + 10}&{x + 14}\end{array}\,} \right| = $
$\left| {\,\begin{array}{*{20}{c}}{x + 1}&{x + 2}&{x + 4}\\{x + 3}&{x + 5}&{x + 8}\\{x + 7}&{x + 10}&{x + 14}\end{array}\,} \right| = $