જો $a,b,c$ એ અસમાન હોય અને $\left| {\,\begin{array}{*{20}{c}}a&{{a^2}}&{{a^3} - 1}\\b&{{b^2}}&{{b^3} - 1}\\c&{{c^2}}&{{c^3} - 1}\end{array}\,} \right| = 0$ તો . . .
જો $a,b,c$ એ અસમાન હોય અને $\left| {\,\begin{array}{*{20}{c}}a&{{a^2}}&{{a^3} - 1}\\b&{{b^2}}&{{b^3} - 1}\\c&{{c^2}}&{{c^3} - 1}\end{array}\,} \right| = 0$ તો . . .
- A
$a + b + c = 0$
- B
$abc = 1$
- C
$a + b + c = 1$
- D
$ab + bc + ca = 0$
Similar Questions
સમીકરણ $\left|\begin{array}{ccc}1+\sin ^{2} x & \sin ^{2} x & \sin ^{2} x \\ \cos ^{2} x & 1+\cos ^{2} x & \cos ^{2} x \\ 4 \sin 2 x & 4 \sin 2 x & 1+4 \sin 2 x\end{array}\right|=0,(0< x< \pi) $ નો ઉકેલ મેળવો.
સમીકરણ $\left|\begin{array}{ccc}1+\sin ^{2} x & \sin ^{2} x & \sin ^{2} x \\ \cos ^{2} x & 1+\cos ^{2} x & \cos ^{2} x \\ 4 \sin 2 x & 4 \sin 2 x & 1+4 \sin 2 x\end{array}\right|=0,(0< x< \pi) $ નો ઉકેલ મેળવો.
- [JEE MAIN 2021]
$\left| {\,\begin{array}{*{20}{c}}{265}&{240}&{219}\\{240}&{225}&{198}\\{219}&{198}&{181}\end{array}\,} \right|$ =
$\left| {\,\begin{array}{*{20}{c}}{265}&{240}&{219}\\{240}&{225}&{198}\\{219}&{198}&{181}\end{array}\,} \right|$ =
સાબિત કરો કે $\left|\begin{array}{ccc}1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c\end{array}\right|=a b c\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=a b c+b c+c a+a b$
સાબિત કરો કે $\left|\begin{array}{ccc}1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c\end{array}\right|=a b c\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=a b c+b c+c a+a b$
$\left| {\,\begin{array}{*{20}{c}}{{a^2}}&{{b^2}}&{{c^2}}\\{{{(a + 1)}^2}}&{{{(b + 1)}^2}}&{{{(c + 1)}^2}}\\{{{(a - 1)}^2}}&{{{(b - 1)}^2}}&{{{(c - 1)}^2}}\end{array}\,} \right| = $
$\left| {\,\begin{array}{*{20}{c}}{{a^2}}&{{b^2}}&{{c^2}}\\{{{(a + 1)}^2}}&{{{(b + 1)}^2}}&{{{(c + 1)}^2}}\\{{{(a - 1)}^2}}&{{{(b - 1)}^2}}&{{{(c - 1)}^2}}\end{array}\,} \right| = $
$\left| {\,\begin{array}{*{20}{c}}4&{ - 6}&1\\{ - 1}&{ - 1}&1\\{ - 4}&{11}&{ - 1\,}\end{array}} \right|= . . . $
$\left| {\,\begin{array}{*{20}{c}}4&{ - 6}&1\\{ - 1}&{ - 1}&1\\{ - 4}&{11}&{ - 1\,}\end{array}} \right|= . . . $