MCQ
જો$\begin{vmatrix}a&a^2&1+a^3\\b&b^2&1+b^3\\c&c^2&1+c^3\end{vmatrix}=0$ અને સદિશો$( {1,a,{a^2}} );( {1,b,{b^2}} )$અને$( {1,c,{c^2}} )$ એ અસમતલીય હોય, તો $abc = ..........$
- ✓$ - 1$
- B$1$
- C$0$
- D$2$
$\begin{vmatrix} {a}& a^2& {1+a^3} \\ b & b^2 & 1+b^3 \\ c&c^2 & 1+c^3 \end{vmatrix} = \begin{vmatrix} {a} & {a^2} & {1} \\ b & b^2 & 1 \\ c&c^2 & 1 \end{vmatrix}+\begin{vmatrix} {a} & {a^2} & {a^3} \\b & b^2 & b^3 \\ c&c^2 & c^3 \end{vmatrix}$
$ \underrightarrow {\ \ \ \ C_{13}\ \ \ \ }\begin{vmatrix} {1} & {a^2} & {a} \\ 1 & b^2 & b \\ 1 &c^2 & c \end{vmatrix} +abc \begin{vmatrix} {1} & {a} & {a^2} \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix}$
$\underrightarrow {\ \ \ \ C_{23}\ \ \ \ } \begin{vmatrix} {1} & {a} & {a^2} \\ 1 & b & b^2 \\ 1 &c & c^2 \end{vmatrix} +abc \begin{vmatrix} {1} & {a} & {a^2} \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix}$
$\therefore (abc+1)\begin{vmatrix}{1} & {a} & {a^2} \\1 & b & b^2 \\ 1 & c & c^2\end{vmatrix} ={{0}}$
$ \therefore abc+1={{0}}$
$ abc = -1$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.