નિશ્ચાયકના ગુણધર્મનો ઉપયોગ કરીને સાબિત કરો : $\left|\begin{array}{lll}1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2}\end{array}\right|=(a+b)(b-c)(c-a)$
Let $\Delta=\left|\begin{array}{lll}1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2}\end{array}\right|$
Applying $R_{1} \rightarrow R_{1}-R_{3}$ and $R_{2} \rightarrow R_{2}-R_{3},$ we have:
$\Delta=\left|\begin{array}{ccc}0 & a-c & a^{2}-c^{2} \\ 0 & b-c & b^{2}-c^{2} \\ 1 & c & c^{2}\end{array}\right|$
$=(c-a)(b-c)\left|\begin{array}{ccc}0 & -1 & -a-c \\ 0 & 1 & b+c \\ 1 & c & c^{2}\end{array}\right|$
Applying $R_{1} \rightarrow R_{1}+R_{2},$ we have:
$\Delta=(b-c)(c-a)\left|\begin{array}{ccc}0 & 0 & -a+b \\ 0 & 1 & b+c \\ 1 & c & c^{2}\end{array}\right|$
$=(a-b)(b-c)(c-a)\left|\begin{array}{ccc}0 & 0 & -1 \\ 0 & 1 & b+c \\ 1 & c & c^{2}\end{array}\right|$
Expanding along $\mathrm{C}_{1},$ we have:
$\Delta=(a-b)(b-c)(c-a)\left|\begin{array}{cc}0 & -1 \\ 1 & b+c\end{array}\right|=(a-b)(b-c)(c-a)$
Hence, the given result is proved.
જો ${D_r} = \left| {\begin{array}{*{20}{c}}{{2^{r - 1}}}&{{{2.3}^{r - 1}}}&{{{4.5}^{r - 1}}}\\x&y&z\\{{2^n} - 1}&{{3^n} - 1}&{{5^n} - 1}\end{array}} \right|$, તો $\sum\limits_{r = 1}^n {{D_r} = } $
સાબિત કરો કે $\left|\begin{array}{ccc}a & b & c \\ a+2 x & b+2 y & c+2 z \\ x & y & z\end{array}\right|=0$
જો $\mathrm{a, b, c}$ પૈકી પ્રત્યેક બે અસમાન અને પ્રત્યેક ધન હોય, તો સાબિત કરો કે નિશ્ચાયક $\Delta=\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|$ નું મૂલ્ય ઋણ છે.
$\left|\begin{array}{ccc}\cos \alpha \cos \beta & \cos \alpha \operatorname{csin} \beta & -\sin \alpha \\ -\sin \beta & \cos \beta & 0 \\ \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha\end{array}\right|$ નું મૂલ્ય શોધો.
જો $\left| {\,\begin{array}{*{20}{c}}{x + 1}&{x + 2}&{x + 3}\\{x + 2}&{x + 3}&{x + 4}\\{x + a}&{x + b}&{x + c}\end{array}\,} \right| = 0$, તો $a,b,c$ એ . . . શ્રેણીમાં છે.