જો $x, y, z$ ભિન્ન હોય અને $\Delta=\left|\begin{array}{lll}x & x^{2} & 1+x^{2} \\ y & y^{2} & 1+y^{2} \\ z & z^{2} & 1+z^{2}\end{array}\right|=0$ હોય, તો સાબિત કરો કે $1+x y z=0$.
જો $x, y, z$ ભિન્ન હોય અને $\Delta=\left|\begin{array}{lll}x & x^{2} & 1+x^{2} \\ y & y^{2} & 1+y^{2} \\ z & z^{2} & 1+z^{2}\end{array}\right|=0$ હોય, તો સાબિત કરો કે $1+x y z=0$.
We have $\Delta=\left|\begin{array}{lll}x & x^{2} & 1+x^{3} \\ y & y^{2} & 1+y^{3} \\ z & z^{2} & 1+z^{3}\end{array}\right|$
$=\left|\begin{array}{lll}x & x^{2} & 1 \\ y & y^{2} & 1 \\ z & z^{2} & 1\end{array}\right|+\left|\begin{array}{lll}x & x^{2} & x^{3} \\ y & y^{2} & y^{3} \\ z & z^{2} & z^{3}\end{array}\right|$
$=(-1)^{2}\left|\begin{array}{ccc}1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2}\end{array}\right|+x y z\left|\begin{array}{ccc}1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2}\end{array}\right| \quad\left(\text { Using } \mathrm{C}_{3} \leftrightarrow \mathrm{C}_{2} \text { and then } \mathrm{C}_{1} \leftrightarrow \mathrm{C}_{2}\right)$
$=\left|\begin{array}{lll}1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2}\end{array}\right|(1+x y z)$
$=(1+x y z)\left|\begin{array}{ccc}
1 & x & x^{2} \\
0 & y-x & y^{2}-x^{2} \\
0 & z-x & z^{2}-x^{2}
\end{array}\right| $ $( { Using } \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1} \text { and } \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1})$
Taking out common factor $(y-x)$ from $\mathrm{R}_{2}$ and $(z-x)$ from $\mathrm{R}_{3},$ we get
$\Delta=(1+x y z)(y-x)(z-x)\left|\begin{array}{ccc}
1 & x & x^{2} \\
0 & 1 & y+x \\
0 & 1 & z+x
\end{array}\right|$
$=(1+x y z)(y-x)(z-x)(z-y)$ (on expanding along $\mathrm{C}_{1}$ )
since $\Delta=0$ and $x, y, z$ are all different, i.e., $x-y \neq 0, y-z \neq 0, z-x \neq 0,$ we get $1+x y z=0$
Similar Questions
$\left| {\,\begin{array}{*{20}{c}}1&1&1\\{\cos (nx)}&{\cos (n + 1)x}&{\cos (n + 2)x}\\{\sin (nx)}&{\sin (n + 1)x}&{\sin (n + 2)x}\end{array}\,} \right|$ એ . . . પર આધારિત નથી .
$\left| {\,\begin{array}{*{20}{c}}1&1&1\\{\cos (nx)}&{\cos (n + 1)x}&{\cos (n + 2)x}\\{\sin (nx)}&{\sin (n + 1)x}&{\sin (n + 2)x}\end{array}\,} \right|$ એ . . . પર આધારિત નથી .
$\left| {\,\begin{array}{*{20}{c}}{{b^2} + {c^2}}&{{a^2}}&{{a^2}}\\{{b^2}}&{{c^2} + {a^2}}&{{b^2}}\\{{c^2}}&{{c^2}}&{{a^2} + {b^2}}\end{array}\,} \right| = $
$\left| {\,\begin{array}{*{20}{c}}{{b^2} + {c^2}}&{{a^2}}&{{a^2}}\\{{b^2}}&{{c^2} + {a^2}}&{{b^2}}\\{{c^2}}&{{c^2}}&{{a^2} + {b^2}}\end{array}\,} \right| = $
- [IIT 1980]
$\left| {\,\begin{array}{*{20}{c}}1&1&1\\{bc}&{ca}&{ab}\\{b + c}&{c + a}&{a + b}\end{array}\,} \right|$ =
$\left| {\,\begin{array}{*{20}{c}}1&1&1\\{bc}&{ca}&{ab}\\{b + c}&{c + a}&{a + b}\end{array}\,} \right|$ =
જો $\left| {\,\begin{array}{*{20}{c}}{x + \alpha }&\beta &\gamma \\\gamma &{x + \beta }&\alpha \\\alpha &\beta &{x + \gamma }\end{array}\,} \right| = 0$ તો $x$ મેળવો.
જો $\left| {\,\begin{array}{*{20}{c}}{x + \alpha }&\beta &\gamma \\\gamma &{x + \beta }&\alpha \\\alpha &\beta &{x + \gamma }\end{array}\,} \right| = 0$ તો $x$ મેળવો.
નિશ્ચાયકના ગુણધર્મનો ઉપયોગ કરી સાબિત કરો કે, $\left| {\begin{array}{*{20}{l}}
{\sin \alpha }&{\cos \alpha }&{\cos (\alpha + \delta )} \\
{\sin \beta }&{\cos \beta }&{\cos (\beta + \delta )} \\
{\sin \gamma }&{\cos \gamma }&{\cos (\gamma + \delta )}
\end{array}} \right| = 0$
નિશ્ચાયકના ગુણધર્મનો ઉપયોગ કરી સાબિત કરો કે, $\left| {\begin{array}{*{20}{l}}
{\sin \alpha }&{\cos \alpha }&{\cos (\alpha + \delta )} \\
{\sin \beta }&{\cos \beta }&{\cos (\beta + \delta )} \\
{\sin \gamma }&{\cos \gamma }&{\cos (\gamma + \delta )}
\end{array}} \right| = 0$