નિશ્ચાયકના ગુણધર્મનો ઉપયોગ કરીને સાબિત કરો : $\left|\begin{array}{lll}x & x^{2} & y z \\ y & y^{2} & z x \\ z & z^{2} & x y\end{array}\right|=(x-y)(y-z)(z-x)(x y+y z+z x)$
નિશ્ચાયકના ગુણધર્મનો ઉપયોગ કરીને સાબિત કરો : $\left|\begin{array}{lll}x & x^{2} & y z \\ y & y^{2} & z x \\ z & z^{2} & x y\end{array}\right|=(x-y)(y-z)(z-x)(x y+y z+z x)$
Let $\Delta=\left|\begin{array}{lll}x & x^{2} & y z \\ y & y^{2} & z x \\ z & z^{2} & x y\end{array}\right|$
Applying $R_{2} \rightarrow R_{2}-R_{1},$ and $R_{3} \rightarrow R_{3}-R_{1},$ we have:
$\Delta=\left|\begin{array}{ccc}x & x^{2} & y z \\ y-x & y^{2}-x^{2} & z x-y z \\ z-x & z^{2}-x^{2} & x y-y z\end{array}\right|$
$=\left|\begin{array}{ccc}x & x^{2} & y z \\ -(x-y) & -(x-y)(x+y) & z(x-y) \\ (z-x) & (z-x)(z+x) & -y(z-x)\end{array}\right|$
$=(x-y)(z-x)\left|\begin{array}{ccc}x & x^{2} & y z \\ -1 & -x-y & z \\ 1 & z-y & z-y\end{array}\right|$
Applying $R_{3} \rightarrow R_{3}+R_{2},$ we have:
$\Delta=(x-y)(z-x)\left|\begin{array}{ccc}x & x^{2} & y z \\ -1 & -x-y & z \\ 1 & z-y & z-y\end{array}\right|$
$=(x-y)(z-x)(z-y)\left|\begin{array}{ccc}x & x^{2} & y z \\ -1 & -x-y & z \\ 0 & 1 & 1\end{array}\right|$
Expanding along $R_{3},$ we have:
$\Delta=[(x-y)(z-x)(z-y)]\left[(-1)\left|\begin{array}{cc}x & y z \\ -1 & z\end{array}\right|+1\left|\begin{array}{cc}x & x^{2} \\ -1 & -x-y\end{array}\right|\right]$
$=(x-y)(z-x)(z-y)\left[(-x z-y z)+\left(-x^{2}-x y+x^{2}\right)\right]$
$=-(x-y)(z-x)(z-y)(x y+y z+z x)$
$=(x-y)(y-z)(z-x)(x y+y z+z x)$
Hence, the given result is proved.
Similar Questions
$x$ ની . . . કિમત માટે $\left| {\,\begin{array}{*{20}{c}}{x + {\omega ^2}}&\omega &1\\\omega &{{\omega ^2}}&{1 + x}\\1&{x + \omega }&{{\omega ^2}}\end{array}\,} \right| = 0$ થાય.
$x$ ની . . . કિમત માટે $\left| {\,\begin{array}{*{20}{c}}{x + {\omega ^2}}&\omega &1\\\omega &{{\omega ^2}}&{1 + x}\\1&{x + \omega }&{{\omega ^2}}\end{array}\,} \right| = 0$ થાય.
$\left|\begin{array}{ccc}\cos \alpha \cos \beta & \cos \alpha \operatorname{csin} \beta & -\sin \alpha \\ -\sin \beta & \cos \beta & 0 \\ \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha\end{array}\right|$ નું મૂલ્ય શોધો.
$\left|\begin{array}{ccc}\cos \alpha \cos \beta & \cos \alpha \operatorname{csin} \beta & -\sin \alpha \\ -\sin \beta & \cos \beta & 0 \\ \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha\end{array}\right|$ નું મૂલ્ય શોધો.
નિશ્ચાયકના ગુણધર્મનો ઉપયોગ કરીને સાબિત કરો : $\left|\begin{array}{ccc}x+4 & 2 x & 2 x \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4\end{array}\right|=(5 x+4)(4-x)^{2}$
નિશ્ચાયકના ગુણધર્મનો ઉપયોગ કરીને સાબિત કરો : $\left|\begin{array}{ccc}x+4 & 2 x & 2 x \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4\end{array}\right|=(5 x+4)(4-x)^{2}$
જો $a,b,c$ એ ભિન્ન અને સંમેય સંખ્યા હોય તો $\left| {\begin{array}{*{20}{c}}
{\left( {{a^2} + {b^2} + {c^2}} \right)}&{ab + bc + ca}&{ab + bc + ca}\\
{ab + bc + ca}&{\left( {{a^2} + {b^2} + {c^2}} \right)}&{\left( {bc + ca + ab} \right)}\\
{ab + bc + ca}&{\left( {ab + bc + ca} \right)}&{\left( {{a^2} + {b^2} + {c^2}} \right)}
\end{array}} \right|$ એ હંમેશા..
જો $a,b,c$ એ ભિન્ન અને સંમેય સંખ્યા હોય તો $\left| {\begin{array}{*{20}{c}}
{\left( {{a^2} + {b^2} + {c^2}} \right)}&{ab + bc + ca}&{ab + bc + ca}\\
{ab + bc + ca}&{\left( {{a^2} + {b^2} + {c^2}} \right)}&{\left( {bc + ca + ab} \right)}\\
{ab + bc + ca}&{\left( {ab + bc + ca} \right)}&{\left( {{a^2} + {b^2} + {c^2}} \right)}
\end{array}} \right|$ એ હંમેશા..
નિશ્ચાયકના ગુણધર્મનો ઉપયોગ કરીને સાબિત કરો : $\left|\begin{array}{ccc}a^{2}+1 & a b & a c \\ a b & b^{2}+1 & b c \\ c a & c b & c^{2}+1\end{array}\right|=1+a^{2}+b^{2}+c^{2}$
નિશ્ચાયકના ગુણધર્મનો ઉપયોગ કરીને સાબિત કરો : $\left|\begin{array}{ccc}a^{2}+1 & a b & a c \\ a b & b^{2}+1 & b c \\ c a & c b & c^{2}+1\end{array}\right|=1+a^{2}+b^{2}+c^{2}$