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નિશ્ચાયકના ગુણધર્મનો ઉપયોગ કરી સાબિત કરો કે, $\left|\begin{array}{lll}x & x^{2} & 1+p x^{3} \\ y & y^{2} & 1+p y^{3} \\ z & z^{2} & 1+p z^{3}\end{array}\right|=(1+p x y z)(x-y)(y-z)(z-x),$ $p$ અચળ છે.
Solution
$\Delta=\left|\begin{array}{lll}x & x^{2} & 1+p x^{3} \\ y & y^{2} & 1+p y^{3} \\ z & z^{2} & 1+p z^{3}\end{array}\right|$
Applying $R_{2} \rightarrow R_{2}-R_{1}$ and $R_{3} \rightarrow R_{3}-R_{1},$ we have:
$\Delta=\left|\begin{array}{ccc}x & x^{2} & 1+p x^{3} \\ y-x & y^{2}-x^{2} & p\left(y^{3}-x^{3}\right) \\ z-x & z^{2}-x^{2} & p\left(z^{3}-x^{3}\right)\end{array}\right|$
$=(y-x)(z-x)\left|\begin{array}{ccc}
x & x^{2} & 1+p x^{3} \\
1 & y+x & p\left(y^{2}+x^{2}+x y\right) \\
1 & z+x & p\left(z^{2}+x^{2}+x z\right)
\end{array}\right|$
Applying $R_{3} \rightarrow R_{3}-R_{2},$ we have:
${\Delta = (y – x)(z – x)\left| {\begin{array}{*{20}{c}}
x&{{x^2}}&{1 + p{x^3}} \\
1&{y + x}&{p\left( {{y^2} + {x^2} + xy} \right)} \\
0&{z – y}&{p(z – y)(x + y + z)}
\end{array}} \right|}$
${ = (y – x)(z – x)(z – y)\left| {\begin{array}{*{20}{c}}
x&{{x^2}}&{1 + p{x^3}} \\
1&{y + x}&{p\left( {{y^2} + {x^2} + xy} \right)} \\
0&1&{p(x + y + z)}
\end{array}} \right|}$
Expanding along $R_{3},$ we have:
$\Delta=(x-y)(y-z)(z-x)\left[(-1)(p)\left(x y^{2}+x^{3}+x^{2} y\right)\right.$ $\left. { + 1 + p{x^3} + p(x + y + z)(xy)} \right]$
$ = (x – y)(y – z)(z – x)$ $\left[ { – px{y^2} – p{x^3} – p{x^2}y} \right.\left. { + 1 + p{x^3} + p{x^2}y + px{y^2} + pxyz} \right]$
$ = (x – y)(y – z)(z – x)(1 + pxyz)$
Hence, the given result is proved.