3 and 4 .Determinants and MatricesMedium

सारणिकों के गुणधर्मों का प्रयोग करके सिद्ध कीजिए :

$\left|\begin{array}{ccc}-a^{2} & a b & a c \\ b a & -b^{2} & b c \\ c a & c b & -c^{2}\end{array}\right|=4 a^{2} b^{2} c^{2}$

Solution

$\Delta=\left|\begin{array}{ccc}-a^{2} & a b & a c \\ b a & -b^{2} & b c \\ c a & c b & -c^{2}\end{array}\right|$

$=a b c\left|\begin{array}{ccc}-a & b & c \\ a & -b & c \\ a & b & -c\end{array}\right|$

Taking out factors $ a,\ b,\ c $ from $ R_{1}, R_{2}$ and $R_{3}$

$=a^{2} b^{2} c^{2}\left|\begin{array}{ccc}-1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1\end{array}\right| $

[ Taking out factors $ a,\ b,\ c$ from $ C_{1}, C_{2} $ and $C_{3}]$

Applying $R_{2} \rightarrow R_{2}+R_{1}$ and $R_{3} \rightarrow R_{3}+R_{1},$ we have:

$\Delta=a^{2} b^{2} c^{2}\left|\begin{array}{ccc}-1 & 1 & 1 \\ 0 & 0 & 2 \\ 0 & 2 & 0\end{array}\right|$

$ = {a^2}{b^2}{c^2}( - 1)\left| {\begin{array}{*{20}{c}}
0&2 \\
2&0
\end{array}} \right|$

$ = - {a^2}{b^2}{c^2}(0 - 4) = 4{a^2}{b^2}{c^2}$

यदि $f(x) = \left| {\begin{array}{*{20}{c}}{x - 3}&{2{x^2} - 18}&{3{x^3} - 81}\\{x - 5}&{2{x^2} - 50}&{4{x^3} - 500}\\1&2&3\end{array}} \right|$ then $f(1).f(3) + f(3).f(5) + f(5).f(1)$=

Difficult

Medium

Easy

Easy

Medium