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By using properties of determinants, show that:
$\left|\begin{array}{ccc}-a^{2} & a b & a c \\ b a & -b^{2} & b c \\ c a & c b & -c^{2}\end{array}\right|=4 a^{2} b^{2} c^{2}$
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}$
