By using properties of determinants, show that:
$\left|\begin{array}{ccc}-a^{2} & a b & a c \\ b a & -b^{2} & b c \\ c a & c b & -c^{2}\end{array}\right|=4 a^{2} b^{2} c^{2}$
By using properties of determinants, show that:
$\left|\begin{array}{ccc}-a^{2} & a b & a c \\ b a & -b^{2} & b c \\ c a & c b & -c^{2}\end{array}\right|=4 a^{2} b^{2} c^{2}$
$\Delta=\left|\begin{array}{ccc}-a^{2} & a b & a c \\ b a & -b^{2} & b c \\ c a & c b & -c^{2}\end{array}\right|$
$=a b c\left|\begin{array}{ccc}-a & b & c \\ a & -b & c \\ a & b & -c\end{array}\right|$
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}$
Similar Questions
Using properties of determinants, prove that:
$\left|\begin{array}{ccc}3 a & -a+b & -a+c \\ -b+a & 3 b & -b+c \\ -c+a & -c+b & 3 c\end{array}\right|=3(a+b+c)(a b+b c+c a)$
Using properties of determinants, prove that:
$\left|\begin{array}{ccc}3 a & -a+b & -a+c \\ -b+a & 3 b & -b+c \\ -c+a & -c+b & 3 c\end{array}\right|=3(a+b+c)(a b+b c+c a)$
If $\mathrm{a, b, c}$ are in $\mathrm{A.P}$, find value of
$\left|\begin{array}{ccc}
2 y+4 & 5 y+7 & 8 y+a \\
3 y+5 & 6 y+8 & 9 y+b \\
4 y+6 & 7 y+9 & 10 y+c
\end{array}\right|$
If $\mathrm{a, b, c}$ are in $\mathrm{A.P}$, find value of
$\left|\begin{array}{ccc}
2 y+4 & 5 y+7 & 8 y+a \\
3 y+5 & 6 y+8 & 9 y+b \\
4 y+6 & 7 y+9 & 10 y+c
\end{array}\right|$
If $\left| {\begin{array}{*{20}{c}} {a - b}&{b - c}&{c - a} \\ {b - c}&{c - a}&{a - b} \\ {c - a + 1}&{a - b}&{b - c} \end{array}} \right| = 0$ ,$\left( {a,b,c \in R - \left\{ 0 \right\}} \right),$ then
If $\left| {\begin{array}{*{20}{c}} {a - b}&{b - c}&{c - a} \\ {b - c}&{c - a}&{a - b} \\ {c - a + 1}&{a - b}&{b - c} \end{array}} \right| = 0$ ,$\left( {a,b,c \in R - \left\{ 0 \right\}} \right),$ then
The value of the determinant $\left| {\,\begin{array}{*{20}{c}}{31}&{37}&{92}\\{31}&{58}&{71}\\{31}&{105}&{24}\end{array}\,} \right|$ is
The value of the determinant $\left| {\,\begin{array}{*{20}{c}}{31}&{37}&{92}\\{31}&{58}&{71}\\{31}&{105}&{24}\end{array}\,} \right|$ is
Let $\alpha $, $\beta$ $\gamma$, $\delta$ are distinct imaginary roots of
$z^5=1$ then value of $\left| {\begin{array}{*{20}{c}}
{{e^\alpha }}&{{e^{2\alpha }}}&{{e^{3\alpha + 1}}}&{ - {e^{ - \delta }}} \\
{{e^\beta }}&{{e^{2\beta }}}&{{e^{3\beta + 1}}}&{ - {e^{ - \delta }}} \\
{{e^\gamma }}&{{e^{2\gamma }}}&{{e^{3\gamma + 1}}}&{ - {e^{ - \delta }}}
\end{array}} \right|$
Let $\alpha $, $\beta$ $\gamma$, $\delta$ are distinct imaginary roots of
$z^5=1$ then value of $\left| {\begin{array}{*{20}{c}}
{{e^\alpha }}&{{e^{2\alpha }}}&{{e^{3\alpha + 1}}}&{ - {e^{ - \delta }}} \\
{{e^\beta }}&{{e^{2\beta }}}&{{e^{3\beta + 1}}}&{ - {e^{ - \delta }}} \\
{{e^\gamma }}&{{e^{2\gamma }}}&{{e^{3\gamma + 1}}}&{ - {e^{ - \delta }}}
\end{array}} \right|$