# By using properties of determinants, show that:

Question:

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:

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

$=a b c\left|\begin{array}{lll}-a & b & c \\ a & -b & c \\ a & b & -c\end{array}\right| \quad$ [Taking out factors $a, b, c$ from $\mathrm{R}_{1}, \mathrm{R}_{2}$, and $\left.\mathrm{R}_{3}\right]$

$=a^{2} b^{2} c^{2}\left|\begin{array}{lll}-1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1\end{array}\right| \quad\left[\right.$ Taking out factors $a, b, c$ from $\mathrm{C}_{1}, \mathrm{C}_{2}$, and $\left.\mathrm{C}_{3}\right]$

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

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

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

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