Prove the following using properties of determinants: (i) $\begin{vmatrix} -a^2 & ab & ac \ ab & -b^2 & bc \ ac & bc & -c^2 \end{vmatrix} = 4a^2b^2c^2$ (ii) $\begin{vmatrix} 1 & 1 & 1 \ 1 & 1+x & 1 \ 1 & 1 & 1+y \end{vmatrix} = xy$ (iii) $\begin{vmatrix} 1 & a & a^2 \ 1 & b

Question

Prove the following using properties of determinants: (i) $\begin{vmatrix} -a^2 & ab & ac \ ab & -b^2 & bc \ ac & bc & -c^2 \end{vmatrix} = 4a^2b^2c^2$ (ii) $\begin{vmatrix} 1 & 1 & 1 \ 1 & 1+x & 1 \ 1 & 1 & 1+y \end{vmatrix} = xy$ (iii) $\begin{vmatrix} 1 & a & a^2 \ 1 & b & b^2 \ 1 & c & c^2 \end{vmatrix} = (a-b)(b-c)(c-a)$ (iv) $\begin{vmatrix} \alpha & \beta & \gamma \ \alpha^2 & \beta^2 & \gamma^2 \ \beta+\gamma & \gamma+\alpha & \alpha+\beta \end{vmatrix} = (\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)(\alpha+\beta+\gamma)$ (v) $\begin{vmatrix} 1 & 1+p & 1+p+q \ 2 & 3+2p & 1+3p+2q \ 3 & 6+3p & 1+6p+3q \end{vmatrix} = 1$.

Accepted Answer

> [!TLDR]
> Each identity is established by taking out common factors and applying elementary row/column operations $R_i	o R_i-kR_j$.

> [!FORMULA]
> Key properties used:
> - A common factor of a row/column can be taken outside the determinant.
> - $R_i	o R_i+kR_j$ (or $C_i	o C_i+kC_j$) leaves a determinant unchanged.
> - Vandermonde: $\begin{vmatrix}1&a&a^2\1&b&b^2\1&c&c^2\end{vmatrix}=(a-b)(b-c)(c-a)$.

**(i)** $\begin{vmatrix} -a^2 & ab & ac \ ab & -b^2 & bc \ ac & bc & -c^2 \end{vmatrix}=4a^2b^2c^2$

1. Take $a,b,c$ common from $R_1,R_2,R_3$:
$$=abc\begin{vmatrix} -a & b & c \ a & -b & c \ a & b & -c \end{vmatrix}$$
2. Take $a,b,c$ common from $C_1,C_2,C_3$:
$$=a^2b^2c^2\begin{vmatrix} -1 & 1 & 1 \ 1 & -1 & 1 \ 1 & 1 & -1 \end{vmatrix}$$
3. Expand the numeric determinant:
$$=-1(1-1)-1(-1-1)+1(1+1)=0+2+2=4.$$
4. Hence the value $=4a^2b^2c^2$. $\blacksquare$

**(ii)** $\begin{vmatrix} 1 & 1 & 1 \ 1 & 1+x & 1 \ 1 & 1 & 1+y \end{vmatrix}=xy$

1. Apply $R_2	o R_2-R_1$ and $R_3	o R_3-R_1$:
$$=\begin{vmatrix} 1 & 1 & 1 \ 0 & x & 0 \ 0 & 0 & y \end{vmatrix}$$
2. Expand along $C_1$ (upper-triangular): $=1\cdot(x\cdot y-0)=xy$. $\blacksquare$

**(iii)** $\begin{vmatrix} 1 & a & a^2 \ 1 & b & b^2 \ 1 & c & c^2 \end{vmatrix}=(a-b)(b-c)(c-a)$

1. Apply $R_1	o R_1-R_2$ and $R_2	o R_2-R_3$:
$$=\begin{vmatrix} 0 & a-b & a^2-b^2 \ 0 & b-c & b^2-c^2 \ 1 & c & c^2 \end{vmatrix}$$
2. Take $(a-b)$ from $R_1$ and $(b-c)$ from $R_2$ (since $a^2-b^2=(a-b)(a+b)$):
$$=(a-b)(b-c)\begin{vmatrix} 0 & 1 & a+b \ 0 & 1 & b+c \ 1 & c & c^2 \end{vmatrix}$$
3. Expand along $C_1$: $=(a-b)(b-c)\big[1\cdot\big((b+c)-(a+b)\big)\big]=(a-b)(b-c)(c-a)$. $\blacksquare$

**(iv)** $\begin{vmatrix} \alpha & \beta & \gamma \ \alpha^2 & \beta^2 & \gamma^2 \ \beta+\gamma & \gamma+\alpha & \alpha+\beta \end{vmatrix}=(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)(\alpha+\beta+\gamma)$

1. Apply $R_3	o R_3+R_1$. Each entry becomes $\alpha+\beta+\gamma$:
$$=\begin{vmatrix} \alpha & \beta &

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