Question

The value of the determinant $\begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{vmatrix} =$

• A. (a-b)-(b-c)-(c-a)-(1-1-1)
• B. (a-b)-(b-c)-(c-a)
• C. (a-b)-(b-c)-(c-a)-0
• D. (a-b)-(b-c)-(c-a)-(1+1+1)
• E. (a-b)-(b-c)-(c-a)

Answer 1

The options provided are incorrect. The value of the determinant is $(b-a)(c-a)(c-b)$.

Answer 2

The determinant given is a Vandermonde determinant. The general form of a $3 \times 3$ Vandermonde determinant is:

$D = \begin{vmatrix} 1 & 1 & 1 \\ x & y & z \\ x^2 & y^2 & z^2 \end{vmatrix}$

To evaluate this determinant, we can perform column operations:

Apply $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - C_1$:

$D = \begin{vmatrix} 1 & 1-1 & 1-1 \\ a & b-a & c-a \\ a^2 & b^2-a^2 & c^2-a^2 \end{vmatrix} = \begin{vmatrix} 1 & 0 & 0 \\ a & b-a & c-a \\ a^2 & (b-a)(b+a) & (c-a)(c+a) \end{vmatrix}$

Now, expand the determinant along the first row ($R_1$):

$D = 1 \cdot \begin{vmatrix} b-a & c-a \\ (b-a)(b+a) & (c-a)(c+a) \end{vmatrix} - 0 + 0$

Factor out $(b-a)$ from the first column and $(c-a)$ from the second column of the $2 \times 2$ determinant:

$D = (b-a)(c-a) \begin{vmatrix} 1 & 1 \\ b+a & c+a \end{vmatrix}$

Now, evaluate the $2 \times 2$ determinant:

$D = (b-a)(c-a) [1 \cdot (c+a) - 1 \cdot (b+a)] = (b-a)(c-a) [c+a-b-a] = (b-a)(c-a)(c-b)$

Thus, the value of the determinant is $(b-a)(c-a)(c-b)$. The options provided are in the form of sums/differences, not products, and therefore do not represent the correct value of the determinant. For example, option (b) is $(a-b)-(b-c)-(c-a) = 2a-2b$, which is incorrect. All options are mathematically incorrect representations of the determinant's value.