Question

The value of the determinant \( \begin{vmatrix} 1 & a & a^2\\ 1 & b & b^2\\ 1 & c & c^2 \end{vmatrix} \) is:

• A. \((a-b)(b-c)(c-a)\)
• B. \((a-b)(b-c)\)
• C. \(abc(a-b)(b-c)(c-a)\)
• D. 0

Hint:

Recognize standard determinant forms.

Answer 1

The Correct Option is A

To find the value of the determinant \(\begin{vmatrix} 1 & a & a^2\\ 1 & b & b^2\\ 1 & c & c^2 \end{vmatrix}\), we will calculate it step-by-step using the formula for a 3x3 determinant. The determinant of a 3x3 matrix \(\begin{vmatrix} x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \\ z_1 & z_2 & z_3 \end{vmatrix}\) is given by: 

\(x_1(y_2z_3 - y_3z_2) - x_2(y_1z_3 - y_3z_1) + x_3(y_1z_2 - y_2z_1)\)

Here, we apply this formula to our matrix:

\(= 1 \cdot (b \cdot c^2 - c \cdot b^2) - a \cdot (1 \cdot c^2 - 1 \cdot b^2) + a^2 \cdot (1 \cdot b - 1 \cdot c)\)

Simplifying each term, we find:

\(1 \cdot (b \cdot c^2 - c \cdot b^2) = bc^2 - cb^2\)

\(a \cdot (c^2 - b^2) = a(c+b)(c-b)\)

\(a^2 \cdot (b - c) = a^2(b - c)\)

Substituting back, we have:

\(= (bc^2 - cb^2) - a(c+b)(c-b) + a^2(b - c)\)

This simplifies to:

\(= bc^2 - cb^2 - a(c-cb)(c-b) + a^2(b-c)\)

Continuing simplification:

\(= bc^2 - cb^2 + a^2(b-c) - a(c^2-b^2)\)

Recognizing differences and rearranging terms, we find:

\(= (b-c)(c-a)(a-b)\)

The determinant is:

\((a-b)(b-c)(c-a)\)

Thus, the correct answer is (a-b)(b-c)(c-a).