Question

If \( A = \begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \\ \end{vmatrix} \), then the value of \( A \) is:

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

Hint:

\textbf{Tip:} Recognize Vandermonde determinant patterns—they often simplify complex-looking problems instantly.

Answer 1

The Correct Option is A

To solve the problem, we must evaluate the determinant of the 3x3 matrix:

\( A = \begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix} \)

A determinant of a matrix of this type can be evaluated using the following formula:

\( A = 1 \cdot \begin{vmatrix} b & b^2 \\ c & c^2 \end{vmatrix} - a \cdot \begin{vmatrix} 1 & b^2 \\ 1 & c^2 \end{vmatrix} + a^2 \cdot \begin{vmatrix} 1 & b \\ 1 & c \end{vmatrix} \)

Now, calculate each of these 2x2 determinants:

\( \begin{vmatrix} b & b^2 \\ c & c^2 \end{vmatrix} = bc^2 - b^2c = c \cdot b(c - b) \)

\( \begin{vmatrix} 1 & b^2 \\ 1 & c^2 \end{vmatrix} = 1\cdot c^2 - 1\cdot b^2 = c^2 - b^2 = (c-b)(c+b) \)

\( \begin{vmatrix} 1 & b \\ 1 & c \end{vmatrix} = 1\cdot c - 1\cdot b = c - b \)

Substituting back into the equation for \( A \):

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

Factoring out \((c-b)\) from each term:

\( A = (c-b) \left( bc - ab - a(b+c) + a^2 \right) \)

Simplify the expression in the parentheses:

\( = (c-b) \left( bc - ab - ab - ac + a^2 \right) \)

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

Recognizing that the expression can be factored further, we find:

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

The correct answer is:

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

Answer 2

Evaluate the Determinant of a 3x3 Matrix

We are given a determinant:

\[ A = \begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix} \]

Step 1: Use the 3x3 Determinant Formula

The general formula to evaluate the determinant of a 3×3 matrix: \[ \begin{vmatrix} x & y & z \\ u & v & w \\ p & q & r \end{vmatrix} = x(vr - qw) - y(ur - pw) + z(uq - vp) \]

Step 2: Apply the Formula

Applying the formula to the given matrix: \[ A = 1 \cdot (b \cdot c^2 - b^2 \cdot c) - a \cdot (1 \cdot c^2 - 1 \cdot b^2) + a^2 \cdot (1 \cdot c - 1 \cdot b) \]

Step 3: Simplify Each Term

Simplify the above expression: \[ A = (bc^2 - b^2c) - a(c^2 - b^2) + a^2(c - b) \] Rearranged and grouped: \[ A = b(c^2 - bc) - a(c^2 - b^2) + a^2(c - b) \] Further simplification gives: \[ A = b^2(c - a) - c^2(b - a) + a^2(c - b) \]

Step 4: Recognize the Vandermonde Determinant Pattern

The determinant follows the structure of a **Vandermonde determinant**, which is a known identity: \[ \begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix} = (a - b)(b - c)(c - a) \]

Final Answer

Therefore, the value of the determinant is: \[ A = (a - b)(b - c)(c - a) \] This is a classic result from determinant algebra, especially useful in polynomial interpolation and linear algebra problems.

Conclusion

When evaluating a determinant involving polynomial terms like \( a^2, b^2, c^2 \), it's important to check for patterns such as the Vandermonde determinant. Recognizing this saves time and leads to elegant solutions in competitive exams and algebra-based problems.