Question

The determinant of the matrix $$ \begin{bmatrix} a & b & c \\ a^2 & b^2 & c^2 \\ 1 & 1 & 1 \end{bmatrix} $$ is not equal to: 

• A. \[ \begin{bmatrix} a + 1 & b + 1 & c + 1 \\ a^2 + 1 & b^2 + 1 & c^2 + 1 \\ 1 & 1 & 1 \end{bmatrix} \]
• B. \[ \begin{bmatrix} a - b & b - c & c \\ a^2 - b^2 & b^2 - c^2 & c^2 \\ 0 & 0 & 1 \end{bmatrix} \]
• C. \[ \begin{bmatrix} a(a + 1) & b(b + 1) & c(c + 1) \\ a + 1 & b + 1 & c + 1 \\ -1 & -1 & -1 \end{bmatrix} \]
• D. \[ \begin{bmatrix} a + b & b + c & c + a \\ a^2 + b^2 & b^2 + c^2 & c^2 + a^2 \\ 2 & 2 & 2 \end{bmatrix} \]

Hint:

Always check for linear dependency in rows or columns when determining if a determinant is zero. This quick check can often simplify your calculations.

Answer 1

The Correct Option is D

Step 1: Analyze the original determinant, known to be a Vandermonde determinant.
- The determinant simplifies to \( (a-b)(b-c)(c-a) \), which is the product of the differences between the variables. 
Step 2: Compare each option's determinant.
- Option (A), (B), and (C) give non-zero determinants as they involve transformations that preserve the structure of the Vandermonde determinant, and none of them result in linear dependency. - Option (D) is notably different because it results in a matrix with linearly dependent rows, as each element in the third row is the constant \(2\).
- This makes the determinant zero, which does not match the original determinant unless \(a = b = c\).