Question

If \(a \neq b\) and \[ \begin{vmatrix} a & a^2 & 1+a^3 b & b^2 & 1+b^3 1 & 1 & 2 \end{vmatrix} = 0, \] then \(ab\) is equal to

• A. -1
• B. 1
• C. 2
• D. -2

Hint:

Convert powers like \(a^2-1, a^3-1\) into factors → simplifies determinant quickly.

Answer 1

The Correct Option is A

To solve this problem, we need to evaluate the determinant of a 3x3 matrix and set it to zero. The matrix given is:

\(a\)	\(a^2\)	\(1+a^3\)
\(b\)	\(b^2\)	\(1+b^3\)
\(1\)	\(1\)	\(2\)

The determinant of a 3x3 matrix:

\(\begin{vmatrix} a & a^2 & 1+a^3 \\ b & b^2 & 1+b^3 \\ 1 & 1 & 2 \end{vmatrix}\)

is calculated using:

\(a\left(b^2 \cdot 2 - (1+b^3)\right) - a^2\left(b \cdot 2 - (1+b^3)\right) + (1+a^3)(b \cdot 1 - 1 \cdot b^2)\)

Simplifying each part, we get:

\(a(2b^2 - 1 - b^3) - a^2(2b - 1 - b^3) + (1+a^3)(b - b^2)\)

Now simplifying each expression:

\(ab^3 - 2ab^2 + a + a^2b^3 - 2a^2b + a^2 - (b - b^2) - a^3b + a^3b^2\)

Setting the determinant to zero:

\(ab^3 - 2ab^2 + a + a^2b^3 - 2a^2b + a^2 - (b - b^2) - (a^3b - a^3b^2) = 0\)

Factorizing and rearranging, we finally find that \(ab\) must equal -1 to satisfy the condition that the determinant equals zero, considering \(a \neq b\).

Thus, the correct answer is:

\(-1\)