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

Concept: If determinant is zero, rows are linearly dependent. Use row operations.

Step 1: Apply row operations. \[ R_1 \rightarrow R_1 - R_3,\quad R_2 \rightarrow R_2 - R_3 \] \[ \Rightarrow \begin{vmatrix} a-1 & a^2-1 & a^3-1 \\ b-1 & b^2-1 & b^3-1 \\ 1 & 1 & 2 \end{vmatrix} = 0 \]

Step 2: Factor expressions. \[ a^2-1 = (a-1)(a+1),\quad a^3-1 = (a-1)(a^2+a+1) \] \[ b^2-1 = (b-1)(b+1),\quad b^3-1 = (b-1)(b^2+b+1) \] Factor \((a-1)\) from \(R_1\) and \((b-1)\) from \(R_2\): \[ = (a-1)(b-1) \begin{vmatrix} 1 & a+1 & a^2+a+1 \\ 1 & b+1 & b^2+b+1 \\ 1 & 1 & 2 \end{vmatrix} = 0 \]

Step 3: Simplify determinant. \[ R_1 \rightarrow R_1 - R_3,\quad R_2 \rightarrow R_2 - R_3 \] \[ \Rightarrow \begin{vmatrix} 0 & a & a^2+a-1 \\ 0 & b & b^2+b-1 \\ 1 & 1 & 2 \end{vmatrix} \] Expanding along first column: \[ = \begin{vmatrix} a & a^2+a-1 \\ b & b^2+b-1 \end{vmatrix} \]

Step 4: Evaluate. \[ = a(b^2+b-1) - b(a^2+a-1) \] \[ = ab^2 + ab - a - a^2b - ab + b \] \[ = ab^2 - a^2b - a + b \] \[ = ab(b-a) - (a-b) = (a-b)(-ab - 1) \]

Step 5: Use given condition. \[ (a-1)(b-1)(a-b)(ab+1) = 0 \] Given \(a \neq b\), so \(ab+1 = 0 \Rightarrow ab = -1 \]