Question

Let \( A = \left[ {\matrix{ 1 & { - 2} & \alpha \cr \alpha & 2 & { - 1} \cr } } \right] \) and \( B = \left[ {\matrix{ 2 & \alpha \cr { - 1} & 2 \cr 4 & { - 5} \cr } } \right],\,\alpha \in C \). Then the absolute value of the sum of all values of \( \alpha \) for which det(AB) = 0 is :

• A. 3
• B. 4
• C. 2
• D. 5

Hint:

A is 2x3 and B is 3x2, so AB is 2x2; compute AB and set its determinant to zero, then use the sum of roots of the resulting polynomial.

Answer 1

The Correct Option is A

Step 1: \(AB=\begin{bmatrix}1\cdot2+(-2)(-1)+4\alpha & \alpha-4-5\alpha\\ 2\alpha-2-4 & \alpha^2+4+5\end{bmatrix}=\begin{bmatrix}4+4\alpha & -4-4\alpha\\ 2\alpha-6 & \alpha^2+9\end{bmatrix}\).
Step 2: \(\det(AB)=(4+4\alpha)(\alpha^2+9)+(4+4\alpha)(2\alpha-6)=4(1+\alpha)(\alpha^2+2\alpha+3)\).
Step 3: Roots: \(\alpha=-1\), and the roots of \(\alpha^2+2\alpha+3=0\) with sum \(-2\) (complex roots allowed as \(\alpha\in C\)).
Step 4: Sum of all values \(=-1-2=-3\); absolute value = 3.
Answer: (A) 3