Match List-I with List-II
| List-I | List-II |
| Determinant | Value |
| (A) \( \begin{vmatrix}6 & 2\\4 & 3\end{vmatrix}= \) | (I) 0 |
| (B) \( \begin{vmatrix}x & x+1\\x-1 & x\end{vmatrix}= \) | (II) -1 |
| (C) \( \begin{vmatrix}1 & \log_a b\\\log_b a & 1\end{vmatrix}= \) | (III) 10 |
| (D) \( \begin{vmatrix}x-1 & 1\\x^3 & x^2+x+1\end{vmatrix}= \) | (IV) 1 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) - (III), (B) - (IV), (C) - (I), (D) - (II)
(A) \( \begin{vmatrix}6 & 2\\4 & 3\end{vmatrix} \)
This matches (III) 10.
\( \displaystyle 6\times 3-2\times 4=18-8=10 \)
(B) \( \begin{vmatrix}x & x+1\\x-1 & x\end{vmatrix} \)
This matches (IV) 1.
\( \displaystyle x^2-(x^2-1)=1 \)
(C) \( \begin{vmatrix}1 & \log_a b\\\log_b a & 1\end{vmatrix} \)
This matches (I) 0, since \( \log_a b\cdot\log_b a=1 \):
\( \displaystyle 1-1=0 \)
(D) \( \begin{vmatrix}x-1 & 1\\x^3 & x^2+x+1\end{vmatrix} \)
This matches (II) −1, since \( (x-1)(x^2+x+1)=x^3-1 \):
\( \displaystyle x^3-1-x^3=-1 \)
So the matching is (A) - (III), (B) - (IV), (C) - (I), (D) - (II).