Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| Determinant | Value | ||
| (A) | \(\begin{vmatrix}x + 1 & x\\ x & x + 1\end{vmatrix}\) | (I) | 4 |
| (B) | \(\begin{vmatrix}x - 1 & x + 1\\ x - 3 & x - 1\end{vmatrix}\) | (II) | \(-3\) |
| (C) | \(\begin{vmatrix}x + 2 & x - 1\\ x + 1 & x - 2\end{vmatrix}\) | (III) | \(2x + 1\) |
| (D) | \(\begin{vmatrix}x + 3 & x\\ x + 1 & x - 2\end{vmatrix}\) | (IV) | \(-6\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(III), (B)-(I), (C)-(II), (D)-(IV)
A \(2 \times 2\) determinant is the product of the main diagonal minus the product of the other diagonal. In each case the \(x^{2}\) terms cancel.
(A) \(\begin{vmatrix}x + 1 & x\\ x & x + 1\end{vmatrix}\)
This matches (III) \(2x + 1\).
\((x + 1)^{2} - x^{2} = x^{2} + 2x + 1 - x^{2} = 2x + 1\)
(B) \(\begin{vmatrix}x - 1 & x + 1\\ x - 3 & x - 1\end{vmatrix}\)
This matches (I) 4.
\((x - 1)^{2} - (x + 1)(x - 3) = (x^{2} - 2x + 1) - (x^{2} - 2x - 3) = 4\)
(C) \(\begin{vmatrix}x + 2 & x - 1\\ x + 1 & x - 2\end{vmatrix}\)
This matches (II) \(-3\).
\((x + 2)(x - 2) - (x - 1)(x + 1) = (x^{2} - 4) - (x^{2} - 1) = -3\)
(D) \(\begin{vmatrix}x + 3 & x\\ x + 1 & x - 2\end{vmatrix}\)
This matches (IV) \(-6\).
\((x + 3)(x - 2) - x(x + 1) = (x^{2} + x - 6) - (x^{2} + x) = -6\)