Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| Determinant | Value | ||
| (A) | \(\begin{vmatrix}1 & 2 & 3\\ 2 & 4 & 6\\ 5 & 7 & 9\end{vmatrix}\) | (I) | \(-6\) |
| (B) | \(\begin{vmatrix}2 & 5 & 7\\ 0 & 3 & 8\\ 0 & 0 & -1\end{vmatrix}\) | (II) | 12 |
| (C) | \(\begin{vmatrix}a + 3 & a + 2\\ a + 2 & a + 1\end{vmatrix}\) | (III) | 0 |
| (D) | \(\begin{vmatrix}0 & 1 & 2\\ 1 & 0 & 3\\ 2 & 3 & 0\end{vmatrix}\) | (IV) | \(-1\) |
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)-(IV), (D)-(II)
(A) \(\begin{vmatrix}1 & 2 & 3\\ 2 & 4 & 6\\ 5 & 7 & 9\end{vmatrix}\)
This matches (III) 0.
The second row is 2 times the first row. A determinant with two proportional rows is 0.
(B) \(\begin{vmatrix}2 & 5 & 7\\ 0 & 3 & 8\\ 0 & 0 & -1\end{vmatrix}\)
This matches (I) \(-6\).
All entries below the diagonal are 0, so the value is the product of the diagonal entries:
\(2 \times 3 \times (-1) = -6\)
(C) \(\begin{vmatrix}a + 3 & a + 2\\ a + 2 & a + 1\end{vmatrix}\)
This matches (IV) \(-1\).
\(\Delta = (a + 3)(a + 1) - (a + 2)^{2}\)
\(\Delta = (a^{2} + 4a + 3) - (a^{2} + 4a + 4)\)
\(\Delta = -1\)
(D) \(\begin{vmatrix}0 & 1 & 2\\ 1 & 0 & 3\\ 2 & 3 & 0\end{vmatrix}\)
This matches (II) 12.
Expanding along the first row:
\(0(0 - 9) - 1(0 - 6) + 2(3 - 0) = 6 + 6 = 12\)