Match List-I with List-II
| List-I (Determinant) | List-II (Value) | ||
| (A) | \(\begin{vmatrix}2 & -3\\ 4 & 1\end{vmatrix}\) | (I) | 0 |
| (B) | \(\begin{vmatrix}5 & 4\\ 3 & 1\end{vmatrix}\) | (II) | \(-7\) |
| (C) | \(\begin{vmatrix}2 & 7 & 1\\ 0 & 3 & 5\\ 0 & 0 & -1\end{vmatrix}\) | (III) | 14 |
| (D) | \(\begin{vmatrix}1 & 2 & 3\\ 2 & 4 & 6\\ 1 & 0 & 1\end{vmatrix}\) | (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)-(II), (C)-(IV), (D)-(I)
(A) \(\begin{vmatrix}2 & -3\\ 4 & 1\end{vmatrix}\)
This matches (III) 14.
\( 2 \times 1 - (-3) \times 4 = 2 + 12 = 14 \)
(B) \(\begin{vmatrix}5 & 4\\ 3 & 1\end{vmatrix}\)
This matches (II) \(-7\).
\( 5 \times 1 - 4 \times 3 = 5 - 12 = -7 \)
(C) \(\begin{vmatrix}2 & 7 & 1\\ 0 & 3 & 5\\ 0 & 0 & -1\end{vmatrix}\)
This matches (IV) \(-6\).
All entries below the main diagonal are zero (an upper triangular determinant), so the value is the product of the diagonal entries:
\( 2 \times 3 \times (-1) = -6 \)
(D) \(\begin{vmatrix}1 & 2 & 3\\ 2 & 4 & 6\\ 1 & 0 & 1\end{vmatrix}\)
This matches (I) 0.
The second row (2, 4, 6) is twice the first row (1, 2, 3). A determinant with proportional rows is zero.