Let \(\Delta = \begin{vmatrix}a_1 & b_1 & c_1\\ a_2 & b_2 & c_2\\ a_3 & b_3 & c_3\end{vmatrix} = 5\).
Match List-I with List-II
| List-I (Determinant obtained from \(\Delta\)) | List-II (Value) | ||
| (A) | By interchanging the first and second rows | (I) | 15 |
| (B) | By multiplying every element of the first row by 3 | (II) | 5 |
| (C) | By multiplying every element of \(\Delta\) by 2 | (III) | 40 |
| (D) | By the operation \(R_1 \to R_1 + 2R_3\) | (IV) | \(-5\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(IV), (B)-(I), (C)-(III), (D)-(II)
(A) By interchanging the first and second rows
This matches (IV) \(-5\).
Interchanging two rows changes the sign of a determinant: \(-\Delta = -5\).
(B) By multiplying every element of the first row by 3
This matches (I) 15.
Multiplying one row by 3 multiplies the determinant by 3: \(3 \times 5 = 15\).
(C) By multiplying every element of \(\Delta\) by 2
This matches (III) 40.
Each of the three rows gets a factor 2, so the determinant is multiplied by \(2^{3}\): \(8 \times 5 = 40\).
(D) By the operation \(R_1 \to R_1 + 2R_3\)
This matches (II) 5.
Adding a multiple of one row to another row does not change the value: it stays 5.