Match List-I with List-II
| List-I (Matrix) | List-II (Determinant) |
|---|---|
| (A) \(\begin{bmatrix}0 & 1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 1\end{bmatrix}\) | (I) \(-6\) |
| (B) \(\begin{bmatrix}0 & 1 & 0\\ 0 & 0 & 1\\ 1 & 0 & 0\end{bmatrix}\) | (II) \(-1\) |
| (C) \(\begin{bmatrix}1 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 3\end{bmatrix}\) | (III) 1 |
| (D) \(\begin{bmatrix}0 & 0 & 1\\ 0 & 2 & 0\\ 3 & 0 & 0\end{bmatrix}\) | (IV) 6 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
Each matrix has only one non-zero entry in every row, so each expansion has a single non-zero term.
(A) Expanding along the third row:
\(\displaystyle 1 \times \begin{vmatrix}0 & 1\\ 1 & 0\end{vmatrix} = 1 \times (0 - 1) = -1\) → (II)
This is the identity matrix with its first two rows interchanged, so the value 1 changes sign.
(B) Expanding along the first column:
\(\displaystyle 1 \times (-1)^{3 + 1}\begin{vmatrix}1 & 0\\ 0 & 1\end{vmatrix} = 1\) → (III)
(C) A diagonal matrix has determinant equal to the product of the diagonal entries:
\(1 \times 2 \times 3 = 6\) → (IV)
(D) Expanding along the first row:
\(\displaystyle 1 \times \begin{vmatrix}0 & 2\\ 3 & 0\end{vmatrix} = 1 \times (0 - 6) = -6\) → (I)