Let \(A = \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix}\) and \(B = \begin{bmatrix} 1 & 0 \\ 2 & -2 \end{bmatrix}\).
Match List-I with List-II
| List-I | List-II | ||
| (A) | Element in row 1, column 2 of AB | (I) | 7 |
| (B) | Element in row 2, column 2 of BA | (II) | \(-14\) |
| (C) | \(|AB|\) | (III) | 2 |
| (D) | Element in row 2, column 1 of AB | (IV) | \(-8\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
The element in row i, column j of a product is row i of the first matrix times column j of the second.
(A) Element in row 1, column 2 of AB
This matches (III) 2.
Row 1 of A times column 2 of B: \(2 \times 0 + (-1) \times (-2) = 2\)
(B) Element in row 2, column 2 of BA
This matches (IV) \(-8\).
Row 2 of B times column 2 of A: \(2 \times (-1) + (-2) \times 3 = -8\)
(C) \(|AB|\)
This matches (II) \(-14\).
\(|A| = 6 + 1 = 7\), \(|B| = -2 - 0 = -2\)
\(|AB| = |A| \times |B| = 7 \times (-2) = -14\)
(D) Element in row 2, column 1 of AB
This matches (I) 7.
Row 2 of A times column 1 of B: \(1 \times 1 + 3 \times 2 = 7\)