Let \(A = \begin{bmatrix}2&-1\\3&4\end{bmatrix}\) and \(B = \begin{bmatrix}1&5\\-2&0\end{bmatrix}\).
Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| (A) | The element in the first row and second column of \(A + B\) | (I) | 8 |
| (B) | The element in the second row and first column of \(2A - B\) | (II) | 12 |
| (C) | The element in the first row and second column of \(A'\) | (III) | 4 |
| (D) | The element in the second row and second column of \(3A - B\) | (IV) | 3 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
Sums and scalar multiples are worked out element by element, so only the element in the asked position is needed. In the transpose, the element in row i and column j of A' is the element in row j and column i of A.
(A) The element in the first row and second column of \(A + B\)
This matches (III) 4.
\(a_{12} + b_{12} = -1 + 5 = 4\)
(B) The element in the second row and first column of \(2A - B\)
This matches (I) 8.
\(2a_{21} - b_{21} = 2(3) - (-2) = 8\)
(C) The element in the first row and second column of \(A'\)
This matches (IV) 3.
It equals \(a_{21} = 3\)
(D) The element in the second row and second column of \(3A - B\)
This matches (II) 12.
\(3a_{22} - b_{22} = 3(4) - 0 = 12\)