Let \(A = \begin{bmatrix}1&2\\3&-1\end{bmatrix}\) and \(B = \begin{bmatrix}2&0\\1&3\end{bmatrix}\).
Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| (A) | The element in the second row and first column of AB | (I) | 1 |
| (B) | The value of \(k\) for which \(A^2 = kI\) | (II) | 5 |
| (C) | The sum of the diagonal elements of BA | (III) | 7 |
| (D) | The element in the first row and second column of \(A'B\) | (IV) | 9 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(II), (B)-(III), (C)-(I), (D)-(IV)
The element in row i, column j of a product is row i of the first matrix times column j of the second, so only the elements asked for need to be worked out.
(A) The element in the second row and first column of AB
This matches (II) 5.
\(3(2) + (-1)(1) = 6 - 1 = 5\)
(B) The value of \(k\) for which \(A^2 = kI\)
This matches (III) 7.
\( \displaystyle A^2 = \begin{bmatrix}1 + 6&2 - 2\\3 - 3&6 + 1\end{bmatrix} = \begin{bmatrix}7&0\\0&7\end{bmatrix} = 7I \)
(C) The sum of the diagonal elements of BA
This matches (I) 1.
\(\text{first diagonal element} = 2(1) + 0(3) = 2\)
\(\text{second diagonal element} = 1(2) + 3(-1) = -1\)
\(\text{sum} = 2 + (-1) = 1\)
(D) The element in the first row and second column of \(A'B\)
This matches (IV) 9. The first row of \(A'\) is the first column of A, that is (1, 3).
\(1(0) + 3(3) = 9\)