Let \(A = \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix}\) and \(B = \begin{bmatrix} 1 & 2 \\ 0 & 4 \end{bmatrix}\).
Match List-I with List-II
| List-I | List-II | ||
| (A) | \(|AB|\) | (I) | 20 |
| (B) | \(|A + B|\) | (II) | 28 |
| (C) | \(|A - B|\) | (III) | 36 |
| (D) | \(|3B|\) | (IV) | 2 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
For square matrices of the same order the product rule \(|AB| = |A||B|\) holds, and for a matrix of order 2, \(|kB| = k^2|B|\). The two determinants are
\(|A| = 2 \times 3 - (-1) \times 1 = 7\)
\(|B| = 1 \times 4 - 2 \times 0 = 4\)
(A) \(|AB|\)
This matches (II) 28.
\(|AB| = |A||B| = 7 \times 4 = 28\)
(B) \(|A + B|\)
This matches (I) 20. There is no sum rule, so A + B is found first:
\( \displaystyle A + B = \begin{bmatrix} 3 & 1 \\ 1 & 7 \end{bmatrix} \)
\(|A + B| = 21 - 1 = 20\)
This is not \(|A| + |B| = 11\).
(C) \(|A - B|\)
This matches (IV) 2.
\( \displaystyle A - B = \begin{bmatrix} 1 & -3 \\ 1 & -1 \end{bmatrix} \)
\(|A - B| = -1 + 3 = 2\)
(D) \(|3B|\)
This matches (III) 36.
\(|3B| = 3^2 \times 4 = 36\)