Match List - I with List - II for Matrix \(A=\begin{bmatrix}1 & -1 & 2\\ 2 & 3 & 5\\ -2 & 0 & 1\end{bmatrix}\).
| List - I | List - II | ||
| (A) | \(|A|\) | (I) | \(\displaystyle\frac{8}{27}\) |
| (B) | \(|A^{-1}|\) | (II) | \(\displaystyle\frac{1}{27}\) |
| (C) | \(|\mathrm{adj}\ A|\) | (III) | 27 |
| (D) | \(|2A^{-1}|\) | (IV) | 729 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(III), (B)-(II), (C)-(IV), (D)-(I)
A is of order n = 3. Every item follows from |A|, using the determinant results \(|A^{-1}| = \frac{1}{|A|}\), \(|\mathrm{adj}\,A| = |A|^{n-1}\) and \(|kB| = k^n|B|\).
(A) \(|A|\)
This matches (III) 27.
Expanding along the first row:
\(|A| = 1(3\times 1-5\times 0)-(-1)(2\times 1-5\times(-2))+2(2\times 0-3\times(-2))\)
\(= 3+12+12 = 27\)
(B) \(|A^{-1}|\)
This matches (II) \(\frac{1}{27}\).
\( \displaystyle |A^{-1}| = \frac{1}{|A|} = \frac{1}{27} \)
(C) \(|\mathrm{adj}\ A|\)
This matches (IV) 729.
\(|\mathrm{adj}\,A| = |A|^{3-1} = 27^2 = 729\)
(D) \(|2A^{-1}|\)
This matches (I) \(\frac{8}{27}\).
Multiplying a matrix of order 3 by 2 multiplies its determinant by \(2^3\):
\( \displaystyle |2A^{-1}| = 2^3\,|A^{-1}| = \frac{8}{27} \)
So the matching is (A)-(III), (B)-(II), (C)-(IV), (D)-(I).