Match the LIST-I with LIST-II
| LIST-I | LIST-II | ||
| Matrix A | |adj A| | ||
| (A) | \( \begin{bmatrix}1 & 2 & 3\\0 & 2 & 5\\0 & 0 & 1\end{bmatrix} \) | (I) | 9 |
| (B) | \( \begin{bmatrix}3 & -1 & 4\\0 & 1 & 2\\0 & 0 & -1\end{bmatrix} \) | (II) | 16 |
| (C) | \( \begin{bmatrix}4 & 2 & -3\\0 & -2 & 5\\0 & 0 & 1\end{bmatrix} \) | (III) | 64 |
| (D) | \( \begin{bmatrix}2 & 5 & 3\\0 & 2 & 1\\0 & 0 & 1\end{bmatrix} \) | (IV) | 4 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A) - (IV), (B) - (I), (C) - (III), (D) - (II)
All four matrices are upper triangular, so each determinant is the product of the diagonal entries. For a matrix of order 3,
\( \displaystyle |\mathrm{adj}\,A|=|A|^{3-1}=|A|^2 \)
(A)
This matches (IV) 4.
\( \displaystyle |A|=1\times 2\times 1=2 \)
\( \displaystyle |\mathrm{adj}\,A|=2^2=4 \)
(B)
This matches (I) 9.
\( \displaystyle |A|=3\times 1\times(-1)=-3 \)
\( \displaystyle |\mathrm{adj}\,A|=(-3)^2=9 \)
(C)
This matches (III) 64.
\( \displaystyle |A|=4\times(-2)\times 1=-8 \)
\( \displaystyle |\mathrm{adj}\,A|=(-8)^2=64 \)
(D)
This matches (II) 16.
\( \displaystyle |A|=2\times 2\times 1=4 \)
\( \displaystyle |\mathrm{adj}\,A|=4^2=16 \)
So the matching is (A)-(IV), (B)-(I), (C)-(III), (D)-(II).