For a matrix \( A=\begin{bmatrix}1&4&5\\3&2&6\\0&1&0\end{bmatrix} \), where \( I \) is identity matrix of order 3
Match List-I with List-II
| List-I | List-II | ||
| (A) | \( \vert A\vert \) | (I) | \( 72 \) |
| (B) | \( (\mathrm{adj}\,A)A \) | (II) | \( 18 \) |
| (C) | \( \vert 2A\vert \) | (III) | \( 9 \) |
| (D) | \( 2\vert A\vert \) | (IV) | \( 9I \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) - (III), (B) - (IV), (C) - (I), (D) - (II)
The third row has only one non-zero entry, so expanding along it:
\( \displaystyle |A|=-1\times\begin{vmatrix}1&5\\3&6\end{vmatrix} \)
\( =-(6-15) \)
\( =9 \)
(A) \( |A| \)
This matches (III) 9, since \( |A|=9 \).
(B) \( (\mathrm{adj}\,A)A \)
This matches (IV) \( 9I \), since \( (\mathrm{adj}\,A)A=|A|\,I=9I \).
(C) \( |2A| \)
This matches (I) 72, since \( |2A|=2^{3}|A|=8\times9=72 \).
(D) \( 2|A| \)
This matches (II) 18, since \( 2|A|=2\times9=18 \).
So the matching is (A) - (III), (B) - (IV), (C) - (I), (D) - (II).
Tip: For a square matrix of order n, \( |kA|=k^{n}|A| \), not \( k|A| \): here \( |2A|=72 \) but \( 2|A|=18 \).