If \( A=\begin{bmatrix}1 & 1 & 1\\1 & 0 & 3\\1 & -2 & 1\end{bmatrix} \), then
Match List-I with List-II
| List-I | List-II | ||
| (A) | |A| | (I) | 48 |
| (B) | |2A| | (II) | 36 |
| (C) | |adj A| | (III) | 12 |
| (D) | 2|A| | (IV) | 6 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) - (IV), (B) - (I), (C) - (II), (D) - (III)
(A) |A|
This matches (IV) 6.
Expanding along R1:
\( \displaystyle |A|=1\begin{vmatrix}0 & 3\\-2 & 1\end{vmatrix}-1\begin{vmatrix}1 & 3\\1 & 1\end{vmatrix}+1\begin{vmatrix}1 & 0\\1 & -2\end{vmatrix} \)
\( \displaystyle =1(0+6)-1(1-3)+1(-2-0) \)
\( \displaystyle =6+2-2 \)
\( \displaystyle =6 \)
(B) |2A|
This matches (I) 48.
\( \displaystyle |2A|=2^3|A|=8\times 6=48 \)
(C) |adj A|
This matches (II) 36.
\( \displaystyle |\mathrm{adj}\,A|=|A|^{3-1}=6^2=36 \)
(D) 2|A|
This matches (III) 12.
\( \displaystyle 2|A|=2\times 6=12 \)
So the matching is (A)-(IV), (B)-(I), (C)-(II), (D)-(III).
Tip: Do not confuse |2A| with 2|A|: for a matrix of order 3 the factor 2 comes out of every row, giving \( 2^3|A| \).