If matrix \( A=\begin{bmatrix}12 & 7 & 9\\6 & 3 & 2\\5 & 1 & 8\end{bmatrix} \), then
Match the List-I with List-II
| List-I | List-II | ||
| (A) | |A| | (I) | 38 |
| (B) | Cofactor of \( \mathrm{a_{31}} \) | (II) | −23 |
| (C) | Minor of \( \mathrm{a_{23}} \) | (III) | −83 |
| (D) | Cofactor of \( \mathrm{a_{12}} \) | (IV) | −13 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) - (III), (B) - (IV), (C) - (II), (D) - (I)
The minor \( M_{ij} \) is the determinant left after deleting row i and column j; the cofactor is \( A_{ij}=(-1)^{i+j}M_{ij} \).
(A) |A|
This matches (III) −83.
Expanding along the first row:
\( \displaystyle |A|=12\begin{vmatrix}3 & 2\\1 & 8\end{vmatrix}-7\begin{vmatrix}6 & 2\\5 & 8\end{vmatrix}+9\begin{vmatrix}6 & 3\\5 & 1\end{vmatrix} \)
\( \displaystyle |A|=12(24-2)-7(48-10)+9(6-15) \)
\( \displaystyle |A|=264-266-81=-83 \)
(B) Cofactor of \( a_{31} \)
This matches (IV) −13.
\( \displaystyle A_{31}=(-1)^{3+1}\begin{vmatrix}7 & 9\\3 & 2\end{vmatrix} \)
\( \displaystyle A_{31}=14-27=-13 \)
(C) Minor of \( a_{23} \)
This matches (II) −23.
\( \displaystyle M_{23}=\begin{vmatrix}12 & 7\\5 & 1\end{vmatrix} \)
\( \displaystyle M_{23}=12-35=-23 \)
(D) Cofactor of \( a_{12} \)
This matches (I) 38, by elimination.
\( \displaystyle A_{12}=(-1)^{1+2}\begin{vmatrix}6 & 2\\5 & 8\end{vmatrix} \)
\( \displaystyle A_{12}=-(48-10)=-38 \)
So the cofactor of \( a_{12} \) is −38, and the printed entry 38 does not satisfy it; 38 is the minor \( M_{12} \).
So the matching is (A)-(III), (B)-(IV), (C)-(II), (D)-(I).
Note: It seems that there is a mistake in the original NTA paper, which prints 38 in List-II for the cofactor of \( a_{12} \) when it should read −38 (38 is the minor of \( a_{12} \)). The question can still be solved because (A), (B) and (C) have only one possible match each, (A)-(III), (B)-(IV) and (C)-(II), and Option (3) is the only option containing them.