Match List-I with List-II
| List-I | List-II | ||
| (A) | \( If\ A=\begin{bmatrix}2 & -5\\3 & 4\end{bmatrix},\ then\ \mathrm{adj}\ A= \) | (I) | \( \begin{bmatrix}4 & -5\\-2 & -3\end{bmatrix} \) |
| (B) | \( If\ A=\begin{bmatrix}2 & 3\\5 & -2\end{bmatrix},then\ A^{-1}= \) | (II) | \( \begin{bmatrix}\displaystyle\frac{1}{17} & \displaystyle\frac{-5}{17}\\\displaystyle\frac{3}{17} & \displaystyle\frac{2}{17}\end{bmatrix} \) |
| (C) | \( If\ A=\begin{bmatrix}-3 & 5\\2 & 4\end{bmatrix},then\ \mathrm{adj}\ A= \) | (III) | \( \begin{bmatrix}4 & 5\\-3 & 2\end{bmatrix} \) |
| (D) | \( If\ A=\begin{bmatrix}2 & 5\\-3 & 1\end{bmatrix},then\ A^{-1}= \) | (IV) | \( \begin{bmatrix}\displaystyle\frac{2}{19} & \displaystyle\frac{3}{19}\\\displaystyle\frac{5}{19} & \displaystyle\frac{-2}{19}\end{bmatrix} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A) - (III), (B) - (IV), (C) - (I), (D) - (II)
For \( A=\begin{bmatrix}a & b\\c & d\end{bmatrix} \): \( \mathrm{adj}\,A=\begin{bmatrix}d & -b\\-c & a\end{bmatrix} \) (swap the diagonal entries, change the signs of the other two) and \( A^{-1}=\frac{1}{|A|}\mathrm{adj}\,A \).
(A) \( A=\begin{bmatrix}2 & -5\\3 & 4\end{bmatrix} \), adj A
This matches (III).
\( \displaystyle \mathrm{adj}\,A=\begin{bmatrix}4 & 5\\-3 & 2\end{bmatrix} \)
(B) \( A=\begin{bmatrix}2 & 3\\5 & -2\end{bmatrix} \), \( A^{-1} \)
This matches (IV).
\( \displaystyle |A|=-4-15=-19 \)
\( \displaystyle \mathrm{adj}\,A=\begin{bmatrix}-2 & -3\\-5 & 2\end{bmatrix} \)
\( \displaystyle A^{-1}=\frac{1}{-19}\begin{bmatrix}-2 & -3\\-5 & 2\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}\frac{2}{19} & \frac{3}{19}\\\frac{5}{19} & \frac{-2}{19}\end{bmatrix} \)
(C) \( A=\begin{bmatrix}-3 & 5\\2 & 4\end{bmatrix} \), adj A
This matches (I).
\( \displaystyle \mathrm{adj}\,A=\begin{bmatrix}4 & -5\\-2 & -3\end{bmatrix} \)
(D) \( A=\begin{bmatrix}2 & 5\\-3 & 1\end{bmatrix} \), \( A^{-1} \)
This matches (II).
\( \displaystyle |A|=2+15=17 \)
\( \displaystyle A^{-1}=\frac{1}{17}\begin{bmatrix}1 & -5\\3 & 2\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}\frac{1}{17} & \frac{-5}{17}\\\frac{3}{17} & \frac{2}{17}\end{bmatrix} \)
So the matching is (A)-(III), (B)-(IV), (C)-(I), (D)-(II).
Shortcut: The denominators settle (B) and (D) at once: |A| = −19 in (B) and 17 in (D). Only Option (4) matches (B) with 19 and (D) with 17.