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Match List-I with List-II

List-IList-II
(A) If A=[2−534]A=\begin{bmatrix}2 & -5\\3 & 4\end{bmatrix}, then adj A =(I) [4−5−2−3]\begin{bmatrix}4 & -5\\-2 & -3\end{bmatrix}
(B) If A=[235−2]A=\begin{bmatrix}2 & 3\\5 & -2\end{bmatrix}, then A−1=A^{-1}=(II) [117−517317217]\begin{bmatrix}\frac{1}{17} & \frac{-5}{17}\\\frac{3}{17} & \frac{2}{17}\end{bmatrix}
(C) If A=[−3524]A=\begin{bmatrix}-3 & 5\\2 & 4\end{bmatrix}, then adj A =(III) [45−32]\begin{bmatrix}4 & 5\\-3 & 2\end{bmatrix}
(D) If A=[25−31]A=\begin{bmatrix}2 & 5\\-3 & 1\end{bmatrix}, then A−1=A^{-1}=(IV) [219319519−219]\begin{bmatrix}\frac{2}{19} & \frac{3}{19}\\\frac{5}{19} & \frac{-2}{19}\end{bmatrix}

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 4

For [abcd]\begin{bmatrix}a&b\\c&d\end{bmatrix}, adj A=[d−b−ca]adj\,A=\begin{bmatrix}d&-b\\-c&a\end{bmatrix}.

(A) gives [45−32]\begin{bmatrix}4&5\\-3&2\end{bmatrix}, so (III).

(C) gives [4−5−2−3]\begin{bmatrix}4&-5\\-2&-3\end{bmatrix}, so (I).

(B) ∣A∣=−19|A|=-19, so A−1=1−19[−2−3−52]A^{-1}=\frac{1}{-19}\begin{bmatrix}-2&-3\\-5&2\end{bmatrix}, matching (IV).

(D) ∣A∣=17|A|=17, giving (II).

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