Match List-I with List-II

Choose the correct answer from the options given below:

(A)-(III), (B)-(I), (C)-(IV), (D)-(II)

The correct answer is Option (3) → (A)-(III), (B)-(I), (C)-(IV), (D)-(II)

For a $2\times2$ matrix $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$,

$A^{-1}=\frac{1}{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$

(A) $A=\begin{pmatrix}1&7\\4&-2\end{pmatrix}$

$|A|=1(-2)-7(4)=-30$

$A^{-1}=\frac{1}{-30}\begin{pmatrix}-2&-7\\-4&1\end{pmatrix} =\begin{pmatrix}\frac{1}{15}&\frac{7}{30}\\\frac{2}{15}&-\frac{1}{30}\end{pmatrix}$

$(A)\rightarrow(III)$

(B) $B=\begin{pmatrix}6&-3\\2&4\end{pmatrix}$

$|B|=6(4)-(-3)(2)=30$

$B^{-1}=\frac{1}{30}\begin{pmatrix}4&3\\-2&6\end{pmatrix} =\begin{pmatrix}\frac{2}{15}&\frac{1}{10}\\-\frac{1}{15}&\frac{1}{5}\end{pmatrix}$

$(B)\rightarrow(I)$

(C) $C=\begin{pmatrix}5&2\\-5&4\end{pmatrix}$

$|C|=5(4)-2(-5)=30$

$C^{-1}=\frac{1}{30}\begin{pmatrix}4&-2\\5&5\end{pmatrix} =\begin{pmatrix}\frac{2}{15}&-\frac{1}{15}\\\frac{1}{6}&\frac{1}{6}\end{pmatrix}$

$(C)\rightarrow(IV)$

(D) $D=\begin{pmatrix}7&4\\3&6\end{pmatrix}$

$|D|=7(6)-4(3)=30$

$D^{-1}=\frac{1}{30}\begin{pmatrix}6&-4\\-3&7\end{pmatrix} =\begin{pmatrix}\frac{1}{5}&-\frac{2}{15}\\-\frac{1}{10}&\frac{7}{30}\end{pmatrix}$

$(D)\rightarrow(II)$

Final Matching: (A)-(III), (B)-(I), (C)-(IV), (D)-(II).