If $A=\left[\begin{array}{lll}2 & 5 & 3 \\ 3 & 1 & 2 \\ 1 & 2 & 1\end{array}\right]$, then $A^{-1}$ is equal to

$\left[\begin{array}{ccc}-3 / 4 & 1 / 4 & 7 / 4 \\ -1 / 4 & -1 / 4 & 5 / 4 \\ 5 / 4 & 1 / 4 & -13 / 4\end{array}\right]$

Since, $|A|=\left|\begin{array}{lll}2 & 5 & 3 \\ 3 & 1 & 2 \\ 1 & 2 & 1\end{array}\right|$

= 2(1 – 4) –5 (3 –2 )+( 6 –1 ) = –6 –5 + 15 =15 – 11 = 4 ≠0

⇒ A is a non – singular matrix.

Now,

$A_{11}=-3, \quad A_{12}=-1, \quad A_{13}=5$

$A_{21}=1 \quad A_{22}=-1, \quad A_{23}=1$

$A_{31}=7, \quad A_{32}=5, \quad A_{33}=-13$

Let,  $B ⥂ ⥂ ⥂ =\left[\begin{array}{ccc}-3 & -1 & 5 \\ 1 & -1 & 1 \\ 7 & 5 & -13\end{array}\right]$

$\Rightarrow adj~(A)=B'=\left[\begin{array}{ccc}-3 & 1 & 7 \\ -1 & -1 & 1 \\ 7 & 5 & -13\end{array}\right]$

Hence, $A^{-1}=\frac{adj.~(A)}{|A|}=\left[\begin{array}{ccc}\frac{-3}{4} & \frac{1}{4} & \frac{7}{4} \\ \frac{-1}{4} & \frac{-1}{4} & \frac{5}{4} \\ \frac{5}{4} & \frac{1}{4} & \frac{-13}{4}\end{array}\right]$

Hence (1) is the correct answer.