$$
\begin{aligned}
x^2-25 x+24 & =0 \\
x^2-x-24 x+24 & =0 \\
x(x-1)-24(x-1) & =0 \\
(x-1)(x-24) & =0 \Rightarrow x=1,24
\end{aligned}
$$
$\because k$ is one of the root of the Eq. (i),
$$
\begin{array}{rlrl}
\therefore & k & =1,24 \\
& & k & =1, \\
\therefore & \quad A & =\left[\begin{array}{lll}
1 & 2 & 1 \\
3 & 2 & 3 \\
1 & 1 & 1
\end{array}\right] \\
|A| & =1(2-3)-2(3-3)+1(3-2) \\
& =-1-0+1=0 \\
|A| & =0
\end{array}
$$
$k=1$, not possible, because given matrix $A$ is singular.
Now,
$$
\begin{aligned}
& k=24 \text {, } \\
& \therefore \quad A=\left[\begin{array}{ccc}
1 & 2 & 1 \\
3 & 2 & 3 \\
1 & 1 & 24
\end{array}\right] \\
& |A|=1(48-3)-2(72-3)+1(3-2) \\
& =45-138+1=-92 \neq 0 \\
& \operatorname{adj} A=\left[\begin{array}{ccc}
45 & -69 & 1 \\
-47 & 23 & 1 \\
4 & 0 & -4
\end{array}\right]^1 \\
& =\left[\begin{array}{ccc}
45 & -47 & 4 \\
-69 & 23 & 0 \\
1 & 1 & -4
\end{array}\right] \\
& \therefore \quad A^{-1}=\frac{1}{|A|} \cdot \operatorname{adj} A \\
& =-\frac{1}{92}\left[\begin{array}{ccc}
45 & -47 & 4 \\
-69 & 23 & 0 \\
1 & 1 & -4
\end{array}\right] \\
&
\end{aligned}
$$