Given, \(Q=A^{-1}\)
\(\begin{aligned}
A & =\left[\begin{array}{ccc}
1 & -1 & 1 \\
2 & 1 & -3 \\
1 & 1 & 1
\end{array}\right] \\
|A| & =1(1+3)+1(2+3)+1(2-1)=4+5+1 \\
|A| & =10
\end{aligned}\)
\(\operatorname{Adj} A=\left[\begin{array}{ccc}4 & -5 & 1 \\ 2 & 0 & -2 \\ 2 & 5 & 3\end{array}\right]^{\mathrm{T}}\)
Adj \(A=\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{array}\right]\)
\(\begin{aligned}
& A^{-1}=\frac{1}{|\mathrm{~A}|} \operatorname{Adj} A \\
& A^{-1}=\frac{1}{10}\left[\begin{array}{rrr}
4 & 2 & 2 \\
-5 & 0 & 5 \\
1 & -2 & 3
\end{array}\right] \\
& 10 Q=\left[\begin{array}{rrr}
4 & 2 & 2 \\
-5 & 0 & x \\
1 & -2 & 3
\end{array}\right] \\
& 10 A^{-1}=\left[\begin{array}{rrr}
4 & 2 & 2 \\
-5 & 0 & x \\
1 & -2 & 3
\end{array}\right] \\
& {\left[\begin{array}{rrr}
4 & 2 & 2 \\
-5 & 0 & 5 \\
1 & -2 & 3
\end{array}\right]=\left[\begin{array}{rrr}
4 & 2 & 2 \\
-5 & 0 & x \\
1 & -2 & 3
\end{array}\right]} \\
& \therefore \quad x=5 \\
\end{aligned}\)
Hence, option (d) is correct.