Let \(A=\left[\begin{array}{cc}2 & -1 \\ 0 & 2\end{array}\right]\).
If \(\mathrm{B}=\mathrm{I}-{ }^3 \mathrm{C}_1(\operatorname{adj} \mathrm{A})+{ }^3 \mathrm{C}_2(\operatorname{adj} \mathrm{A})^2-{ }^3 \mathrm{C}_3(\operatorname{adj} \mathrm{A})^3\), then the sum of all elements of the matrix \(B\) is

\(
\begin{aligned}
\mathrm{B} & =\mathrm{I}-{ }^3 \mathrm{C}_1(\operatorname{adj} \mathrm{A})+{ }^3 \mathrm{C}_2(\operatorname{adj} \mathrm{A})^2-{ }^3 \mathrm{C}_3(\operatorname{adj} \mathrm{A})^3 \\
& =\mathrm{I}-3 \operatorname{adj} \mathrm{A}+3(\operatorname{adj} \mathrm{A})^2-1(\operatorname{adj} \mathrm{A})^3 \\
& =(\mathrm{I}-\operatorname{adj} \mathrm{A})^3 \\
\operatorname{adj} \mathrm{A} & =\left[\begin{array}{ll}
2 & 1 \\
0 & 2
\end{array}\right] \\
\therefore \quad \mathrm{B} & =\left(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]-\left[\begin{array}{ll}
2 & 1 \\
0 & 2
\end{array}\right]\right)^3 \\
& =\left[\begin{array}{rr}
-1 & -1 \\
0 & -1
\end{array}\right] \\
& =\left[\begin{array}{rr}
-1 & -1 \\
0 & -1
\end{array}\right]\left[\begin{array}{rr}
-1 & -1 \\
0 & -1
\end{array}\right]\left[\begin{array}{cc}
-1 & -1 \\
0 & -1
\end{array}\right] \\
& =\left[\begin{array}{rr}
1 & 2 \\
0 & 1
\end{array}\right]\left[\begin{array}{rr}
-1 & -1 \\
0 & -1
\end{array}\right] \\
& =\left[\begin{array}{rr}
-1 & -3 \\
0 & -1
\end{array}\right]
\end{aligned}
\)

Sum of all elements of the matrix B \(=-1-3-1=-5\)