Certainly! Let's go through each part step by step.
### Part (i): Show that [tex]\((AB)^T = B^T A^T\)[/tex]
1. Define Matrices [tex]\(A\)[/tex] and [tex]\(B\)[/tex]:
[tex]\[
A = \begin{bmatrix}
2 & 1 & -1 \\
1 & -2 & 3 \\
-2 & 1 & 2
\end{bmatrix}, \quad
B = \begin{bmatrix}
1 & -2 & 2 \\
-2 & 1 & 3 \\
2 & -1 & 1
\end{bmatrix}
\][/tex]
2. Calculate the Product [tex]\(AB\)[/tex]:
[tex]\[
AB = \begin{bmatrix}
2 & 1 & -1 \\
1 & -2 & 3 \\
-2 & 1 & 2
\end{bmatrix}
\begin{bmatrix}
1 & -2 & 2 \\
-2 & 1 & 3 \\
2 & -1 & 1
\end{bmatrix}
= \begin{bmatrix}
-2 & -2 & 6 \\
11 & -7 & -1 \\
0 & 3 & 1
\end{bmatrix}
\][/tex]
3. Calculate the Transpose of [tex]\(AB\)[/tex]:
[tex]\[
(AB)^T = \begin{bmatrix}
-2 & 11 & 0 \\
-2 & -7 & 3 \\
6 & -1 & 1
\end{bmatrix}
\][/tex]
4. Calculate the Transpose of [tex]\(B\)[/tex] and [tex]\(A\)[/tex]:
[tex]\[
B^T = \begin{bmatrix}
1 & -2 & 2 \\
-2 & 1 & -1 \\
2 & 3 & 1
\end{bmatrix}, \quad
A^T = \begin{bmatrix}
2 & 1 & -2 \\
1 & -2 & 1 \\
-1 & 3 & 2
\end{bmatrix}
\][/tex]
5. Calculate the Product [tex]\(B^T A^T\)[/tex]:
[tex]\[
B^T A^T = \begin{bmatrix}
1 & -2 & 2 \\
-2 & 1 & -1 \\
2 & 3 & 1
\end{bmatrix}
\begin{bmatrix}
2 & 1 & -2 \\
1 & -2 & 1 \\
-1 & 3 & 2
\end{bmatrix}
= \begin{bmatrix}
-2 & -2 & 6 \\
11 & -7 & -1 \\
0 & 3 & 1
\end{bmatrix}
\][/tex]
6. Transpose of [tex]\(B^T A^T\)[/tex]:
[tex]\[
(B^T A^T)^T = \begin{bmatrix}
-2 & 11 & 0 \\
-2 & -7 & 3 \\
6 & -1 & 1
\end{bmatrix}
\][/tex]
Clearly,
[tex]\((AB)^T = B^T A^T\)[/tex].
### Part (ii): Show that [tex]\(\text{trace}(AB) = \text{trace}(BA)\)[/tex]
1. Calculate the Product [tex]\(BA\)[/tex]:
[tex]\[
BA = \begin{bmatrix}
1 & -2 & 2 \\
-2 & 1 & 3 \\
2 & -1 & 1
\end{bmatrix}
\begin{bmatrix}
2 & 1 & -1 \\
1 & -2 & 3 \\
-2 & 1 & 2
\end{bmatrix}
= \begin{bmatrix}
-2 & 11 & 0 \\
-2 & -7 & 3 \\
6 & -1 & 1
\end{bmatrix}
\][/tex]
2. Calculate [tex]\(\text{trace}(AB)\)[/tex]:
[tex]\[
\text{trace}(AB) = -2 + (-7) + 1 = -8
\][/tex]
3. Calculate [tex]\(\text{trace}(BA)\)[/tex]:
[tex]\[
\text{trace}(BA) = -2 + (-7) + 1 = -8
\][/tex]
Thus, [tex]\(\text{trace}(AB) = \text{trace}(BA)\)[/tex].
### Conclusion
We have shown that:
1. [tex]\((AB)^T = B^T A^T\)[/tex]
2. [tex]\(\text{trace}(AB) = \text{trace}(BA)\)[/tex]
Both conditions are satisfied for the given matrices [tex]\(A\)[/tex] and [tex]\(B\)[/tex].
