Let's solve the given problem step-by-step.
### [tex]\( A \)[/tex] and [tex]\( B \)[/tex]:
Given matrices:
[tex]\[ A = \begin{pmatrix} 1 & 1 & 2 \\ 1 & 2 & 1 \\ 3 & 2 & 2 \end{pmatrix} \][/tex]
[tex]\[ B = \begin{pmatrix} 1 & 2 & 1 \\ 2 & 1 & 1 \\ 1 & -1 & 2 \end{pmatrix} \][/tex]
### Part (i): Calculate [tex]\( A + B \)[/tex]
To find the sum [tex]\( A + B \)[/tex], we simply add the corresponding elements of matrices [tex]\( A \)[/tex] and [tex]\( B \)[/tex].
[tex]\[ A + B = \begin{pmatrix} 1+1 & 1+2 & 2+1 \\ 1+2 & 2+1 & 1+1 \\ 3+1 & 2-1 & 2+2 \end{pmatrix} = \begin{pmatrix} 2 & 3 & 3 \\ 3 & 3 & 2 \\ 4 & 1 & 4 \end{pmatrix} \][/tex]
### Part (ii): Calculate [tex]\( (A B)^T \)[/tex]
To find [tex]\( A B \)[/tex], we perform matrix multiplication. Each element of the product matrix [tex]\( AB \)[/tex] is computed as the dot product of the rows of [tex]\( A \)[/tex] with the columns of [tex]\( B \)[/tex]:
[tex]\[ AB = \begin{pmatrix}
1\cdot1 + 1\cdot2 + 2\cdot1 & 1\cdot2 + 1\cdot1 + 2\cdot(-1) & 1\cdot1 + 1\cdot1 + 2\cdot2 \\
1\cdot1 + 2\cdot2 + 1\cdot1 & 1\cdot2 + 2\cdot1 + 1\cdot(-1) & 1\cdot1 + 2\cdot1 + 1\cdot2 \\
3\cdot1 + 2\cdot2 + 2\cdot1 & 3\cdot2 + 2\cdot1 + 2\cdot(-1) & 3\cdot1 + 2\cdot1 + 2\cdot2
\end{pmatrix}
= \begin{pmatrix}
1+2+2 & 2+1-2 & 1+1+4 \\
1+4+1 & 2+2-1 & 1+2+2 \\
3+4+2 & 6+2-2 & 3+2+4
\end{pmatrix}
= \begin{pmatrix}
5 & 1 & 6 \\
6 & 3 & 5 \\
9 & 6 & 9
\end{pmatrix}
\][/tex]
Now, to find [tex]\( (AB)^T \)[/tex], we take the transpose of the [tex]\( AB \)[/tex] matrix we just calculated. Transposing a matrix involves swapping rows and columns:
[tex]\[ (AB)^T = \begin{pmatrix}
5 & 1 & 6 \\
6 & 3 & 5 \\
9 & 6 & 9
\end{pmatrix}^T = \begin{pmatrix}
5 & 6 & 9 \\
1 & 3 & 6 \\
6 & 5 & 9
\end{pmatrix} \][/tex]
### Final Answer:
[tex]\[
A + B = \begin{pmatrix} 2 & 3 & 3 \\ 3 & 3 & 2 \\ 4 & 1 & 4 \end{pmatrix}
\][/tex]
[tex]\[
(AB)^T = \begin{pmatrix} 5 & 6 & 9 \\ 1 & 3 & 6 \\ 6 & 5 & 9 \end{pmatrix}
\][/tex]
