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Sagot :
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].
### 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].
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