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1. Given the following non-singular matrices

[tex]\[A = \left[\begin{array}{ccc}2 & 1 & -1 \\ 1 & -2 & 3 \\ -2 & 1 & 2\end{array}\right] \quad \text{and} \quad B = \left[\begin{array}{ccc}1 & -2 & 2 \\ -2 & 1 & 3 \\ 2 & -1 & 1\end{array}\right]\][/tex]

Show that:
i. [tex]\((AB)^T = B^T A^T\)[/tex]

ii. [tex]\(\operatorname{trace}(AB) = \operatorname{trace}(BA)\)[/tex]

Sagot :

GinnyAnswer
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].
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