Transpose Properties D3018B

1. **Problem Statement:** Verify the transpose properties for matrices \(A\) and \(B\) given as: $$A = \begin{pmatrix} 1 & -1 & 2 \\ 5 & -4 & 3 \\ 1 & -2 & -3 \end{pmatrix}, \quad B = \begin{pmatrix} -2 & 6 & -2 \\ -1 & 0 & 1 \\ -2 & 1 & 0 \end{pmatrix}$$ We will verify the following properties: (i) \((A + B)^t = A^t + B^t\) (ii) \((A^t)^t = A\) (iii) \((kA)^t = k A^t\) for scalar \(k\) (iv) \((AB)^t = B^t A^t\) --- 2. **Property (i):** \((A + B)^t = A^t + B^t\) First, compute \(A + B\): $$A + B = \begin{pmatrix} 1 + (-2) & -1 + 6 & 2 + (-2) \\ 5 + (-1) & -4 + 0 & 3 + 1 \\ 1 + (-2) & -2 + 1 & -3 + 0 \end{pmatrix} = \begin{pmatrix} -1 & 5 & 0 \\ 4 & -4 & 4 \\ -1 & -1 & -3 \end{pmatrix}$$ Now, transpose \(A + B\): $$(A + B)^t = \begin{pmatrix} -1 & 4 & -1 \\ 5 & -4 & -1 \\ 0 & 4 & -3 \end{pmatrix}$$ Next, compute \(A^t\) and \(B^t\): $$A^t = \begin{pmatrix} 1 & 5 & 1 \\ -1 & -4 & -2 \\ 2 & 3 & -3 \end{pmatrix}, \quad B^t = \begin{pmatrix} -2 & -1 & -2 \\ 6 & 0 & 1 \\ -2 & 1 & 0 \end{pmatrix}$$ Sum \(A^t + B^t\): $$A^t + B^t = \begin{pmatrix} 1 + (-2) & 5 + (-1) & 1 + (-2) \\ -1 + 6 & -4 + 0 & -2 + 1 \\ 2 + (-2) & 3 + 1 & -3 + 0 \end{pmatrix} = \begin{pmatrix} -1 & 4 & -1 \\ 5 & -4 & -1 \\ 0 & 4 & -3 \end{pmatrix}$$ We see that \((A + B)^t = A^t + B^t\) holds. --- 3. **Property (ii):** \((A^t)^t = A\) Transpose \(A\) to get \(A^t\) as above. Now transpose \(A^t\): $$ (A^t)^t = \begin{pmatrix} 1 & -1 & 2 \\ 5 & -4 & 3 \\ 1 & -2 & -3 \end{pmatrix} = A $$ Thus, \((A^t)^t = A\) is verified. --- 4. **Property (iii):** \((kA)^t = k A^t\) for scalar \(k\) Let \(k = 3\) (example scalar). Compute \(3A\): $$3A = \begin{pmatrix} 3 & -3 & 6 \\ 15 & -12 & 9 \\ 3 & -6 & -9 \end{pmatrix}$$ Transpose \(3A\): $$(3A)^t = \begin{pmatrix} 3 & 15 & 3 \\ -3 & -12 & -6 \\ 6 & 9 & -9 \end{pmatrix}$$ Compute \(3 A^t\): $$3 A^t = 3 \times \begin{pmatrix} 1 & 5 & 1 \\ -1 & -4 & -2 \\ 2 & 3 & -3 \end{pmatrix} = \begin{pmatrix} 3 & 15 & 3 \\ -3 & -12 & -6 \\ 6 & 9 & -9 \end{pmatrix}$$ Hence, \((kA)^t = k A^t\) holds. --- 5. **Property (iv):** \((AB)^t = B^t A^t\) First, compute \(AB\): $$AB = \begin{pmatrix} 1 & -1 & 2 \\ 5 & -4 & 3 \\ 1 & -2 & -3 \end{pmatrix} \begin{pmatrix} -2 & 6 & -2 \\ -1 & 0 & 1 \\ -2 & 1 & 0 \end{pmatrix}$$ Calculate each element: - First row: - \(1 \times -2 + (-1) \times -1 + 2 \times -2 = -2 + 1 - 4 = -5\) - \(1 \times 6 + (-1) \times 0 + 2 \times 1 = 6 + 0 + 2 = 8\) - \(1 \times -2 + (-1) \times 1 + 2 \times 0 = -2 -1 + 0 = -3\) - Second row: - \(5 \times -2 + (-4) \times -1 + 3 \times -2 = -10 + 4 - 6 = -12\) - \(5 \times 6 + (-4) \times 0 + 3 \times 1 = 30 + 0 + 3 = 33\) - \(5 \times -2 + (-4) \times 1 + 3 \times 0 = -10 -4 + 0 = -14\) - Third row: - \(1 \times -2 + (-2) \times -1 + (-3) \times -2 = -2 + 2 + 6 = 6\) - \(1 \times 6 + (-2) \times 0 + (-3) \times 1 = 6 + 0 - 3 = 3\) - \(1 \times -2 + (-2) \times 1 + (-3) \times 0 = -2 - 2 + 0 = -4\) So, $$AB = \begin{pmatrix} -5 & 8 & -3 \\ -12 & 33 & -14 \\ 6 & 3 & -4 \end{pmatrix}$$ Transpose \(AB\): $$(AB)^t = \begin{pmatrix} -5 & -12 & 6 \\ 8 & 33 & 3 \\ -3 & -14 & -4 \end{pmatrix}$$ Now compute \(B^t A^t\): Recall: $$B^t = \begin{pmatrix} -2 & -1 & -2 \\ 6 & 0 & 1 \\ -2 & 1 & 0 \end{pmatrix}, \quad A^t = \begin{pmatrix} 1 & 5 & 1 \\ -1 & -4 & -2 \\ 2 & 3 & -3 \end{pmatrix}$$ Calculate \(B^t A^t\): - First row: - \((-2) \times 1 + (-1) \times -1 + (-2) \times 2 = -2 + 1 - 4 = -5\) - \((-2) \times 5 + (-1) \times -4 + (-2) \times 3 = -10 + 4 - 6 = -12\) - \((-2) \times 1 + (-1) \times -2 + (-2) \times -3 = -2 + 2 + 6 = 6\) - Second row: - \(6 \times 1 + 0 \times -1 + 1 \times 2 = 6 + 0 + 2 = 8\) - \(6 \times 5 + 0 \times -4 + 1 \times 3 = 30 + 0 + 3 = 33\) - \(6 \times 1 + 0 \times -2 + 1 \times -3 = 6 + 0 - 3 = 3\) - Third row: - \((-2) \times 1 + 1 \times -1 + 0 \times 2 = -2 - 1 + 0 = -3\) - \((-2) \times 5 + 1 \times -4 + 0 \times 3 = -10 - 4 + 0 = -14\) - \((-2) \times 1 + 1 \times -2 + 0 \times -3 = -2 - 2 + 0 = -4\) So, $$B^t A^t = \begin{pmatrix} -5 & -12 & 6 \\ 8 & 33 & 3 \\ -3 & -14 & -4 \end{pmatrix}$$ This matches \((AB)^t\), verifying the property. --- **Final conclusion:** All four transpose properties (i) to (iv) are verified for the given matrices \(A\) and \(B\).