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To solve the given matrix problems, let's define the matrices \( A \) and \( B \) and address each part of the exercise. 1. **Matrix Definitions**: \[ A = \begin{pmatrix} -1 & 2 \\ 3 & -4 \\ 5 & -6 \end{pmatrix}, \quad B = \begin{pmatrix} 2 & -3 \\ 1 & 0 \end{pmatrix} \] 2. **Calculate \( A \cdot B \)**: \[ A \cdot B = \begin{pmatrix} (-1)(2) + (2)(1) & (-1)(-3) + (2)(0) \\ (3)(2) + (-4)(1) & (3)(-3) + (-4)(0) \\ (5)(2) + (-6)(1) & (5)(-3) + (-6)(0) \end{pmatrix} = \begin{pmatrix} -2 + 2 & 3 \\ 6 - 4 & -9 \\ 10 - 6 & -15 \end{pmatrix} = \begin{pmatrix} 0 & 3 \\ 2 & -9 \\ 4 & -15 \end{pmatrix} \] 3. **Calculate \( B \cdot A \) (Not possible: Incompatible dimensions for multiplication)**: Since \( B \) is a 2x2 matrix and \( A \) is a 3x2 matrix, \( B \cdot A \) cannot be calculated due to incompatible dimensions for matrix multiplication. 4. **Calculate \( A^T \cdot A \)**: \[ A^T = \begin{pmatrix} -1 & 3 & 5 \\ 2 & -4 & -6 \end{pmatrix} \] \[ A^T \cdot A = \begin{pmatrix} (-1)(-1) + (3)(3) + (5)(5) & (-1)(2) + (3)(-4) + (5)(-6) \\ (2)(-1) + (-4)(3) + (-6)(5) & (2)(2) + (-4)(-4) + (-6)(-6) \end{pmatrix} = \begin{pmatrix} 1 + 9 + 25 & -2 - 12 - 30 \\ -2 - 12 - 30 & 4 + 16 + 36 \end{pmatrix} = \begin{pmatrix} 35 & -44 \\ -44 & 56 \end{pmatrix} \] 5. **Calculate \( A \cdot A^T \)**: \[ A \cdot A^T = \begin{pmatrix} (-1)(-1) + (2)(2) & (-1)(3) + (2)(-4) & (-1)(5) + (2)(-6) \\ (3)(-1) + (-4)(2) & (3)(3) + (-4)(-4) & (3)(5) + (-4)(-6) \\ (5)(-1) + (-6)(2) & (5)(3) + (-6)(-4) & (5)(5) + (-6)(-6) \end{pmatrix} = \begin{pmatrix} 1 + 4 & -3 - 8 & -5 - 12 \\ -3 - 8 & 9 + 16 & 15 + 24 \\ -5 - 12 & 15 + 24 & 25 + 36 \end{pmatrix} = \begin{pmatrix} 5 & -11 & -17 \\ -11 & 25 & 39 \\ -17 & 39 & 61 \end{pmatrix} \] Feel free to ask if you need further assistance with this problem.