Glven the following matrices, perform the following: Glven the following: \[ A=\left[\begin{array}{ccc} 3 & 2 & -1 \\ 7 & 3 & -2 \\ 6 & 1 & 0 \end{array}\right] \] \[ B=\left[\begin{array}{ccc} -1 & 4 & 2 \\ 6 & 6 & -1 \\ -5 & 2 & 6 \end{array}\right] \] Find \[ A+B \] Find \( A-B \) Find \[ 2 A \] Find 3B Find \( \quad 2 A-3 B \) Find \( \quad 3 B-2 A \) Additive Inverse Matrix (for \( 3 \times 3 \) ) Find \( -A \)

\( A = \begin{bmatrix} 3 & 2 & -1 \\ 7 & 3 & -2 \\ 6 & 1 & 0 \end{bmatrix} \quad \text{and} \quad B = \begin{bmatrix} -1 & 4 & 2 \\ 6 & 6 & -1 \\ -5 & 2 & 6 \end{bmatrix} \) **1. Calculate \( A + B \):** Add corresponding elements: - First row: - \( 3 + (-1) = 2 \) - \( 2 + 4 = 6 \) - \( -1 + 2 = 1 \) - Second row: - \( 7 + 6 = 13 \) - \( 3 + 6 = 9 \) - \( -2 + (-1) = -3 \) - Third row: - \( 6 + (-5) = 1 \) - \( 1 + 2 = 3 \) - \( 0 + 6 = 6 \) Thus, \[ A + B = \begin{bmatrix} 2 & 6 & 1 \\ 13 & 9 & -3 \\ 1 & 3 & 6 \end{bmatrix} \] **2. Calculate \( A - B \):** Subtract corresponding elements: - First row: - \( 3 - (-1) = 4 \) - \( 2 - 4 = -2 \) - \( -1 - 2 = -3 \) - Second row: - \( 7 - 6 = 1 \) - \( 3 - 6 = -3 \) - \( -2 - (-1) = -1 \) - Third row: - \( 6 - (-5) = 11 \) - \( 1 - 2 = -1 \) - \( 0 - 6 = -6 \) Thus, \[ A - B = \begin{bmatrix} 4 & -2 & -3 \\ 1 & -3 & -1 \\ 11 & -1 & -6 \end{bmatrix} \] **3. Calculate \( 2A \):** Multiply every element in \( A \) by 2: - First row: - \( 2 \times 3 = 6 \) - \( 2 \times 2 = 4 \) - \( 2 \times (-1) = -2 \) - Second row: - \( 2 \times 7 = 14 \) - \( 2 \times 3 = 6 \) - \( 2 \times (-2) = -4 \) - Third row: - \( 2 \times 6 = 12 \) - \( 2 \times 1 = 2 \) - \( 2 \times 0 = 0 \) Thus, \[ 2A = \begin{bmatrix} 6 & 4 & -2 \\ 14 & 6 & -4 \\ 12 & 2 & 0 \end{bmatrix} \] **4. Calculate \( 3B \):** Multiply every element in \( B \) by 3: - First row: - \( 3 \times (-1) = -3 \) - \( 3 \times 4 = 12 \) - \( 3 \times 2 = 6 \) - Second row: - \( 3 \times 6 = 18 \) - \( 3 \times 6 = 18 \) - \( 3 \times (-1) = -3 \) - Third row: - \( 3 \times (-5) = -15 \) - \( 3 \times 2 = 6 \) - \( 3 \times 6 = 18 \) Thus, \[ 3B = \begin{bmatrix} -3 & 12 & 6 \\ 18 & 18 & -3 \\ -15 & 6 & 18 \end{bmatrix} \] **5. Calculate \( 2A - 3B \):** Subtract corresponding elements from \( 2A \) and \( 3B \): - First row: - \( 6 - (-3) = 9 \) - \( 4 - 12 = -8 \) - \( -2 - 6 = -8 \) - Second row: - \( 14 - 18 = -4 \) - \( 6 - 18 = -12 \) - \( -4 - (-3) = -1 \) - Third row: - \( 12 - (-15) = 27 \) - \( 2 - 6 = -4 \) - \( 0 - 18 = -18 \) Thus, \[ 2A - 3B = \begin{bmatrix} 9 & -8 & -8 \\ -4 & -12 & -1 \\ 27 & -4 & -18 \end{bmatrix} \] **6. Calculate \( 3B - 2A \):** Subtract corresponding elements from \( 3B \) and \( 2A \): - First row: - \( -3 - 6 = -9 \) - \( 12 - 4 = 8 \) - \( 6 - (-2) = 8 \) - Second row: - \( 18 - 14 = 4 \) - \( 18 - 6 = 12 \) - \( -3 - (-4) = 1 \) - Third row: - \( -15 - 12 = -27 \) - \( 6 - 2 = 4 \) - \( 18 - 0 = 18 \) Thus, \[ 3B - 2A = \begin{bmatrix} -9 & 8 & 8 \\ 4 & 12 & 1 \\ -27 & 4 & 18 \end{bmatrix} \] **7. Find the additive inverse \( -A \):** Multiply every element in \( A \) by \(-1\): - First row: - \( -1 \times 3 = -3 \) - \( -1 \times 2 = -2 \) - \( -1 \times (-1) = 1 \) - Second row: - \( -1 \times 7 = -7 \) - \( -1 \times 3 = -3 \) - \( -1 \times (-2) = 2 \) - Third row: - \( -1 \times 6 = -6 \) - \( -1 \times 1 = -1 \) - \( -1 \times 0 = 0 \) Thus, \[ -A = \begin{bmatrix} -3 & -2 & 1 \\ -7 & -3 & 2 \\ -6 & -1 & 0 \end{bmatrix} \]