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if A[{:(1,3,-1),(2,2,-1),(3,0,-1):}]and ...

`if A[{:(1,3,-1),(2,2,-1),(3,0,-1):}]and B=[{:(-2,3,-1),(-1,2,-1),(-6,9,-4):}],`then show that AB=BA.

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To show that \( AB = BA \) for the given matrices \( A \) and \( B \), we will perform matrix multiplication for both \( AB \) and \( BA \) and compare the results. ### Step 1: Define the matrices Let: \[ A = \begin{pmatrix} 1 & 3 & -1 \\ 2 & 2 & -1 \\ 3 & 0 & -1 \end{pmatrix}, \quad B = \begin{pmatrix} -2 & 3 & -1 \\ -1 & 2 & -1 \\ -6 & 9 & -4 \end{pmatrix} \] ### Step 2: Calculate \( AB \) To find \( AB \), we multiply matrix \( A \) by matrix \( B \): \[ AB = \begin{pmatrix} 1 & 3 & -1 \\ 2 & 2 & -1 \\ 3 & 0 & -1 \end{pmatrix} \begin{pmatrix} -2 & 3 & -1 \\ -1 & 2 & -1 \\ -6 & 9 & -4 \end{pmatrix} \] Calculating each element of the resulting matrix: - First row, first column: \[ 1 \cdot (-2) + 3 \cdot (-1) + (-1) \cdot (-6) = -2 - 3 + 6 = 1 \] - First row, second column: \[ 1 \cdot 3 + 3 \cdot 2 + (-1) \cdot 9 = 3 + 6 - 9 = 0 \] - First row, third column: \[ 1 \cdot (-1) + 3 \cdot (-1) + (-1) \cdot (-4) = -1 - 3 + 4 = 0 \] - Second row, first column: \[ 2 \cdot (-2) + 2 \cdot (-1) + (-1) \cdot (-6) = -4 - 2 + 6 = 0 \] - Second row, second column: \[ 2 \cdot 3 + 2 \cdot 2 + (-1) \cdot 9 = 6 + 4 - 9 = 1 \] - Second row, third column: \[ 2 \cdot (-1) + 2 \cdot (-1) + (-1) \cdot (-4) = -2 - 2 + 4 = 0 \] - Third row, first column: \[ 3 \cdot (-2) + 0 \cdot (-1) + (-1) \cdot (-6) = -6 + 0 + 6 = 0 \] - Third row, second column: \[ 3 \cdot 3 + 0 \cdot 2 + (-1) \cdot 9 = 9 + 0 - 9 = 0 \] - Third row, third column: \[ 3 \cdot (-1) + 0 \cdot (-1) + (-1) \cdot (-4) = -3 + 0 + 4 = 1 \] Thus, we have: \[ AB = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \] ### Step 3: Calculate \( BA \) Now, we calculate \( BA \): \[ BA = \begin{pmatrix} -2 & 3 & -1 \\ -1 & 2 & -1 \\ -6 & 9 & -4 \end{pmatrix} \begin{pmatrix} 1 & 3 & -1 \\ 2 & 2 & -1 \\ 3 & 0 & -1 \end{pmatrix} \] Calculating each element of the resulting matrix: - First row, first column: \[ -2 \cdot 1 + 3 \cdot 2 + (-1) \cdot 3 = -2 + 6 - 3 = 1 \] - First row, second column: \[ -2 \cdot 3 + 3 \cdot 2 + (-1) \cdot 0 = -6 + 6 + 0 = 0 \] - First row, third column: \[ -2 \cdot (-1) + 3 \cdot (-1) + (-1) \cdot (-1) = 2 - 3 + 1 = 0 \] - Second row, first column: \[ -1 \cdot 1 + 2 \cdot 2 + (-1) \cdot 3 = -1 + 4 - 3 = 0 \] - Second row, second column: \[ -1 \cdot 3 + 2 \cdot 2 + (-1) \cdot 0 = -3 + 4 + 0 = 1 \] - Second row, third column: \[ -1 \cdot (-1) + 2 \cdot (-1) + (-1) \cdot (-1) = 1 - 2 + 1 = 0 \] - Third row, first column: \[ -6 \cdot 1 + 9 \cdot 2 + (-4) \cdot 3 = -6 + 18 - 12 = 0 \] - Third row, second column: \[ -6 \cdot 3 + 9 \cdot 2 + (-4) \cdot 0 = -18 + 18 + 0 = 0 \] - Third row, third column: \[ -6 \cdot (-1) + 9 \cdot (-1) + (-4) \cdot (-1) = 6 - 9 + 4 = 1 \] Thus, we have: \[ BA = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \] ### Step 4: Conclusion Since \( AB = BA \), we conclude that: \[ AB = BA = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]
To show that \( AB = BA \) for the given matrices \( A \) and \( B \), we will perform matrix multiplication for both \( AB \) and \( BA \) and compare the results. ### Step 1: Define the matrices Let: \[ A = \begin{pmatrix} 1 & 3 & -1 \\ 2 & 2 & -1 \\ ...
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