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if A=[{:(2,-3),(1,4):}],B=[{:(-1,0),(1,2...

if `A=[{:(2,-3),(1,4):}],B=[{:(-1,0),(1,2):}], C=[{:(3,1),(1,2):}]`, then show that A (BC)=(AB)c.

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To show that \( A(BC) = (AB)C \) for the given matrices \( A \), \( B \), and \( C \), we will follow these steps: ### Step 1: Define the matrices Given: \[ A = \begin{pmatrix} 2 & -3 \\ 1 & 4 \end{pmatrix}, \quad B = \begin{pmatrix} -1 & 0 \\ 1 & 2 \end{pmatrix}, \quad C = \begin{pmatrix} 3 & 1 \\ 1 & 2 \end{pmatrix} \] ### Step 2: Calculate \( BC \) To find \( BC \), we perform matrix multiplication: \[ BC = B \cdot C = \begin{pmatrix} -1 & 0 \\ 1 & 2 \end{pmatrix} \cdot \begin{pmatrix} 3 & 1 \\ 1 & 2 \end{pmatrix} \] Calculating the elements: - First row, first column: \( (-1) \cdot 3 + 0 \cdot 1 = -3 + 0 = -3 \) - First row, second column: \( (-1) \cdot 1 + 0 \cdot 2 = -1 + 0 = -1 \) - Second row, first column: \( 1 \cdot 3 + 2 \cdot 1 = 3 + 2 = 5 \) - Second row, second column: \( 1 \cdot 1 + 2 \cdot 2 = 1 + 4 = 5 \) Thus, \[ BC = \begin{pmatrix} -3 & -1 \\ 5 & 5 \end{pmatrix} \] ### Step 3: Calculate \( A(BC) \) Now, we calculate \( A(BC) \): \[ A(BC) = A \cdot BC = \begin{pmatrix} 2 & -3 \\ 1 & 4 \end{pmatrix} \cdot \begin{pmatrix} -3 & -1 \\ 5 & 5 \end{pmatrix} \] Calculating the elements: - First row, first column: \( 2 \cdot (-3) + (-3) \cdot 5 = -6 - 15 = -21 \) - First row, second column: \( 2 \cdot (-1) + (-3) \cdot 5 = -2 - 15 = -17 \) - Second row, first column: \( 1 \cdot (-3) + 4 \cdot 5 = -3 + 20 = 17 \) - Second row, second column: \( 1 \cdot (-1) + 4 \cdot 5 = -1 + 20 = 19 \) Thus, \[ A(BC) = \begin{pmatrix} -21 & -17 \\ 17 & 19 \end{pmatrix} \] ### Step 4: Calculate \( AB \) Next, we calculate \( AB \): \[ AB = A \cdot B = \begin{pmatrix} 2 & -3 \\ 1 & 4 \end{pmatrix} \cdot \begin{pmatrix} -1 & 0 \\ 1 & 2 \end{pmatrix} \] Calculating the elements: - First row, first column: \( 2 \cdot (-1) + (-3) \cdot 1 = -2 - 3 = -5 \) - First row, second column: \( 2 \cdot 0 + (-3) \cdot 2 = 0 - 6 = -6 \) - Second row, first column: \( 1 \cdot (-1) + 4 \cdot 1 = -1 + 4 = 3 \) - Second row, second column: \( 1 \cdot 0 + 4 \cdot 2 = 0 + 8 = 8 \) Thus, \[ AB = \begin{pmatrix} -5 & -6 \\ 3 & 8 \end{pmatrix} \] ### Step 5: Calculate \( (AB)C \) Now, we calculate \( (AB)C \): \[ (AB)C = AB \cdot C = \begin{pmatrix} -5 & -6 \\ 3 & 8 \end{pmatrix} \cdot \begin{pmatrix} 3 & 1 \\ 1 & 2 \end{pmatrix} \] Calculating the elements: - First row, first column: \( -5 \cdot 3 + (-6) \cdot 1 = -15 - 6 = -21 \) - First row, second column: \( -5 \cdot 1 + (-6) \cdot 2 = -5 - 12 = -17 \) - Second row, first column: \( 3 \cdot 3 + 8 \cdot 1 = 9 + 8 = 17 \) - Second row, second column: \( 3 \cdot 1 + 8 \cdot 2 = 3 + 16 = 19 \) Thus, \[ (AB)C = \begin{pmatrix} -21 & -17 \\ 17 & 19 \end{pmatrix} \] ### Conclusion We have shown that: \[ A(BC) = \begin{pmatrix} -21 & -17 \\ 17 & 19 \end{pmatrix} = (AB)C \] Thus, \( A(BC) = (AB)C \).
To show that \( A(BC) = (AB)C \) for the given matrices \( A \), \( B \), and \( C \), we will follow these steps: ### Step 1: Define the matrices Given: \[ A = \begin{pmatrix} 2 & -3 \\ 1 & 4 \end{pmatrix}, \quad B = \begin{pmatrix} -1 & 0 \\ 1 & 2 \end{pmatrix}, \quad C = \begin{pmatrix} 3 & 1 \\ 1 & 2 \end{pmatrix} \] ...
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