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if A=[{:(1,0),(0,1):}]and B=[{:(a,c),(b,...

`if A=[{:(1,0),(0,1):}]and B=[{:(a,c),(b,d):}],`then show that
`(AB)'=B'A'`

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To prove that \((AB)' = B'A'\), where \(A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\) (the identity matrix) and \(B = \begin{pmatrix} a & c \\ b & d \end{pmatrix}\), we will follow these steps: ### Step 1: Multiply Matrices A and B First, we need to calculate the product \(AB\). \[ AB = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} a & c \\ b & d \end{pmatrix} \] Calculating the elements of the resulting matrix: - The element at position (1,1): \[ 1 \cdot a + 0 \cdot b = a \] - The element at position (1,2): \[ 1 \cdot c + 0 \cdot d = c \] - The element at position (2,1): \[ 0 \cdot a + 1 \cdot b = b \] - The element at position (2,2): \[ 0 \cdot c + 1 \cdot d = d \] Thus, we have: \[ AB = \begin{pmatrix} a & c \\ b & d \end{pmatrix} \] ### Step 2: Find the Transpose of AB Now we will find the transpose of the matrix \(AB\): \[ (AB)' = \begin{pmatrix} a & c \\ b & d \end{pmatrix}' \] The transpose of a matrix is obtained by swapping rows and columns: \[ (AB)' = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \] ### Step 3: Find the Transpose of B Next, we find the transpose of matrix \(B\): \[ B' = \begin{pmatrix} a & c \\ b & d \end{pmatrix}' \] By swapping rows and columns: \[ B' = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \] ### Step 4: Find the Transpose of A Now, we find the transpose of matrix \(A\): \[ A' = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}' \] Since \(A\) is the identity matrix, its transpose remains the same: \[ A' = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \] ### Step 5: Multiply B' and A' Now we need to multiply \(B'\) and \(A'\): \[ B'A' = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \] Calculating the elements of the resulting matrix: - The element at position (1,1): \[ a \cdot 1 + b \cdot 0 = a \] - The element at position (1,2): \[ a \cdot 0 + b \cdot 1 = b \] - The element at position (2,1): \[ c \cdot 1 + d \cdot 0 = c \] - The element at position (2,2): \[ c \cdot 0 + d \cdot 1 = d \] Thus, we have: \[ B'A' = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \] ### Conclusion From Steps 2 and 5, we see that: \[ (AB)' = B'A' \] Hence, we have proved that \((AB)' = B'A'\). ---
To prove that \((AB)' = B'A'\), where \(A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\) (the identity matrix) and \(B = \begin{pmatrix} a & c \\ b & d \end{pmatrix}\), we will follow these steps: ### Step 1: Multiply Matrices A and B First, we need to calculate the product \(AB\). \[ AB = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} a & c \\ b & d \end{pmatrix} \] ...
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