2. \( \left[\begin{array}{cc}0 & -1 \\ 1 & 2\end{array}\right]^{t}-\left(2\left[\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right]\right)^{t} \)

Alright, I need to solve the matrix expression: \( \left[\begin{array}{cc}0 & -1 \\ 1 & 2\end{array}\right]^{t}-\left(2\left[\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right]\right)^{t} \). Let's break this down step by step to ensure accuracy. First, I'll identify the matrices involved: 1. **Matrix A**: \( \left[\begin{array}{cc}0 & -1 \\ 1 & 2\end{array}\right] \) 2. **Matrix B**: \( \left[\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right] \) The expression involves transposing these matrices and then performing subtraction after scaling Matrix B by a factor of 2. **Step 1: Transpose Matrix A** The transpose of a matrix is obtained by flipping it over its diagonal, which means converting its rows to columns and vice versa. So, \( A^{t} = \left[\begin{array}{cc}0 & 1 \\ -1 & 2\end{array}\right] \) **Step 2: Transpose Matrix B** Similarly, the transpose of Matrix B is: \( B^{t} = \left[\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right] \) **Step 3: Scale Matrix B by 2** Multiplying Matrix B by 2: \( 2B = 2 \times \left[\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right] = \left[\begin{array}{ll}2 \times 1 & 2 \times 2 \\ 2 \times 2 & 2 \times 1\end{array}\right] = \left[\begin{array}{ll}2 & 4 \\ 4 & 2\end{array}\right] \) **Step 4: Transpose the Scaled Matrix B** Now, transpose the scaled Matrix B: \( (2B)^{t} = \left[\begin{array}{ll}2 & 4 \\ 4 & 2\end{array}\right]^{t} = \left[\begin{array}{ll}2 & 4 \\ 4 & 2\end{array}\right] \) (Note: The transpose of a matrix is equal to the matrix itself if it's symmetric, which it is in this case.) **Step 5: Subtract the Transposed Scaled Matrix B from the Transposed Matrix A** Now, perform the subtraction: \( A^{t} - (2B)^{t} = \left[\begin{array}{cc}0 & 1 \\ -1 & 2\end{array}\right] - \left[\begin{array}{ll}2 & 4 \\ 4 & 2\end{array}\right] \) To subtract two matrices, subtract corresponding elements: - First row, first column: \( 0 - 2 = -2 \) - First row, second column: \( 1 - 4 = -3 \) - Second row, first column: \( -1 - 4 = -5 \) - Second row, second column: \( 2 - 2 = 0 \) So, the resulting matrix is: \( \left[\begin{array}{cc}-2 & -3 \\ -5 & 0\end{array}\right] \) **Double-Checking the Calculations** It's always good to verify the results to ensure there are no calculation errors. 1. **Transpose of A**: Correctly flipped the rows and columns. 2. **Scaling Matrix B**: Multiplied each element by 2 accurately. 3. **Transpose of Scaled B**: Recognized that the matrix is symmetric, so the transpose is the same as the original. 4. **Subtraction**: Performed element-wise subtraction correctly. Everything checks out, so the final result is accurate.