Let's break down the solution to the problem step by step: **Given:** \[ \left(2I_3 - \begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix}\right)^T = B \] where \[ B = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & -1 & 1 \end{bmatrix} \] and you are to find \(A = B^T\). --- ### **Step 1: Compute \(2I_3 - M\)** Where \(M = \begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix}\) and \(I_3\) is the \(3 \times 3\) identity matrix. \[ 2I_3 = 2 \times \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix} \] So, \[ 2I_3 - M = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix} - \begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 2-1 & 0-0 & 0-(-1) \\ 0-0 & 2-1 & 0-0 \\ 0-(-1) & 0-0 & 2-1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} \] **Explanation:** We subtract each element of \(M\) from \(2I_3\) element-wise to get the new matrix. --- ### **Step 2: Take the Transpose and Set \(A = B^T\)** Now, take the transpose: \[ \left(2I_3 - M\right)^T = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix}^T = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} \] Since \(A = B^T\) and \(B = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & -1 & 1 \end{bmatrix}\): \[ A = B^T = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & -1 \\ 1 & 0 & 1 \end{bmatrix} \] **Explanation:** The transpose of \(B\) is found by swapping rows and columns. --- ## **Final Answer:** \[ \boxed{ A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & -1 \\ 1 & 0 & 1 \end{bmatrix} } \]