If then \( A=\left[\begin{array}{ccc}1 & -1 & 0 \\ -1 & 0 & -1 \\ 0 & -1 & 0\end{array}\right] \) \( A^{-1}=\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 0 & 1\end{array}\right] \)

It seems there is a misunderstanding in the provided information. The matrix \( A \) is a \( 3 \times 3 \) matrix, while the matrix \( A^{-1} \) is given as a \( 2 \times 3 \) matrix. For a matrix to have an inverse, both the original matrix and its inverse must be square matrices of the same size. Let's first check if the matrix \( A \) is invertible by calculating its determinant. The matrix \( A \) is: \[ A = \begin{bmatrix} 1 & -1 & 0 \\ -1 & 0 & -1 \\ 0 & -1 & 0 \end{bmatrix} \] To find the determinant of \( A \), we can use the formula for the determinant of a \( 3 \times 3 \) matrix: \[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \] where the matrix is represented as: \[ \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} \] For our matrix \( A \): - \( a = 1, b = -1, c = 0 \) - \( d = -1, e = 0, f = -1 \) - \( g = 0, h = -1, i = 0 \) Now, we can calculate the determinant: \[ \text{det}(A) = 1(0 \cdot 0 - (-1)(-1)) - (-1)(-1 \cdot 0 - (-1)(0)) + 0(-1 \cdot -1 - 0 \cdot 0) \] Calculating each term: 1. \( 1(0 - 1) = -1 \) 2. \( -(-1)(0 - 0) = 0 \) 3. \( 0 \) (the third term is zero) Thus, we have: \[ \text{det}(A) = -1 + 0 + 0 = -1 \] Since the determinant is non-zero (\(-1\)), the matrix \( A \) is invertible. To find the inverse \( A^{-1} \), we can use the formula for the inverse of a \( 3 \times 3 \) matrix, which involves the adjugate and the determinant. However, the provided \( A^{-1} \) does not seem to be correct as it is not a square matrix. If you need to find the actual inverse of \( A \), please let me know, and I can help you calculate it.