Let \( A=\left[\begin{array}{lll}1 & 0 & 1 \\ 2 & 1 & 0 \\ 1 & 1 & a\end{array}\right] \). The value(s) of \( a \) such that \( A^{-1} \) exists a. \( \mathbb{R}-\{1\} \) b. None b c. 2 d. \( \mathbb{R}-\{-1\} \) e. -1 f. \( \mathbb{R}-\{2\} \)

To determine the values of \( a \) for which the inverse of matrix \( A \) exists, we need to find the determinant of \( A \) and ensure it is not equal to zero. Calculating the determinant: \[ \text{det}(A) = 1 \cdot \begin{vmatrix} 1 & 0 \\ 1 & a \end{vmatrix} - 0 + 1 \cdot \begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix} \] Calculating the \( 2 \times 2 \) determinants: \[ \begin{vmatrix} 1 & 0 \\ 1 & a \end{vmatrix} = 1 \cdot a - 0 \cdot 1 = a \] \[ \begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix} = 2 \cdot 1 - 1 \cdot 1 = 1 \] Thus, \[ \text{det}(A) = 1 \cdot a + 0 + 1 \cdot 1 = a + 1 \] The matrix \( A \) is invertible if \( \text{det}(A) \neq 0 \), which gives us: \[ a + 1 \neq 0 \implies a \neq -1 \] So, the values of \( a \) for which \( A^{-1} \) exists are all real numbers except \( -1 \). Therefore, the correct answer is: **d**. \( \mathbb{R}-\{-1\} \)