if $$ A=\left|\begin{array}{ccc} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{array}\right| $$ then \(\mathrm{A}^{3}=\) (a) 0 (b) \(\mathrm{A}^{\mathrm{T}}\) (c) I (d) \(\mathrm{A}^{-1}\)

First, we need to calculate A^2. To do this, we will multiply matrix A with itself: $$ A^2 = A \times A $$

To perform the matrix multiplication for A^2, follow the normal matrix multiplication procedure: $$ A^2=\left[\begin{array}{ccc} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{array}\right]\times\left[\begin{array}{ccc} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{array}\right] $$ $$ A^2= \left[\begin{array}{ccc} 1(1) + (-1)(2) + 1(1) & 1(-1) + (-1)(-1) + 1(0) & 1(1) + (-1)(0) + 1(0) \\ 2(1) + (-1)(2) + 0(1) & 2(-1) + (-1)(-1) + 0(0) & 2(1) + (-1)(0) + 0(0) \\ 1(1) + 0(2) + 0(1) & 1(-1) + 0(-1) + 0(0) & 1(1) + 0(0) + 0(0) \end{array}\right] $$ $$ A^2=\left[\begin{array}{ccc} 0 & 1 & 1 \\ 0 & 1 & 2 \\ 1 & -1 & 1 \end{array}\right] $$

Now that we have found A^2, we can find A^3 by multiplying A^2 by A: $$ A^3 = A^2 \times A $$

To perform the matrix multiplication, follow the normal matrix multiplication procedure: $$ A^3=\left[\begin{array}{ccc} 0 & 1 & 1 \\ 0 & 1 & 2 \\ 1 & -1 & 1 \end{array}\right]\times\left[\begin{array}{ccc} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{array}\right] $$ $$ A^3= \left[\begin{array}{ccc} 0(1) + 1(2) + 1(1) & 0(-1) + 1(-1) + 1(0) & 0(1) + 1(0) + 1(0) \\ 0(1) + 1(2) + 2(1) & 0(-1) + 1(-1) + 2(0) & 0(1) + 1(0) + 2(0) \\ 1(1) + (-1)(2) + 1(1) & 1(-1) + (-1)(-1) + 1(0) & 1(1) + (-1)(0) + 1(0) \end{array}\right] $$ $$ A^3=\left[\begin{array}{ccc} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{array}\right] $$ Now, we can compare the obtained matrix A^3 with the given options: (a) 0: The matrix A^3 is not a zero matrix. (b) \(A^T\): The transpose of A is: $$ A^T = \left[\begin{matrix} 1 & 2 & 1 \\ -1 & -1 & 0 \\ 1 & 0 & 0 \end{matrix}\right] $$ which is not equal to A^3. (c) I (identity matrix): The identity matrix is $$ I = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] $$ A^3 is not equal to the identity matrix. (d) \(A^{-1}\) (inverse of A): It can be observed that A^3 = A, which implies that A^2 is the inverse of A (i.e., \(A^2 = A^{-1}\)). Therefore, the correct answer is option (d).