If $$ A=\left|\begin{array}{ll} -2 & 3 \\ -1 & 1 \end{array}\right| $$ then \(\mathrm{A}^{3}=\ldots\) (a)I (b) \(\mathrm{O}\) (c) \(-\mathrm{A}\) (d) \(\mathrm{A}+\mathrm{I}\)

Write down the given matrix A: \[ A=\left|\begin{array}{ll} -2 & 3 \\ -1 & 1 \end{array}\right| \]

Multiply matrix A by itself (A*A): \[ A^{2} = A*A =\left|\begin{array}{ll} -2 & 3 \\ -1 & 1 \end{array}\right| * \left|\begin{array}{ll} -2 & 3 \\ -1 & 1 \end{array}\right| \] To multiply these matrices, use the following formula for each element in the result: \((A^2)_{ij} = \sum_{k=1}^{n} (A_{ik} * A_{kj})\) Now perform the multiplication: \[ A^{2} = \left|\begin{array}{ll} (-2)*(-2) + 3*(-1) & (-2)*3 + 3*1\\ (-1)*(-2) + 1*(-1) & (-1)*3 + 1*1 \end{array}\right| \] Calculate the values within the resulting matrix: \[ A^{2} = \left|\begin{array}{ll} 4 - 3 & -6 + 3\\ 2 - 1 & -3 + 1 \end{array}\right| \] And the result becomes: \[ A^{2} = \left|\begin{array}{ll} 1 & -3\\ 1 & -2 \end{array}\right| \]

Now multiply matrix \(A^2\) by matrix A to find \(A^3\): \[ A^{3} = A^{2} * A =\left|\begin{array}{ll} 1 & -3 \\ 1 & -2 \end{array}\right| * \left|\begin{array}{ll} -2 & 3 \\ -1 & 1 \end{array}\right| \] Now perform the multiplication as previously done for A*A: \[ A^{3} = \left|\begin{array}{ll} (1)*(-2) -3*(-1) & 1*3 -3*1\\ (1)*(-2) -2*(-1) & 1*3 -2*1 \end{array}\right| \] Calculate the values within the resulting matrix: \[ A^{3} = \left|\begin{array}{ll} -2 + 3 & 3 - 3\\ -2 + 2 & 3 - 2 \end{array}\right| \] Finally, we have the matrix \(A^3\): \[ A^{3} = \left|\begin{array}{ll} 1 & 0\\ 0 & 1 \end{array}\right| \] Now, compare this result to the given options: (a) I is the identity matrix, \[I=\left|\begin{array}{ll}1 & 0\\0 & 1\end{array}\right|\] (b) O is the zero matrix, \[O=\left|\begin{array}{ll}0 & 0\\0 & 0\end{array}\right|\] (c) -A is the negation of matrix A, \[-A=\left|\begin{array}{ll}2 & -3\\1 & -1\end{array}\right|\] (d) A + I is the sum of matrices A and I, \[ A + I=\left|\begin{array}{ll}-1 & 3\\-1 & 2\end{array}\right|\] Comparing the result with the options, we can see that the correct answer is: \(A^{3} = I\), so the answer is (a).