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

To find the cofactors of the elements in matrix A, we need to find the determinants of the 2x2 matrices formed by removing the row and column of each element. Cofactor of element a11 (in position (1, 1)): \[ C_{11} = \left|\begin{array}{cc} 1 & -2 \\ -2 & 1 \end{array}\right| = (1)(1) - (-2)(-2) = -3 \] Cofactor of element a12 (in position (1, 2)): \[ C_{12} = \left|\begin{array}{cc} 2 & -2 \\ 2 & 1 \end{array}\right| = (2)(1) - (-2)(2) = 4 \] Cofactor of element a13 (in position (1, 3)): \[ C_{13} = \left|\begin{array}{cc} 2 & 1 \\ 2 & -2 \end{array}\right| = (2)(-2) - (1)(2) = -6 \] Cofactor of element a21 (in position (2, 1)): \[ C_{21} = \left|\begin{array}{cc} -2 & -2 \\ -2 & 1 \end{array}\right| = (-2)(1) - (-2)(-2) = -6 \] Cofactor of element a22 (in position (2, 2)): \[ C_{22} = \left|\begin{array}{cc} -1 & -2 \\ 2 & 1 \end{array}\right| = (-1)(1) - (-2)(2) = -1 \] Cofactor of element a23 (in position (2, 3)): \[ C_{23} = \left|\begin{array}{cc} -1 & -2 \\ 2 & -2 \end{array}\right| = (-1)(-2) - (-2)(2) = 6 \] Cofactor of element a31 (in position (3, 1)): \[ C_{31} = \left|\begin{array}{cc} -2 & -2 \\ 1 & -2 \end{array}\right| = (-2)(-2) - (-2)(1) = 6 \] Cofactor of element a32 (in position (3, 2)): \[ C_{32} = \left|\begin{array}{cc} -1 & -2 \\ 2 & -2 \end{array}\right| = (-1)(-2) - (-2)(2) = -4 \] Cofactor of element a33 (in position (3, 3)): \[ C_{33} = \left|\begin{array}{cc} -1 & -2 \\ 2 & 1 \end{array}\right| = (-1)(1) - (-2)(2) = -1 \]

Now we will form a new matrix called the cofactor matrix using the cofactors calculated for each element in matrix A. Cofactor matrix: \[ C = \left|\begin{array}{ccc} -3 & 4 & -6 \\ -6 & -1 & 6 \\ 6 & -4 & -1 \end{array}\right| \]

To find the adjoint of A, we should transpose the cofactor matrix C. \[ \operatorname{adj} A = C^T = \left|\begin{array}{ccc} -3 & -6 & 6 \\ 4 & -1 & -4 \\ -6 & 6 & -1 \end{array}\right| \]

Now that we have the adjoint of A, let us compare it with the options provided. (a) A: \[ \begin{array}{ccc} -1 & -2 & -2 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{array}\right| \] This option does not match our calculated adjoint of A. (b) Aᵀ: \[ \begin{array}{ccc} -1 & 2 & 2 \\ -2 & 1 & -2 \\ -2 & -2 & 1 \end{array}\right| \] This option does not match our calculated adjoint of A. (c) 3A: \[ \begin{array}{ccc} -3 & -6 & -6 \\ 6 & 3 & -6 \\ 6 & -6 & 3 \end{array}\right| \] This option does not match our calculated adjoint of A. (d) 3Aᵀ: \[ \begin{array}{ccc} -3 & 6 & 6 \\ -6 & 3 & 6 \\ -6 & 6 & 3 \end{array}\right| \] This option does not match our calculated adjoint of A. As none of the given options match the calculated adjoint of A, the problem may be incorrect, or our calculations may require a second review.