Let A and B are square matrices of order 2 such that `A+adj(B^(T))=[(3,2),(2,3)] and A^(T)-adj(B)=[(-2,-1),(-1, -1)]`, then `A^(2)+2A^(3)+3A^(4)+5A^(5)` is equal to (where `M^(T)` and adj(M) represent the transpose matrix and adjoint matrix of matrix M respectively and I represents the identity matrix of order 2)

Options

A. 4A
B. 7A
C. 11A
D. 10I

Answer

To solve the problem step by step, we will first analyze the given equations involving matrices \( A \) and \( B \).

### Step 1: Set up the equations
We are given two equations:
1. \( A + \text{adj}(B^T) = \begin{pmatrix} 3 & 2 \\ 2 & 3 \end{pmatrix} \)
2. \( A^T - \text{adj}(B) = \begin{pmatrix} -2 & -1 \\ -1 & -1 \end{pmatrix} \)

### Step 2: Transpose the second equation
Taking the transpose of the second equation:
\[ (A^T - \text{adj}(B))^T = \begin{pmatrix} -2 & -1 \\ -1 & -1 \end{pmatrix}^T \]
This gives:
\[ A - \text{adj}(B)^T = \begin{pmatrix} -2 & -1 \\ -1 & -1 \end{pmatrix} \]

### Step 3: Rearranging the equations
From the first equation, we can express \( A \):
\[ A = \begin{pmatrix} 3 & 2 \\ 2 & 3 \end{pmatrix} - \text{adj}(B^T) \]
From the transposed second equation, we can express \( A \):
\[ A = \begin{pmatrix} -2 & -1 \\ -1 & -1 \end{pmatrix} + \text{adj}(B)^T \]

### Step 4: Equating the two expressions for \( A \)
Setting the two expressions for \( A \) equal to each other:
\[ \begin{pmatrix} 3 & 2 \\ 2 & 3 \end{pmatrix} - \text{adj}(B^T) = \begin{pmatrix} -2 & -1 \\ -1 & -1 \end{pmatrix} + \text{adj}(B)^T \]

### Step 5: Solve for \( \text{adj}(B) \)
Rearranging gives:
\[ \begin{pmatrix} 3 & 2 \\ 2 & 3 \end{pmatrix} + \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix} = \text{adj}(B^T) + \text{adj}(B)^T \]
Calculating the left side:
\[ \begin{pmatrix} 5 & 3 \\ 3 & 4 \end{pmatrix} = \text{adj}(B^T) + \text{adj}(B)^T \]

### Step 6: Finding \( A \)
From the earlier equations, we can find \( A \):
\[ A = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \]

### Step 7: Calculate \( A^2 \), \( A^3 \), \( A^4 \), and \( A^5 \)
Calculating \( A^2 \):
\[ A^2 = A \cdot A = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 2 \\ 2 & 2 \end{pmatrix} \]
Calculating \( A^3 \):
\[ A^3 = A^2 \cdot A = \begin{pmatrix} 2 & 2 \\ 2 & 2 \end{pmatrix} \cdot \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 4 & 4 \\ 4 & 4 \end{pmatrix} \]
Calculating \( A^4 \):
\[ A^4 = A^3 \cdot A = \begin{pmatrix} 4 & 4 \\ 4 & 4 \end{pmatrix} \cdot \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 8 & 8 \\ 8 & 8 \end{pmatrix} \]
Calculating \( A^5 \):
\[ A^5 = A^4 \cdot A = \begin{pmatrix} 8 & 8 \\ 8 & 8 \end{pmatrix} \cdot \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 16 & 16 \\ 16 & 16 \end{pmatrix} \]

### Step 8: Calculate \( A^2 + 2A^3 + 3A^4 + 5A^5 \)
Now we can compute:
\[ A^2 + 2A^3 + 3A^4 + 5A^5 = \begin{pmatrix} 2 & 2 \\ 2 & 2 \end{pmatrix} + 2 \begin{pmatrix} 4 & 4 \\ 4 & 4 \end{pmatrix} + 3 \begin{pmatrix} 8 & 8 \\ 8 & 8 \end{pmatrix} + 5 \begin{pmatrix} 16 & 16 \\ 16 & 16 \end{pmatrix} \]
Calculating each term:
\[ = \begin{pmatrix} 2 & 2 \\ 2 & 2 \end{pmatrix} + \begin{pmatrix} 8 & 8 \\ 8 & 8 \end{pmatrix} + \begin{pmatrix} 24 & 24 \\ 24 & 24 \end{pmatrix} + \begin{pmatrix} 80 & 80 \\ 80 & 80 \end{pmatrix} \]
Adding these matrices together:
\[ = \begin{pmatrix} 2 + 8 + 24 + 80 & 2 + 8 + 24 + 80 \\ 2 + 8 + 24 + 80 & 2 + 8 + 24 + 80 \end{pmatrix} = \begin{pmatrix} 114 & 114 \\ 114 & 114 \end{pmatrix} \]

### Final Answer
Thus, the final result is:
\[ \begin{pmatrix} 114 & 114 \\ 114 & 114 \end{pmatrix} \]