Consider a matrix `A=[a_(ij)[_(3xx3)` where, `a_(ij)={{:(i+2j,ij="even"),(2i-3j,ij="odd"):}`. If `b_(ij)` is the cafactor of `a_(ij)` in matrix A and `C_(ij)=Sigma_(r=1)^(3)a_(ir)b_(jr)`, then `[C_(ij)]_(3xx3)` is

Options

A. `[(1,0,0),(0,1,0),(0,0,1)]`
B. `[(-1,5,-7),(4,6,8),(3,9,-3)]`
C. `[(88,0,0),(0,88,0),(0,0,88)]`
D. `[(2,3,1),(1,6,2),(-1,5,2)]`

Answer

To solve the problem, we need to construct the matrix \( A \) based on the given conditions, find its determinant, and then compute the matrix \( C \) using the cofactors of \( A \).

### Step 1: Construct the matrix \( A \)

The matrix \( A \) is defined as follows:
- \( a_{ij} = i + 2j \) when \( ij \) is even.
- \( a_{ij} = 2i - 3j \) when \( ij \) is odd.

We will construct the \( 3 \times 3 \) matrix \( A \):

\[ \begin{align*} a_{11} & = 1 + 2 \cdot 1 = 3 \quad (1 \cdot 1 \text{ is even}) \\ a_{12} & = 1 + 2 \cdot 2 = 5 \quad (1 \cdot 2 \text{ is even}) \\ a_{13} & = 1 + 2 \cdot 3 = 7 \quad (1 \cdot 3 \text{ is odd}) \\ a_{21} & = 2 + 2 \cdot 1 = 4 \quad (2 \cdot 1 \text{ is even}) \\ a_{22} & = 2 + 2 \cdot 2 = 6 \quad (2 \cdot 2 \text{ is even}) \\ a_{23} & = 2 - 3 \cdot 3 = -7 \quad (2 \cdot 3 \text{ is odd}) \\ a_{31} & = 3 + 2 \cdot 1 = 5 \quad (3 \cdot 1 \text{ is odd}) \\ a_{32} & = 3 - 3 \cdot 2 = -3 \quad (3 \cdot 2 \text{ is even}) \\ a_{33} & = 3 + 2 \cdot 3 = 9 \quad (3 \cdot 3 \text{ is odd}) \\ \end{align*} \]

Thus, the matrix \( A \) is:

\[ A = \begin{bmatrix} 3 & 5 & 7 \\ 4 & 6 & -7 \\ 5 & -3 & 9 \end{bmatrix} \]

### Step 2: Calculate the determinant of \( A \)

To find the determinant of the matrix \( A \), we can use the formula for the determinant of a \( 3 \times 3 \) matrix:

\[ \text{det}(A) = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31}) \]

Calculating each term:

1. \( a_{22}a_{33} - a_{23}a_{32} = 6 \cdot 9 - (-7)(-3) = 54 - 21 = 33 \)
2. \( a_{21}a_{33} - a_{23}a_{31} = 4 \cdot 9 - (-7)(5) = 36 + 35 = 71 \)
3. \( a_{21}a_{32} - a_{22}a_{31} = 4 \cdot (-3) - 6 \cdot 5 = -12 - 30 = -42 \)

Now substituting back into the determinant formula:

\[ \text{det}(A) = 3 \cdot 33 - 5 \cdot 71 + 7 \cdot (-42) \]
\[ = 99 - 355 - 294 = 99 - 649 = -550 \]

### Step 3: Calculate the matrix \( C \)

The matrix \( C \) is defined as:

\[ C_{ij} = \sum_{r=1}^{3} a_{ir} b_{jr} \]

Where \( b_{ij} \) is the cofactor of \( a_{ij} \). Since \( C_{ij} \) is constructed using the cofactors, we can note that \( C \) will have a specific structure based on the determinant of \( A \).

From the properties of determinants and cofactors, we know that:

\[ C_{ij} = \text{det}(A) \cdot \delta_{ij} \]

Where \( \delta_{ij} \) is the Kronecker delta (1 if \( i = j \) and 0 otherwise). Thus, we have:

\[ C = \begin{bmatrix} \text{det}(A) & 0 & 0 \\ 0 & \text{det}(A) & 0 \\ 0 & 0 & \text{det}(A) \end{bmatrix} = \begin{bmatrix} -550 & 0 & 0 \\ 0 & -550 & 0 \\ 0 & 0 & -550 \end{bmatrix} \]

### Final Answer

The matrix \( C \) is:

\[ C = \begin{bmatrix} -550 & 0 & 0 \\ 0 & -550 & 0 \\ 0 & 0 & -550 \end{bmatrix} \]