Consider a matrix `A=[a_(ij)]_(3xx3)`, where `a_(ij)={{:(i+j","if","ij=even),(i-j","if","ij=odd):}` if `b_(ij)` is cofactor of `a_(ij)` in matrix A and `c_(ij)=sum_(r=1)^(3)a_(ir)b_(jr)`, then value of `root3(det[c_(ij)]_(3xx3))` is

Options

A. `4`
B. `3`
C. `16`
D. `5`

Explanation

To solve the problem step by step, we will first construct the matrix \( A \) based on the given conditions, then find its determinant, and finally determine the matrix \( C \) and its determinant.

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

The matrix \( A = [a_{ij}]_{3 \times 3} \) is defined as follows:
- If \( i+j \) is even, \( a_{ij} = i + j \)
- If \( i+j \) is odd, \( a_{ij} = i - j \)

Let's fill in the elements of the matrix \( A \):

- For \( i = 1 \):
- \( j = 1 \): \( a_{11} = 1 + 1 = 2 \) (even)
- \( j = 2 \): \( a_{12} = 1 + 2 = 3 \) (even)
- \( j = 3 \): \( a_{13} = 1 - 3 = -2 \) (odd)

- For \( i = 2 \):
- \( j = 1 \): \( a_{21} = 2 - 1 = 1 \) (odd)
- \( j = 2 \): \( a_{22} = 2 + 2 = 4 \) (even)
- \( j = 3 \): \( a_{23} = 2 + 3 = 5 \) (even)

- For \( i = 3 \):
- \( j = 1 \): \( a_{31} = 3 - 1 = 2 \) (odd)
- \( j = 2 \): \( a_{32} = 3 + 2 = 5 \) (even)
- \( j = 3 \): \( a_{33} = 3 + 3 = 6 \) (even)

Thus, the matrix \( A \) is:

\[ A = \begin{bmatrix} 2 & 3 & -2 \\ 1 & 4 & 5 \\ 2 & 5 & 6 \end{bmatrix} \]

### Step 2: Calculate the Determinant of Matrix \( A \)

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

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

Calculating this step-by-step:

1. Calculate \( a_{22}a_{33} - a_{23}a_{32} = 4 \cdot 6 - 5 \cdot 5 = 24 - 25 = -1 \)
2. Calculate \( a_{21}a_{33} - a_{23}a_{31} = 1 \cdot 6 - 5 \cdot 2 = 6 - 10 = -4 \)
3. Calculate \( a_{21}a_{32} - a_{22}a_{31} = 1 \cdot 5 - 4 \cdot 2 = 5 - 8 = -3 \)

Now substituting back into the determinant formula:

\[ \det(A) = 2(-1) - 3(-4) - 2(-3) = -2 + 12 + 6 = 16 \]

### Step 3: Calculate the Cofactor Matrix \( B \)

The cofactor \( b_{ij} \) of \( a_{ij} \) in matrix \( A \) is calculated by taking the determinant of the \( 2 \times 2 \) matrix that remains after removing the \( i \)-th row and \( j \)-th column from \( A \).

However, we will not compute \( B \) explicitly as we will use its properties in the next step.

### Step 4: Construct the Matrix \( C \)

The matrix \( C = [c_{ij}]_{3 \times 3} \) is defined as:

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

From the properties of cofactors, we know that \( C \) can be expressed as:

\[ C = \det(A) I \]

where \( I \) is the identity matrix. Therefore, since \( \det(A) = 16 \):

\[ C = 16 I = \begin{bmatrix} 16 & 0 & 0 \\ 0 & 16 & 0 \\ 0 & 0 & 16 \end{bmatrix} \]

### Step 5: Calculate the Determinant of Matrix \( C \)

The determinant of a diagonal matrix is the product of its diagonal elements:

\[ \det(C) = 16 \cdot 16 \cdot 16 = 16^3 \]

### Step 6: Final Calculation

We need to find \( \sqrt[3]{\det(C)} \):

\[ \sqrt[3]{\det(C)} = \sqrt[3]{16^3} = 16 \]

### Final Answer

The value of \( \sqrt[3]{\det[C_{ij}]_{3 \times 3}} \) is \( 16 \).

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