Let `A=[a_(ij)]_(3xx3)` be a matrix such that `A A^T=4T` and `a_(ij)+2c_(ij)=0` , where `c_(ij)` is the cofactor of `a_(ij)` and I is the unit matrix of order 3 .
`|(a_11 +4 , a_12 , a_13),(a_21 , a_(22)+4 ,a_23),(a_31,a_32,a_(33)+4)|+5lambda|(a_(11)+1 , a_12,a_13),(a_21,a_(22)+1, a_23),(a_31, a_32 , a_(33)+1)|=0` then the value of `10lambda` is ____

Answer

To solve the given problem step by step, we start with the information provided in the question.

### Step 1: Understand the given conditions
We are given that:
1. \( A A^T = 4I \)
2. \( a_{ij} + 2c_{ij} = 0 \) where \( c_{ij} \) is the cofactor of \( a_{ij} \) and \( I \) is the identity matrix of order 3.

### Step 2: Determine the determinant of matrix \( A \)
From the equation \( A A^T = 4I \), we can take the determinant on both sides:
\[ \text{det}(A A^T) = \text{det}(4I) \]
Using the property of determinants, we have:
\[ \text{det}(A) \cdot \text{det}(A^T) = 4^3 \]
Since \( \text{det}(A^T) = \text{det}(A) \), we can write:
\[ (\text{det}(A))^2 = 64 \]
Thus,
\[ \text{det}(A) = \pm 8 \]

### Step 3: Use the relationship between \( a_{ij} \) and \( c_{ij} \)
From the condition \( a_{ij} + 2c_{ij} = 0 \), we can express \( a_{ij} \) in terms of \( c_{ij} \):
\[ a_{ij} = -2c_{ij} \]

### Step 4: Substitute \( a_{ij} \) in the determinant expression
Now, we substitute \( a_{ij} \) in the determinant expressions we need to evaluate:
1. The first determinant:
\[ \begin{vmatrix} a_{11} + 4 & a_{12} & a_{13} \\ a_{21} & a_{22} + 4 & a_{23} \\ a_{31} & a_{32} & a_{33} + 4 \end{vmatrix} \]
2. The second determinant:
\[ \begin{vmatrix} a_{11} + 1 & a_{12} & a_{13} \\ a_{21} & a_{22} + 1 & a_{23} \\ a_{31} & a_{32} & a_{33} + 1 \end{vmatrix} \]

### Step 5: Set up the equation
The problem states that:
\[ \text{det}(A + 4I) + 5\lambda \cdot \text{det}(A + I) = 0 \]
From our previous steps, we can denote:
- \( \text{det}(A + 4I) = -2 \) (as derived)
- \( \text{det}(A + I) = -2 \)

Substituting these into the equation gives:
\[ -2 + 5\lambda(-2) = 0 \]

### Step 6: Solve for \( \lambda \)
This simplifies to:
\[ -2 - 10\lambda = 0 \]
Thus,
\[ 10\lambda = -2 \implies \lambda = -\frac{1}{5} \]

### Step 7: Calculate \( 10\lambda \)
Finally, we find:
\[ 10\lambda = -2 \]

### Final Answer
The value of \( 10\lambda \) is \( \boxed{-2} \).