Let `A=[a_(ij)]_(3xx3)` be a square matrix such that `A A^(T)=4I, |A| lt 0`. If `|(a_(11)+4,a_(12),a_(13)),(a_(21),a_(22)+4,a_(23)),(a_(31),a_(32),a_(33)+4)|=5lambda|A+I|.` Then `lambda` is equal to

Options

A. `(4)/(5)`
B. `-(4)/(5)`
C. `(8)/(5)`
D. `-(8)/(5)`

Answer

To solve the given problem, we will follow a systematic approach step-by-step.

### Step 1: Understand the given conditions
We have a square matrix \( A \) such that:
1. \( A A^T = 4I \)
2. \( |A| < 0 \)

### Step 2: Analyze the determinant condition
From \( A A^T = 4I \), we can take the determinant on both sides:
\[ |A A^T| = |4I| \]
Using the property of determinants, we have:
\[ |A A^T| = |A| |A^T| = |A|^2 \]
And,
\[ |4I| = 4^3 = 64 \]
Thus, we can equate:
\[ |A|^2 = 64 \]
Taking the square root gives:
\[ |A| = 8 \quad \text{(since } |A| < 0, \text{ we take } |A| = -8\text{)} \]

### Step 3: Set up the determinant expression
We need to evaluate the determinant:
\[ \left| \begin{array}{ccc} a_{11}+4 & a_{12} & a_{13} \\ a_{21} & a_{22}+4 & a_{23} \\ a_{31} & a_{32} & a_{33}+4 \end{array} \right| \]
This can be rewritten using the property of determinants:
\[ = \left| A + 4I \right| \]
where \( I \) is the identity matrix.

### Step 4: Calculate \( |A + 4I| \)
Using the property of determinants:
\[ |A + 4I| = |A + 4I| = |A| + 4|I| + \text{(terms involving other elements)} \]
Since we know \( |A| = -8 \) and \( |I| = 1 \):
\[ |A + 4I| = -8 + 4 \cdot 3 = -8 + 12 = 4 \]

### Step 5: Relate the determinant to \( 5\lambda |A + I| \)
We are given:
\[ |(a_{11}+4, a_{12}, a_{13}), (a_{21}, a_{22}+4, a_{23}), (a_{31}, a_{32}, a_{33}+4)| = 5\lambda |A + I| \]
From our previous calculations:
\[ |A + 4I| = |A| + 4|I| = -8 + 12 = 4 \]

### Step 6: Calculate \( |A + I| \)
We also need to compute \( |A + I| \):
\[ |A + I| = |A| + |I| = -8 + 1 = -7 \]

### Step 7: Substitute and solve for \( \lambda \)
Now substituting back into the equation:
\[ 4 = 5\lambda (-7) \]
Solving for \( \lambda \):
\[ 4 = -35\lambda \implies \lambda = -\frac{4}{35} \]

### Conclusion
Thus, the value of \( \lambda \) is:
\[ \lambda = -\frac{4}{35} \]