Let `A=[(1,0,3),(0,b,5),(-(1)/(3),0,c)]`, where a, b, c are positive integers. If `tr(A)=7`, then the greatest value of `|A|` is (where tr (A) denotes the trace of matric A i.e. the sum of principal diagonal elements of matrix A)

Options

A. 6
B. 12
C. 15
D. 10

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the matrix and the trace condition
The matrix \( A \) is given as:
\[ A = \begin{pmatrix} 1 & 0 & 3 \\ 0 & b & 5 \\ -\frac{1}{3} & 0 & c \end{pmatrix} \]
The trace of matrix \( A \), denoted as \( \text{tr}(A) \), is the sum of its diagonal elements:
\[ \text{tr}(A) = 1 + b + c \]
We are given that \( \text{tr}(A) = 7 \). Therefore, we can set up the equation:
\[ 1 + b + c = 7 \]
This simplifies to:
\[ b + c = 6 \]

### Step 2: Express the determinant of the matrix
The determinant \( |A| \) of the matrix \( A \) can be calculated using the formula for the determinant of a 3x3 matrix:
\[ |A| = a(ei - fh) - b(di - fg) + c(dh - eg) \]
For our matrix:
\[ |A| = 1 \cdot (b \cdot c - 0 \cdot 5) - 0 \cdot (0 \cdot c - 5 \cdot (-\frac{1}{3})) + 3 \cdot (0 \cdot 0 - b \cdot (-\frac{1}{3})) \]
This simplifies to:
\[ |A| = 1 \cdot (b \cdot c) + 3 \cdot \left(\frac{b}{3}\right) = bc + b \]
Thus, we have:
\[ |A| = b(c + 1) \]

### Step 3: Substitute \( c \) in terms of \( b \)
From the equation \( b + c = 6 \), we can express \( c \) as:
\[ c = 6 - b \]
Substituting this into the determinant expression gives:
\[ |A| = b((6 - b) + 1) = b(7 - b) \]

### Step 4: Find the maximum value of \( |A| \)
To maximize \( |A| = b(7 - b) \), we can rewrite it as:
\[ |A| = 7b - b^2 \]
This is a quadratic function in \( b \) which opens downwards (since the coefficient of \( b^2 \) is negative). The maximum value occurs at the vertex, given by:
\[ b = -\frac{B}{2A} = -\frac{7}{-2} = \frac{7}{2} = 3.5 \]
Since \( b \) must be a positive integer, we check the integer values around \( 3.5 \), which are \( b = 3 \) and \( b = 4 \).

### Step 5: Calculate \( |A| \) for integer values of \( b \)
1. If \( b = 3 \):
\[ c = 6 - 3 = 3 \quad \Rightarrow \quad |A| = 3(7 - 3) = 3 \cdot 4 = 12 \]
2. If \( b = 4 \):
\[ c = 6 - 4 = 2 \quad \Rightarrow \quad |A| = 4(7 - 4) = 4 \cdot 3 = 12 \]

Both cases yield the same determinant value.

### Conclusion
The greatest value of \( |A| \) is:
\[ \boxed{12} \]