` A = [{:( a,b,c) ,( b,c,a),(c,a,b) :}]` if trace (A)= 9 and a,b , c are positive integers such that ab + bc + ca = 26 Let` A_1 `denotes the adjoint of matrix A , `A_2 ` represent adjoint of `A_1 ….` and so on if value of det` ( A_4) ` is M, then

Options

A. `M= 3^(48)`
B. ` M= 3^(24)`
C. last two digits of M is 61
D. last three digits of M is 861

Answer

To solve the problem step by step, we will analyze the given matrix \( A \) and the conditions provided.

### Step 1: Understanding the Matrix and Its Properties
The matrix \( A \) is given as:
\[ A = \begin{pmatrix} a & b & c \\ b & c & a \\ c & a & b \end{pmatrix} \]
We know that the trace of \( A \) is the sum of its diagonal elements:
\[ \text{trace}(A) = a + c + b = 9 \]

### Step 2: Using the Given Condition
We also have the condition:
\[ ab + bc + ca = 26 \]

### Step 3: Finding \( a^2 + b^2 + c^2 \)
Using the identity:
\[ a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + ac + bc) \]
Substituting the known values:
\[ a^2 + b^2 + c^2 = 9^2 - 2 \cdot 26 = 81 - 52 = 29 \]

### Step 4: Finding the Determinant of Matrix \( A \)
The determinant of a 3x3 matrix can be calculated using the formula:
\[ \text{det}(A) = a(bc - ac) - b(bc - ab) + c(ab - ac) \]
However, for this specific symmetric matrix, we can use the identity:
\[ \text{det}(A) = a^3 + b^3 + c^3 - 3abc \]
We can express \( a^3 + b^3 + c^3 \) using the identity:
\[ a^3 + b^3 + c^3 = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) \]
Substituting the known values:
\[ a^3 + b^3 + c^3 = 9(29 - 26) = 9 \cdot 3 = 27 \]
Thus,
\[ \text{det}(A) = 27 - 3abc \]

### Step 5: Finding \( abc \)
We need to find \( abc \) to calculate \( \text{det}(A) \). We can use the equation:
\[ x^3 - (a+b+c)x^2 + (ab + ac + bc)x - abc = 0 \]
This gives us:
\[ x^3 - 9x^2 + 26x - abc = 0 \]
We can find \( abc \) through trial and error with positive integers \( a, b, c \) that satisfy both conditions.

Assuming \( a = 1, b = 2, c = 6 \):
- \( a + b + c = 1 + 2 + 6 = 9 \)
- \( ab + ac + bc = 1 \cdot 2 + 2 \cdot 6 + 1 \cdot 6 = 2 + 12 + 6 = 20 \) (not valid)

Assuming \( a = 2, b = 3, c = 4 \):
- \( a + b + c = 2 + 3 + 4 = 9 \)
- \( ab + ac + bc = 2 \cdot 3 + 3 \cdot 4 + 2 \cdot 4 = 6 + 12 + 8 = 26 \) (valid)

Thus, \( a = 2, b = 3, c = 4 \):
\[ abc = 2 \cdot 3 \cdot 4 = 24 \]

### Step 6: Final Calculation of Determinant
Substituting \( abc \) back into the determinant:
\[ \text{det}(A) = 27 - 3 \cdot 24 = 27 - 72 = -45 \]

### Step 7: Finding \( \text{det}(A_4) \)
The determinant of the adjoint matrix can be calculated as:
\[ \text{det}(A_n) = (\text{det}(A))^{n-1} \]
For \( n = 4 \):
\[ \text{det}(A_4) = (\text{det}(A))^{3} = (-45)^{3} = -91125 \]
Thus, the value of \( M \) is \( -91125 \).