Let a,b,c `in` R be such that a+b+c `gt` 0 and abc = 2 . Let
`A = [(a,b,c),(b,c,a),(c,a,b)]`
If `A^(2)=I` then value of `a^(3)+b^(3)+c^(3)` is

Options

A. 7
B. 2
C. 0
D. `-1`

Answer

To solve the problem, we need to analyze the matrix \( A \) and the conditions given. Let's break it down step by step.

### Step 1: Define the Matrix \( A \)
The matrix \( A \) is defined as:
\[ A = \begin{pmatrix} a & b & c \\ b & c & a \\ c & a & b \end{pmatrix} \]

### Step 2: Calculate \( A^2 \)
To find \( A^2 \), we multiply matrix \( A \) by itself:
\[ A^2 = A \cdot A = \begin{pmatrix} a & b & c \\ b & c & a \\ c & a & b \end{pmatrix} \cdot \begin{pmatrix} a & b & c \\ b & c & a \\ c & a & b \end{pmatrix} \]

Calculating the elements of \( A^2 \):
- First row, first column:
\[ a^2 + b^2 + c^2 \]
- First row, second column:
\[ ab + bc + ca \]
- First row, third column:
\[ ac + ba + cb \]
- Second row, first column:
\[ ba + cb + ac \]
- Second row, second column:
\[ b^2 + c^2 + a^2 \]
- Second row, third column:
\[ bc + ca + ab \]
- Third row, first column:
\[ ca + ab + bc \]
- Third row, second column:
\[ cb + ac + ba \]
- Third row, third column:
\[ c^2 + a^2 + b^2 \]

Thus, we have:
\[ A^2 = \begin{pmatrix} a^2 + b^2 + c^2 & ab + bc + ca & ac + ba + cb \\ ba + cb + ac & b^2 + c^2 + a^2 & bc + ca + ab \\ ca + ab + bc & cb + ac + ba & c^2 + a^2 + b^2 \end{pmatrix} \]

### Step 3: Set \( A^2 = I \)
Given that \( A^2 = I \), where \( I \) is the identity matrix, we have:
\[ \begin{pmatrix} a^2 + b^2 + c^2 & ab + bc + ca & ac + ba + cb \\ ba + cb + ac & b^2 + c^2 + a^2 & bc + ca + ab \\ ca + ab + bc & cb + ac + ba & c^2 + a^2 + b^2 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

From this, we can derive the following equations:
1. \( a^2 + b^2 + c^2 = 1 \)
2. \( ab + bc + ca = 0 \)

### Step 4: Use the Identity for Cubes
We can use the identity for the sum of cubes:
\[ a^3 + b^3 + c^3 = (a+b+c)(a^2 + b^2 + c^2 - ab - ac - bc) \]

### Step 5: Substitute Known Values
From the equations we derived:
- \( a^2 + b^2 + c^2 = 1 \)
- \( ab + ac + bc = 0 \)

Thus,
\[ a^2 + b^2 + c^2 - ab - ac - bc = 1 - 0 = 1 \]

Now, substituting into the identity:
\[ a^3 + b^3 + c^3 = (a + b + c)(1) \]

### Step 6: Find \( a + b + c \)
Given \( a + b + c > 0 \) and \( abc = 2 \), we can use the fact that \( a, b, c \) are the roots of the polynomial \( x^3 - (a+b+c)x^2 + (ab+ac+bc)x - abc = 0 \). Since \( ab + ac + bc = 0 \) and \( abc = 2 \), we have:
\[ x^3 - (a+b+c)x^2 - 2 = 0 \]

### Step 7: Conclusion
Since we don't have specific values for \( a, b, c \), we can conclude that:
\[ a^3 + b^3 + c^3 = (a + b + c) \]

Thus, the value of \( a^3 + b^3 + c^3 \) is dependent on \( a + b + c \), which is greater than 0.

### Final Answer
The value of \( a^3 + b^3 + c^3 \) is \( a + b + c \), which is greater than 0.