If matrix A `=[{:( a,b,c),(b,c,a),(c,a,b) :}]` where a,b,c are real positive number ,abc =1 and `A^(T) A = I ` then
Which of the following is/are true

Options

A. `a+b+c= 1`
B. `a^(2)+b^(2) +c^(2) =1`
C. ` ab+bc+cane0`
D. `a^(3) +b^(3) +c^(3)= 4`

Answer

To solve the problem, we need to analyze the given matrix \( A \) and the conditions provided. The matrix \( A \) is defined as:

\[ A = \begin{pmatrix} a & b & c \\ b & c & a \\ c & a & b \end{pmatrix} \]

with the conditions that \( abc = 1 \) and \( A^T A = I \).

### Step 1: Compute \( A^T \)

The transpose of matrix \( A \) is obtained by swapping rows with columns:

\[ A^T = \begin{pmatrix} a & b & c \\ b & c & a \\ c & a & b \end{pmatrix} \]

### Step 2: Compute \( A^T A \)

Now, we will multiply \( A^T \) and \( A \):

\[ A^T A = \begin{pmatrix} a & b & c \\ b & c & a \\ c & a & b \end{pmatrix} \begin{pmatrix} a & b & c \\ b & c & a \\ c & a & b \end{pmatrix} \]

Calculating the elements of \( A^T A \):

- The (1,1) entry:
\[ a^2 + b^2 + c^2 \]

- The (1,2) entry:
\[ ab + bc + ca \]

- The (1,3) entry:
\[ ac + ba + cb \]

- The (2,1) entry:
\[ ba + cb + ac \]

- The (2,2) entry:
\[ b^2 + c^2 + a^2 \]

- The (2,3) entry:
\[ bc + ca + ab \]

- The (3,1) entry:
\[ ca + ab + bc \]

- The (3,2) entry:
\[ cb + ac + ba \]

- The (3,3) entry:
\[ c^2 + a^2 + b^2 \]

Thus, we have:

\[ A^T A = \begin{pmatrix} a^2 + b^2 + c^2 & ab + bc + ca & ac + ba + cb \\ ba + cb + ac & b^2 + c^2 + a^2 & bc + ca + ab \\ ca + ab + bc & cb + ac + ba & c^2 + a^2 + b^2 \end{pmatrix} \]

### Step 3: Set \( A^T A = I \)

Since \( A^T A = I \), we know that:

\[ A^T A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

From this, we can derive the following equations:

1. \( a^2 + b^2 + c^2 = 1 \)
2. \( ab + bc + ca = 0 \)

### Step 4: Analyze the conditions

Given \( abc = 1 \), we can use the identity:

\[ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) \]

Substituting \( abc = 1 \) and \( ab + ac + bc = 0 \):

\[ a^3 + b^3 + c^3 - 3 = (a + b + c)(1 - 0) \]

This simplifies to:

\[ a^3 + b^3 + c^3 = a + b + c + 3 \]

### Step 5: Solve for \( a + b + c \)

From the earlier equation \( a^2 + b^2 + c^2 = 1 \) and using the identity \( (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + ac + bc) \):

\[ (a + b + c)^2 = 1 + 2(0) = 1 \]

Thus, \( a + b + c = 1 \).

### Conclusion

Now we can evaluate the options based on our findings:

1. **True**: \( a + b + c = 1 \)
2. **False**: \( ab + bc + ca \neq 0 \) (it equals 0)
3. **True**: \( a^3 + b^3 + c^3 = 3 \) (from our derived equation)
4. **True**: \( a^2 + b^2 + c^2 = 1 \)