Let `A= [[a,b,c],[b,c,a],[c,a,b]]` is an orthogonal matrix and `abc = lambda (lt0).`
The value of ` a^(3) + b^(3)+c^(3)` is

Options

A. `lambda`
B. `2lambda`
C. `3lambda`
D. None of these

Explanation

To solve the problem, we need to find the value of \( a^3 + b^3 + c^3 \) given that the matrix

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

is an orthogonal matrix and \( abc = \lambda \) (where \( \lambda < 0 \)).

### Step 1: Understand Orthogonal Matrix Condition

An orthogonal matrix \( A \) satisfies the condition:

\[ A A^T = I \]

where \( A^T \) is the transpose of \( A \) and \( I \) is the identity matrix.

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

First, we compute \( A A^T \):

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

Now, multiplying \( A \) by \( A^T \):

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

Calculating each element:

- The (1,1) entry:
\[ a^2 + b^2 + c^2 \]
- The (1,2) entry:
\[ ab + bc + ca \]
- The (2,1) entry:
\[ ab + bc + ca \]
- The (2,2) entry:
\[ b^2 + c^2 + a^2 \]
- The (3,3) entry:
\[ c^2 + a^2 + b^2 \]

Thus, we have:

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

### Step 3: Set Equal to Identity Matrix

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

\[ \begin{bmatrix} a^2 + b^2 + c^2 & ab + bc + ca & ab + bc + ca \\ ab + bc + ca & b^2 + c^2 + a^2 & ab + bc + ca \\ ab + bc + ca & ab + bc + ca & c^2 + a^2 + b^2 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \]

From this, we get the equations:

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

### Step 4: Find \( a + b + c \)

Using the identity:

\[ (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) \]

Substituting the known values:

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

Thus,

\[ a + b + c = \pm 1 \]

### Step 5: Use the Identity for Cubes

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 the known values:

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

Thus,

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

### Step 6: Solve for \( a^3 + b^3 + c^3 \)

Now we can express \( a^3 + b^3 + c^3 \):

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

Substituting \( a + b + c = \pm 1 \):

\[ a^3 + b^3 + c^3 = \pm 1 + 3\lambda \]

### Final Answer

Thus, the value of \( a^3 + b^3 + c^3 \) is:

\[ \boxed{\pm 1 + 3\lambda} \]