Let `A= [[a,b,c],[b,c,a],[c,a,b]]` is an orthogonal matrix and `abc = lambda (lt0).`
The value ` a^(2) b^(2) + b^(2) c^(2) + c^(2) a^(2)`, is

Options

A. `2lambda`
B. `-2lambda`
C. `lambda^(2)`
D. `-lambda`

Explanation

To solve the problem, we start with the given orthogonal matrix \( A \) defined as:

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

Since \( A \) is an orthogonal matrix, we have:

\[ A A^T = I \]

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

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

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

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

### Step 2: Multiply \( A \) and \( A^T \)

Now, we compute the product \( A A^T \):

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

Calculating the elements of the resulting matrix:

- 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:

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

### Step 3: Set the product equal to the identity matrix

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

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

From this, we can derive the following equations:

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

### Step 4: Find \( a^2b^2 + b^2c^2 + c^2a^2 \)

We need to find the value of \( a^2b^2 + b^2c^2 + c^2a^2 \). We can use 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 squares

We can express \( a^2b^2 + b^2c^2 + c^2a^2 \) in terms of \( (ab + bc + ca)^2 \):

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

Since \( ab + bc + ca = 0 \):

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

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

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

Thus, we have:

\[ a^2b^2 + b^2c^2 + c^2a^2 = -2abc \]

### Step 6: Substitute \( abc = \lambda \)

Given \( abc = \lambda < 0 \):

\[ a^2b^2 + b^2c^2 + c^2a^2 = -2\lambda \]

### Final Answer

The value of \( a^2b^2 + b^2c^2 + c^2a^2 \) is:

\[ -2\lambda \]