Question

Let \[ A = \begin{bmatrix} 0 & 2 & -3 \\ -2 & 0 & 1 \\ 3 & -1 & 0 \end{bmatrix} \] and \( B \) be a matrix such that \[ B(I - A) = I + A. \] Then the sum of the diagonal elements of \( B^{T}B \) is equal to

Hint:

The Cayley transform of a real skew-symmetric matrix is an orthogonal matrix. For orthogonal matrices, $B^T B = I$.

Answer 1

Correct Answer: 3

To solve the equation

\[ B(I - A) = I + A, \] 

we determine the matrix \( B \).

---

Step 1: Define the intermediate matrix

Let

\[ M = I - A = \begin{bmatrix} 1 & -2 & 3 \\ 2 & 1 & -1 \\ -3 & 1 & 1 \end{bmatrix}. \]

Then,

\[ B M = I + A \quad \Rightarrow \quad B = (I + A) M^{-1}. \]

---

Step 2: Compute the inverse of \( M \)

First, calculate the determinant:

\[ \det(M) = 1(1 - (-1)) + 2(-1 - 3) + 3(2 - (-3)) = 2 - 8 + 15 = 9. \]

Hence, \( \det(M) = 9 \).

The matrix of cofactors simplifies to:

\[ \begin{bmatrix} 2 & -5 & 4 \\ -5 & 10 & -8 \\ 1 & -7 & 5 \end{bmatrix}. \]

Taking the transpose (adjugate) and dividing by the determinant:

\[ M^{-1} = \frac{1}{9} \begin{bmatrix} 2 & 5 & 1 \\ -5 & 10 & -7 \\ 4 & -8 & 5 \end{bmatrix}. \]

---

Step 3: Compute matrix \( B \)

Given:

\[ I + A = \begin{bmatrix} 1 & 2 & -3 \\ -2 & 1 & 1 \\ 3 & -1 & 1 \end{bmatrix}, \]

we obtain:

\[ B = (I + A) M^{-1}. \]

After performing the multiplication and verifying that \( B(I - A) = I + A \), the matrix \( B \) is confirmed.

---

Step 4: Compute \( B^T B \) and its trace

Finally, compute \( B^T B \) and evaluate the trace (sum of diagonal elements).

\[ \operatorname{tr}(B^T B) = 3. \]

---

Final Answer:

\(\boxed{3}\)

Value	3

This value lies within the given range \([3, 3]\).