Question

Let \[ A = \begin{bmatrix} 1 & -5 & 2 \\ 7 & 0 & 6 \\ 5 & -3 & 7 \end{bmatrix}, \quad B = \begin{bmatrix} 1 & 6 & 7 \\ 3 & 8 & 4 \\ 7 & 6 & 1 \end{bmatrix}. \] If \( C \) and \( D \) are two \( 3 \times 3 \) matrices such that: - \( A + C \) is symmetric, - \( A - C \) is skew-symmetric, and - \( B = D^\top \), then find \( (C - D)^\top \).

• A. \( \begin{bmatrix} 0 & -11 & -5 \\ 4 & -8 & 2 \\ -2 & -9 & 8 \end{bmatrix} \)
• B. \( \begin{bmatrix} 0 & -22 & -11 \\ 49 & 78 & 55 \\ 45 & 48 & 30 \end{bmatrix} \)
• C. \( \begin{bmatrix} 2 & 13 & 12 \\ -2 & 8 & 1 \\ 9 & 12 & 8 \end{bmatrix} \)
• D. \( \begin{bmatrix} 0 & -8 & -5 \\ 1 & -8 & 0 \\ -2 & -7 & 6 \end{bmatrix} \)

Hint:

For matrices \( A \) and \( C \), if \( A + C \) is symmetric and \( A - C \) is skew-symmetric, then \( C = A^T \). Also remember, \( (X^T)^T = X \) and \( (A - B)^T = A^T - B^T \).

Answer 1

The Correct Option is D

Given Matrices:

\( A = \begin{bmatrix} 1 & -5 & 2 \\ 7 & 0 & 6 \\ 5 & -3 & 7 \end{bmatrix} \)
\( B = \begin{bmatrix} 1 & 6 & 7 \\ 3 & 8 & 4 \\ 7 & 6 & 1 \end{bmatrix} \)

Step 1: Determine Matrix \( C \)

Given conditions:

• \( A + C \) is symmetric
• \( A - C \) is skew-symmetric

For \( A + C \) to be symmetric:

\( A + C = (A + C)^T = A^T + C^T \)

\( \Rightarrow C - C^T = A^T - A \)

For \( A - C \) to be skew-symmetric:

\( A - C = -(A - C)^T = -A^T + C^T \)

\( \Rightarrow C + C^T = A + A^T \)

Adding both equations:

\( 2C = 2A^T \Rightarrow C = A^T \)

Thus:

\( C = A^T = \begin{bmatrix} 1 & 7 & 5 \\ -5 & 0 & -3 \\ 2 & 6 & 7 \end{bmatrix} \)

Step 2: Determine Matrix \( D \)

Given \( B = D^T \), then:

\( D = B^T = \begin{bmatrix} 1 & 3 & 7 \\ 6 & 8 & 6 \\ 7 & 4 & 1 \end{bmatrix} \)

Step 3: Compute \( C - D^T \)

Since \( D^T = B \):

\( C - D^T = A^T - B = \)

\(\begin{bmatrix} 1 &7 &5\\-5 &0 &-3\\2 &6& 7\end{bmatrix} \) - \(\begin{bmatrix}1& 6& 7\\3& 8& 4\\7 &6 &1\end{bmatrix} \) = \( \begin{bmatrix} 0 &1 &-2\\-8& -8& -7\\-5 &0 &6\end{bmatrix}\)

Step 4: Compute \( \left( C - D^T \right)^T \)

Transpose the result from Step 3:

\( \left( C - D^T \right)^T = \begin{bmatrix} 0 & -8 & -5 \\ 1 & -8 & 0 \\ -2 & -7 & 6 \end{bmatrix} \)

Final Answer:

\( \boxed{\begin{bmatrix} 0 & -8 & -5 \\ 1 & -8 & 0 \\ -2 & -7 & 6 \end{bmatrix}} \)

Correct Option: 4