Question

A, B, C, D are square matrices such that A + B is symmetric, A - B is skew-symmetric, and D is the transpose of C.

If

\[ A = \begin{bmatrix} -1 & 2 & 3 \\ 4 & 3 & -2 \\ 3 & -4 & 5 \end{bmatrix} \]

and

\[ C = \begin{bmatrix} 0 & 1 & -2 \\ 2 & -1 & 0 \\ 0 & 2 & 1 \end{bmatrix} \]

then the matrix \( B + D \) is:

• A. \[ \begin{bmatrix} -1 & 6 & 3 \\ 6 & 2 & -2 \\ 3 & -2 & 6 \end{bmatrix} \]
• B. \[ \begin{bmatrix} -1 & 6 & 3 \\ 3 & 2 & -2 \\ 1 & -2 & 6 \end{bmatrix} \]
• C. \[ \begin{bmatrix} 3 & 2 & -2 \\ 2 & 6 & 3 \\ -2 & 3 & 2 \end{bmatrix} \]
• D. \[ \begin{bmatrix} 1 & -2 & 6 \\ -2 & 3 & 2 \\ 6 & 2 & 1 \end{bmatrix} \]

Hint:

In problems involving symmetric and skew-symmetric matrices, leveraging properties of transpose can simplify calculations significantly.

Answer 1

The Correct Option is B

To solve the problem, we analyze the given conditions and matrices step by step.

Step 1: Understand the conditions

• \( A + B \) is symmetric: \( (A + B)^T = A + B \).
• \( A - B \) is skew-symmetric: \( (A - B)^T = -(A - B) \).
• \( D \) is the transpose of \( C \), so \( D = C^T \).

Step 2: Use the properties of symmetric and skew-symmetric matrices

From the conditions:

\( A + B \) is symmetric:

\[ (A + B)^T = A + B \]

Taking the transpose:

\[ A^T + B^T = A + B \quad \text{(1)} \]

\( A - B \) is skew-symmetric:

\[ (A - B)^T = -(A - B) \]

Taking the transpose:

\[ A^T - B^T = -A + B \quad \text{(2)} \]

Step 3: Solve for \( B \)

Adding equations (1) and (2):

\[ (A^T + B^T) + (A^T - B^T) = (A + B) + (-A + B) \]

Simplify:

\[ 2A^T = 2B \]

Divide by 2:

\[ B = A^T \]

Step 4: Compute \( B \) and \( D \)

Given:

\[ A = \begin{bmatrix} -1 & 2 & 3 \\ 4 & 3 & -2 \\ 3 & -4 & 5 \end{bmatrix} \]

The transpose of \( A \) is:

\[ B = A^T = \begin{bmatrix} -1 & 4 & 3 \\ 2 & 3 & -4 \\ 3 & -2 & 5 \end{bmatrix} \]

Given:

\[ C = \begin{bmatrix} 0 & 1 & -2 \\ 2 & -1 & 0 \\ 0 & 2 & 1 \end{bmatrix} \]

The transpose of \( C \) is:

\[ D = C^T = \begin{bmatrix} 0 & 2 & 0 \\ 1 & -1 & 2 \\ -2 & 0 & 1 \end{bmatrix} \]

Step 5: Compute \( B + D \)

Add \( B \) and \( D \):

\[ B + D = \begin{bmatrix} -1 & 4 & 3 \\ 2 & 3 & -4 \\ 3 & -2 & 5 \end{bmatrix} + \begin{bmatrix} 0 & 2 & 0 \\ 1 & -1 & 2 \\ -2 & 0 & 1 \end{bmatrix} \] \[ B + D = \begin{bmatrix} -1+0 & 4+2 & 3+0 \\ 2+1 & 3+(-1) & -4+2 \\ 3+(-2) & -2+0 & 5+1 \end{bmatrix} \] \[ B + D = \begin{bmatrix} -1 & 6 & 3 \\ 3 & 2 & -2 \\ 1 & -2 & 6 \end{bmatrix} \]

Final Answer:

The matrix \( B + D \) is:

\[ \boxed{\begin{bmatrix} -1 & 6 & 3 \\ 3 & 2 & -2 \\ 1 & -2 & 6 \end{bmatrix}} \]